[{"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "In the last video, we established that if we say the rate of change of a population with respect to time is going to be proportional to the population, we were able to solve that differential equation and find a general solution which involves an exponential, that the population is going to be equal to some constant times e to some other constant times time. In the last video, we assumed time was days. So let's just apply this just to feel good that we can truly model population in this way. So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K?"}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right?"}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50. And we're done. We can now write the particular solution that meets these conditions. So we can now write that our population, and I can even write our population as a function of time, is going to be equal to, is going to be equal to 100, 100 times E to the, now K is natural log of two over 50, so I'll write that."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "What I want to do in this video is do an exercise that takes us the other way. Start with a slope field and figure out which differential equation is the slope field describing the solutions for. And so I encourage you to look at each of these options and think about which of these differential equations is being described by this slope field. I encourage you to pause the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "I encourage you to pause the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one. Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one. Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out. Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out. Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative five and negative seven."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative five and negative seven. This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "I could have picked negative five and negative seven. This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here. So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "We have a positive slope here. So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here. So rule that one out. And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And we've already seen the slope is not zero here. So rule that one out. And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting. It looks like this slope right over here could be two. This looks like one, this looks like two."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting. It looks like this slope right over here could be two. This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases. And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases. And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one. So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one. So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here. And it looks like if you had, if you had your, in a solution, if you had a, if the solution contained, say this point right over here, you would actually, that would actually be a solution to the differential equation. Y is equal to, y is equal to negative x, whoops, y is equal to negative x minus one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here. And it looks like if you had, if you had your, in a solution, if you had a, if the solution contained, say this point right over here, you would actually, that would actually be a solution to the differential equation. Y is equal to, y is equal to negative x, whoops, y is equal to negative x minus one. And you can verify that. If y is equal to negative x minus one, then the x and negative x cancel out, you're just left with dy dx is equal to negative one, which is exactly what's being described by this slope field. Anyway, hopefully you found that interesting."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "In the first part of this problem, we just had this fairly straightforward differential equation. And I know it's a little bit frustrating right now, because this is such an easy one to solve using the characteristic equation. Why are we doing Laplace transforms? Well, I just want to show you that they can solve even these problems. But later on, there are going to be classes of problems that frankly, our traditional methods aren't as good as the Laplace transform. But anyway, how did we solve this? We just took the Laplace transform of both sides of this equation."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, I just want to show you that they can solve even these problems. But later on, there are going to be classes of problems that frankly, our traditional methods aren't as good as the Laplace transform. But anyway, how did we solve this? We just took the Laplace transform of both sides of this equation. We got all this hairy mess. We used the property of the derivatives of functions when you take the Laplace transform. And we ended up, after doing a lot of algebra, essentially, we got this."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We just took the Laplace transform of both sides of this equation. We got all this hairy mess. We used the property of the derivatives of functions when you take the Laplace transform. And we ended up, after doing a lot of algebra, essentially, we got this. We got the Laplace transform of y is equal to this thing. We just took the Laplace transform of both sides and manipulated algebraically. So now our task in this video is to figure out what y's Laplace transform is this thing."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we ended up, after doing a lot of algebra, essentially, we got this. We got the Laplace transform of y is equal to this thing. We just took the Laplace transform of both sides and manipulated algebraically. So now our task in this video is to figure out what y's Laplace transform is this thing. And essentially, what we're trying to do is we're trying to take the inverse Laplace transform of both sides of this equation. So another way to say it, we could say that y, we could take the inverse Laplace transform of both sides. We could say that y is equal to the inverse Laplace transform of this thing."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So now our task in this video is to figure out what y's Laplace transform is this thing. And essentially, what we're trying to do is we're trying to take the inverse Laplace transform of both sides of this equation. So another way to say it, we could say that y, we could take the inverse Laplace transform of both sides. We could say that y is equal to the inverse Laplace transform of this thing. 2s plus 13 over s squared plus 5s plus 6. Now we'll eventually actually learn the formal definition of the inverse Laplace transform. How do you go from the s domain to the t domain?"}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We could say that y is equal to the inverse Laplace transform of this thing. 2s plus 13 over s squared plus 5s plus 6. Now we'll eventually actually learn the formal definition of the inverse Laplace transform. How do you go from the s domain to the t domain? Or how do you go from the frequency domain to the time domain? We're not going to worry about that right now. What we're going to do is we're going to get this into a form that we recognize and say, oh, I know those functions."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "How do you go from the s domain to the t domain? Or how do you go from the frequency domain to the time domain? We're not going to worry about that right now. What we're going to do is we're going to get this into a form that we recognize and say, oh, I know those functions. That's the Laplace transform of whatever and whatever. And then we will know what y is. So let's try to do that."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What we're going to do is we're going to get this into a form that we recognize and say, oh, I know those functions. That's the Laplace transform of whatever and whatever. And then we will know what y is. So let's try to do that. So what we're going to use is something that you probably haven't used since algebra 2, which is, I think, when it's taught in eighth, ninth, or tenth grade, depending. And you finally see it now in differential equations that it actually has some use. We're going to use partial fraction expansion."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's try to do that. So what we're going to use is something that you probably haven't used since algebra 2, which is, I think, when it's taught in eighth, ninth, or tenth grade, depending. And you finally see it now in differential equations that it actually has some use. We're going to use partial fraction expansion. And I'll do a little primer on that in case you don't remember it. So anyway, let's just factor the bottom part right here. And you'll see where I'm going with this."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We're going to use partial fraction expansion. And I'll do a little primer on that in case you don't remember it. So anyway, let's just factor the bottom part right here. And you'll see where I'm going with this. So if I factor the bottom, I get s plus 2 times s plus 3. And what we want to do is we want to rewrite this fraction as the sum of 2 partial fractions. I think that's why it's called partial fraction expansion."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And you'll see where I'm going with this. So if I factor the bottom, I get s plus 2 times s plus 3. And what we want to do is we want to rewrite this fraction as the sum of 2 partial fractions. I think that's why it's called partial fraction expansion. So we want to write this as a sum of a over s plus 2 plus b over s plus 3. And if we can do this, then you might already, bells might already be ringing in your head. We know that these things that look like this are the Laplace transform of functions that we've already solved for."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I think that's why it's called partial fraction expansion. So we want to write this as a sum of a over s plus 2 plus b over s plus 3. And if we can do this, then you might already, bells might already be ringing in your head. We know that these things that look like this are the Laplace transform of functions that we've already solved for. And I'll do a little review on that in a second. But anyway, how do we figure out a and b? Well, if we were to actually add a and b, if we were to, let's do a little side right here."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We know that these things that look like this are the Laplace transform of functions that we've already solved for. And I'll do a little review on that in a second. But anyway, how do we figure out a and b? Well, if we were to actually add a and b, if we were to, let's do a little side right here. So if we said that a, so if we were to give them a common denominator, which is this, s plus 2 times s plus 3, then what would a become? We'd have to multiply a times s plus 3, right? So we'd get a s plus 3a."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, if we were to actually add a and b, if we were to, let's do a little side right here. So if we said that a, so if we were to give them a common denominator, which is this, s plus 2 times s plus 3, then what would a become? We'd have to multiply a times s plus 3, right? So we'd get a s plus 3a. Right? This, as I've written it right now, is the same thing as a over s plus 2. You could cancel out an s plus 3 on the top and the bottom."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we'd get a s plus 3a. Right? This, as I've written it right now, is the same thing as a over s plus 2. You could cancel out an s plus 3 on the top and the bottom. And now we're going to add the b to it. So plus, I'll do that in a different color. Plus, well, if we have this as the denominator, we could multiply the numerator and the denominator by s plus 2, right?"}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You could cancel out an s plus 3 on the top and the bottom. And now we're going to add the b to it. So plus, I'll do that in a different color. Plus, well, if we have this as the denominator, we could multiply the numerator and the denominator by s plus 2, right? To get b times s plus 2b. And that's going to equal this thing. That's going to equal, all I did is I added these two fractions."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Plus, well, if we have this as the denominator, we could multiply the numerator and the denominator by s plus 2, right? To get b times s plus 2b. And that's going to equal this thing. That's going to equal, all I did is I added these two fractions. Nothing fancier than there. That was algebra 2. Actually, I think I should do an actual video on that as well."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That's going to equal, all I did is I added these two fractions. Nothing fancier than there. That was algebra 2. Actually, I think I should do an actual video on that as well. But that's going to equal this thing. 2s plus 13, all of that over s plus 2 times s plus 3. Notice, in all differential equations, the hairiest part's always the algebra."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Actually, I think I should do an actual video on that as well. But that's going to equal this thing. 2s plus 13, all of that over s plus 2 times s plus 3. Notice, in all differential equations, the hairiest part's always the algebra. So now what we do is we match up. We say, well, let's add the s terms here. And we could just say that the numerators have to equal each other because the denominators are equal."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Notice, in all differential equations, the hairiest part's always the algebra. So now what we do is we match up. We say, well, let's add the s terms here. And we could just say that the numerators have to equal each other because the denominators are equal. So we have a plus bs plus 3a plus 2b is equal to 2s plus b. So the coefficient on s on the right-hand side is 2. The coefficient on the left-hand side is a plus b."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we could just say that the numerators have to equal each other because the denominators are equal. So we have a plus bs plus 3a plus 2b is equal to 2s plus b. So the coefficient on s on the right-hand side is 2. The coefficient on the left-hand side is a plus b. So we know that a plus b is equal to 2. And then on the right-hand side, we see 3a plus 2b must be equal to, oh, this is a 13. Did I say b?"}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "The coefficient on the left-hand side is a plus b. So we know that a plus b is equal to 2. And then on the right-hand side, we see 3a plus 2b must be equal to, oh, this is a 13. Did I say b? This is a 13. It looks just like a b, right? That was 2s plus 13."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Did I say b? This is a 13. It looks just like a b, right? That was 2s plus 13. Anyway, so on the right-hand side, I get 3a plus 2b is equal to 13. Now we have two equations with two unknowns. And what do we get?"}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That was 2s plus 13. Anyway, so on the right-hand side, I get 3a plus 2b is equal to 13. Now we have two equations with two unknowns. And what do we get? I know this is very tiresome, but it'll be satisfying in the end because you'll actually solve something with the Laplace transform. Let's multiply the top equation by 2. Let's just say minus 2."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And what do we get? I know this is very tiresome, but it'll be satisfying in the end because you'll actually solve something with the Laplace transform. Let's multiply the top equation by 2. Let's just say minus 2. So we get minus 2a minus 2b equals minus 4. And then we add the two equations. You get a is equal to, these cancel out, a is equal to 9."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's just say minus 2. So we get minus 2a minus 2b equals minus 4. And then we add the two equations. You get a is equal to, these cancel out, a is equal to 9. Great. If a is equal to 9, what is b equal to? b is equal to 9 plus what is equal to 2?"}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You get a is equal to, these cancel out, a is equal to 9. Great. If a is equal to 9, what is b equal to? b is equal to 9 plus what is equal to 2? Or 2 minus 9 is minus 7. And we have done some serious simplification because now we can rewrite this whole expression as the Laplace transform of y is equal to a over s plus 2 is equal to 9 over s plus 2 minus 7 over s plus 3. Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2 minus 7 times 1 over s plus 3."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "b is equal to 9 plus what is equal to 2? Or 2 minus 9 is minus 7. And we have done some serious simplification because now we can rewrite this whole expression as the Laplace transform of y is equal to a over s plus 2 is equal to 9 over s plus 2 minus 7 over s plus 3. Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2 minus 7 times 1 over s plus 3. Why did I take the trouble to do this? Well, hopefully you'll recognize this was actually the second Laplace transform we figured out. This was the second Laplace transform we figured out."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2 minus 7 times 1 over s plus 3. Why did I take the trouble to do this? Well, hopefully you'll recognize this was actually the second Laplace transform we figured out. This was the second Laplace transform we figured out. What was that? I'll write it down here just so you remember it. It was the Laplace transform of e to the at was equal to 1 over s minus a."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This was the second Laplace transform we figured out. What was that? I'll write it down here just so you remember it. It was the Laplace transform of e to the at was equal to 1 over s minus a. That was the second Laplace transform we figured out. So this is interesting. So if we were to take, this is the Laplace transform of what?"}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It was the Laplace transform of e to the at was equal to 1 over s minus a. That was the second Laplace transform we figured out. So this is interesting. So if we were to take, this is the Laplace transform of what? So if we were to take the inverse Laplace transform, actually let me just stay consistent. So that means that this is the Laplace transform of y is equal to 9 times the Laplace transform of what? If we just do pattern matching, if this is s minus a, then a is minus 2."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if we were to take, this is the Laplace transform of what? So if we were to take the inverse Laplace transform, actually let me just stay consistent. So that means that this is the Laplace transform of y is equal to 9 times the Laplace transform of what? If we just do pattern matching, if this is s minus a, then a is minus 2. So 9 times the Laplace transform of e to the minus 2t. Does that make sense? Take this, put it in this, which we figured out, and you get 1 over s plus 2."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If we just do pattern matching, if this is s minus a, then a is minus 2. So 9 times the Laplace transform of e to the minus 2t. Does that make sense? Take this, put it in this, which we figured out, and you get 1 over s plus 2. And let me clean this up a little bit because I'm going to need that real estate. I'll write this. I'll leave that there because we'll still use that."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Take this, put it in this, which we figured out, and you get 1 over s plus 2. And let me clean this up a little bit because I'm going to need that real estate. I'll write this. I'll leave that there because we'll still use that. And then we have minus 7 times, this is the Laplace transform of what? This is the Laplace transform of e to the minus 3t. This is just pattern matching."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'll leave that there because we'll still use that. And then we have minus 7 times, this is the Laplace transform of what? This is the Laplace transform of e to the minus 3t. This is just pattern matching. You're like, wow, if you saw this, you would go to your Laplace transform table if you didn't remember it. You'd see this. You're like, wow, that looks a lot like that."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is just pattern matching. You're like, wow, if you saw this, you would go to your Laplace transform table if you didn't remember it. You'd see this. You're like, wow, that looks a lot like that. It's just I just have to figure out what a is. I have s plus 3, I have s minus a. So in this case, a is equal to minus 3."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You're like, wow, that looks a lot like that. It's just I just have to figure out what a is. I have s plus 3, I have s minus a. So in this case, a is equal to minus 3. So if a is equal to minus 3, this is the Laplace transform of e to the minus 3t. So now we can take the inverse Laplace transform. Actually, before we do that, we know that this, because the Laplace transform is a linear operator, and actually now I can delete this down here."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So in this case, a is equal to minus 3. So if a is equal to minus 3, this is the Laplace transform of e to the minus 3t. So now we can take the inverse Laplace transform. Actually, before we do that, we know that this, because the Laplace transform is a linear operator, and actually now I can delete this down here. We know that the Laplace transform is a linear operator, so we can write this. And you normally wouldn't go through all of these steps. I just really want to make you understand what we're doing."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Actually, before we do that, we know that this, because the Laplace transform is a linear operator, and actually now I can delete this down here. We know that the Laplace transform is a linear operator, so we can write this. And you normally wouldn't go through all of these steps. I just really want to make you understand what we're doing. So we could say that this is the same thing as the Laplace transform of 9e to the minus 2t minus 7e to the minus 3t. Now we have something interesting. The Laplace transform of y is equal to the Laplace transform of this."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I just really want to make you understand what we're doing. So we could say that this is the same thing as the Laplace transform of 9e to the minus 2t minus 7e to the minus 3t. Now we have something interesting. The Laplace transform of y is equal to the Laplace transform of this. Well, if that's the case, then y must be equal to 9e to the minus 2t minus 7e to the minus 3t. And I never proved to you, but the Laplace transform is actually a one-to-one transformation. That if a function's Laplace transform, if I take a function and get its Laplace transform, and then if I were to take the inverse Laplace transform, the only function whose Laplace transform that is is that original function."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "The Laplace transform of y is equal to the Laplace transform of this. Well, if that's the case, then y must be equal to 9e to the minus 2t minus 7e to the minus 3t. And I never proved to you, but the Laplace transform is actually a one-to-one transformation. That if a function's Laplace transform, if I take a function and get its Laplace transform, and then if I were to take the inverse Laplace transform, the only function whose Laplace transform that is is that original function. It's not like two different functions can have the same Laplace transform. Anyway, a couple of things to think about here. Notice we had that thing that kind of looked like a characteristic equation popped up here and there, and we still have to solve a system of two equations with two unknowns."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That if a function's Laplace transform, if I take a function and get its Laplace transform, and then if I were to take the inverse Laplace transform, the only function whose Laplace transform that is is that original function. It's not like two different functions can have the same Laplace transform. Anyway, a couple of things to think about here. Notice we had that thing that kind of looked like a characteristic equation popped up here and there, and we still have to solve a system of two equations with two unknowns. Those are both things that we had to do when we solved an initial value problem when we used just traditional, the characteristic equation. But here, it happened all at once. And frankly, it was a little bit hairier because we had to do all this partial fraction expansion."}, {"video_title": "Laplace transform solves an equation 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Notice we had that thing that kind of looked like a characteristic equation popped up here and there, and we still have to solve a system of two equations with two unknowns. Those are both things that we had to do when we solved an initial value problem when we used just traditional, the characteristic equation. But here, it happened all at once. And frankly, it was a little bit hairier because we had to do all this partial fraction expansion. But it's pretty neat. The Laplace transform got us something useful. In the next video, I'll actually do a non-homogeneous equation and show you that the Laplace transform applies equally well there."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's try to fill in our Laplace transform table a little bit more. And a good place to start is just to write our definition of the Laplace transform. The Laplace transform of some function f of t is equal to the integral from 0 to infinity of e to the minus st times our function f of t dt. That's our definition. And the very first one we solved for, we can even do it on the side right here, was the Laplace transform of 1. We can almost do that as t to the 0. And that was equal to the integral from 0 to infinity."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That's our definition. And the very first one we solved for, we can even do it on the side right here, was the Laplace transform of 1. We can almost do that as t to the 0. And that was equal to the integral from 0 to infinity. f of t was just 1, so it's e to the minus st dt, which is equal to the antiderivative of e to the minus st is minus 1 over s, minus 1 over s, e to the minus st. And then you have to evaluate that from 0 to infinity. When you take the limit as this term approaches infinity, this e to the minus, this becomes e to the minus infinity. If we assume s is greater than 0, this whole term goes to 0."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And that was equal to the integral from 0 to infinity. f of t was just 1, so it's e to the minus st dt, which is equal to the antiderivative of e to the minus st is minus 1 over s, minus 1 over s, e to the minus st. And then you have to evaluate that from 0 to infinity. When you take the limit as this term approaches infinity, this e to the minus, this becomes e to the minus infinity. If we assume s is greater than 0, this whole term goes to 0. So you end up with 0 minus this thing evaluated at 0. When you evaluate t is equal to 0, this term right here becomes 1, e to the 0 becomes 1. So it's minus minus 1 over s, which is the same thing as plus 1 over s. So the Laplace transform of 1, of just the constant function 1, is 1 over s. We already solved that."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If we assume s is greater than 0, this whole term goes to 0. So you end up with 0 minus this thing evaluated at 0. When you evaluate t is equal to 0, this term right here becomes 1, e to the 0 becomes 1. So it's minus minus 1 over s, which is the same thing as plus 1 over s. So the Laplace transform of 1, of just the constant function 1, is 1 over s. We already solved that. Now let's increment it a little bit. Let's see if we can figure out the Laplace transform of t. So we could view this as t to the 0, now this is t to the 1. This is going to be equal to the integral from 0 to infinity of e to the minus st times t dt."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So it's minus minus 1 over s, which is the same thing as plus 1 over s. So the Laplace transform of 1, of just the constant function 1, is 1 over s. We already solved that. Now let's increment it a little bit. Let's see if we can figure out the Laplace transform of t. So we could view this as t to the 0, now this is t to the 1. This is going to be equal to the integral from 0 to infinity of e to the minus st times t dt. Now, I can tell you right now that this isn't, I don't have the antiderivative of this memorized, I don't know what it is, but there's a sense that the integration by parts could be useful, because integration by parts kind of decomposes this into a simpler problem. I always forget integration by parts, so I'll re-derive it here in this purple color. So if we have u times v, if we take the derivative with respect to, say, t of that, that's equal to the derivative of the first times the second function, plus the first function times the derivative of the second function."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is going to be equal to the integral from 0 to infinity of e to the minus st times t dt. Now, I can tell you right now that this isn't, I don't have the antiderivative of this memorized, I don't know what it is, but there's a sense that the integration by parts could be useful, because integration by parts kind of decomposes this into a simpler problem. I always forget integration by parts, so I'll re-derive it here in this purple color. So if we have u times v, if we take the derivative with respect to, say, t of that, that's equal to the derivative of the first times the second function, plus the first function times the derivative of the second function. It's just the product rule. Now if we take the integral of both sides of this equation, we get uv is equal to the antiderivative of u prime v plus the antiderivative of u v prime. Now, since we want to apply this to an integral, maybe let's make this what we want to solve for."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if we have u times v, if we take the derivative with respect to, say, t of that, that's equal to the derivative of the first times the second function, plus the first function times the derivative of the second function. It's just the product rule. Now if we take the integral of both sides of this equation, we get uv is equal to the antiderivative of u prime v plus the antiderivative of u v prime. Now, since we want to apply this to an integral, maybe let's make this what we want to solve for. So we can get the integral of u v prime is equal to, we can just subtract this from that side of the equation, so it's equal, I'm just switching, swapping the sides, so I'm just solving for this, I just subtract this from that, so it's equal to u v minus the integral of u prime v. So there you go. Even though I have trouble memorizing this formula, it's not too hard to re-derive as long as you remember the product rule right there. So if we're going to do integration by parts, it's good to define our v prime to be something it's easy to take the antiderivative of, because we're going to have to figure out v later on, and it's good to take u to be something that's easy to take the derivative of."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, since we want to apply this to an integral, maybe let's make this what we want to solve for. So we can get the integral of u v prime is equal to, we can just subtract this from that side of the equation, so it's equal, I'm just switching, swapping the sides, so I'm just solving for this, I just subtract this from that, so it's equal to u v minus the integral of u prime v. So there you go. Even though I have trouble memorizing this formula, it's not too hard to re-derive as long as you remember the product rule right there. So if we're going to do integration by parts, it's good to define our v prime to be something it's easy to take the antiderivative of, because we're going to have to figure out v later on, and it's good to take u to be something that's easy to take the derivative of. So let's make t is equal to our u, and let's make e to the minus st as being our v prime. If that's the case, then what is v? Well, v is just the antiderivative of that."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if we're going to do integration by parts, it's good to define our v prime to be something it's easy to take the antiderivative of, because we're going to have to figure out v later on, and it's good to take u to be something that's easy to take the derivative of. So let's make t is equal to our u, and let's make e to the minus st as being our v prime. If that's the case, then what is v? Well, v is just the antiderivative of that. In fact, we've done it before. It's minus 1 over s, e to the minus st. That's v. And then if we want to figure out u prime, because we're going to have to figure out that later anyway, u prime is just the derivative of t. That's just equal to 1. So let's apply this."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, v is just the antiderivative of that. In fact, we've done it before. It's minus 1 over s, e to the minus st. That's v. And then if we want to figure out u prime, because we're going to have to figure out that later anyway, u prime is just the derivative of t. That's just equal to 1. So let's apply this. Let's see. So the Laplace transform of t is equal to u v. u is t. v is this right here. It's times minus 1 over s, e to the minus st. That's the u v term right there."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's apply this. Let's see. So the Laplace transform of t is equal to u v. u is t. v is this right here. It's times minus 1 over s, e to the minus st. That's the u v term right there. And this is a definite integral, right? So we're going to evaluate this term right here from 0 to infinity. And then it's minus the integral from 0 to infinity of u prime, which is just 1, times v. v, we just figured out here, is minus 1 over s. This is my v right here."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It's times minus 1 over s, e to the minus st. That's the u v term right there. And this is a definite integral, right? So we're going to evaluate this term right here from 0 to infinity. And then it's minus the integral from 0 to infinity of u prime, which is just 1, times v. v, we just figured out here, is minus 1 over s. This is my v right here. So minus 1 over s, e to the minus st dt. And this, let's see if we can simplify this. So this is equal to minus t over s, e to the minus st, evaluated from 0 to infinity."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then it's minus the integral from 0 to infinity of u prime, which is just 1, times v. v, we just figured out here, is minus 1 over s. This is my v right here. So minus 1 over s, e to the minus st dt. And this, let's see if we can simplify this. So this is equal to minus t over s, e to the minus st, evaluated from 0 to infinity. And let's see, we could take this. Well, this is just 1. This 1 times anything is 1."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this is equal to minus t over s, e to the minus st, evaluated from 0 to infinity. And let's see, we could take this. Well, this is just 1. This 1 times anything is 1. We can just not write that. And then bring the minus 1 over s out. So if we bring the minus 1 over s out, this becomes plus 1 over s times the integral from 0 to infinity of e to the minus st dt."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This 1 times anything is 1. We can just not write that. And then bring the minus 1 over s out. So if we bring the minus 1 over s out, this becomes plus 1 over s times the integral from 0 to infinity of e to the minus st dt. And this should look familiar to you. This is exactly what we solved for right here. It was the Laplace transform of 1."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if we bring the minus 1 over s out, this becomes plus 1 over s times the integral from 0 to infinity of e to the minus st dt. And this should look familiar to you. This is exactly what we solved for right here. It was the Laplace transform of 1. So let's keep that in mind. So this right here is the Laplace transform of 1. And I want to write it that way, because we're going to see a pattern of this in the next video."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It was the Laplace transform of 1. So let's keep that in mind. So this right here is the Laplace transform of 1. And I want to write it that way, because we're going to see a pattern of this in the next video. I'm going to write that as Laplace transform of 1. But what is this equal to? So we're going to evaluate this as it added infinity, and then subtract from that, evaluate it at 0."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And I want to write it that way, because we're going to see a pattern of this in the next video. I'm going to write that as Laplace transform of 1. But what is this equal to? So we're going to evaluate this as it added infinity, and then subtract from that, evaluate it at 0. So I'm just going to make, you can kind of view it as a substitution. So this is equal to, well, let me write it this way. As a approaches infinity of minus a over se to the minus sa."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we're going to evaluate this as it added infinity, and then subtract from that, evaluate it at 0. So I'm just going to make, you can kind of view it as a substitution. So this is equal to, well, let me write it this way. As a approaches infinity of minus a over se to the minus sa. So that's this evaluated at infinity. And then from that, we're going to subtract this evaluated at 0. So minus all of this, but we already have a minus sign here, so we could write a plus."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "As a approaches infinity of minus a over se to the minus sa. So that's this evaluated at infinity. And then from that, we're going to subtract this evaluated at 0. So minus all of this, but we already have a minus sign here, so we could write a plus. Plus, we could write 0 over s times e to the minus s times 0. And then, of course, we have this term right here. So let me write that term."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So minus all of this, but we already have a minus sign here, so we could write a plus. Plus, we could write 0 over s times e to the minus s times 0. And then, of course, we have this term right here. So let me write that term. I'll do it in yellow. Or let me do it in blue. Plus 1 over s, that's this right there, times the Laplace transform of 1."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let me write that term. I'll do it in yellow. Or let me do it in blue. Plus 1 over s, that's this right there, times the Laplace transform of 1. And what do we get? So what's the limit as this, as a approaches infinity? You might say, wow, as a approaches infinity right here, this becomes a really big number."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Plus 1 over s, that's this right there, times the Laplace transform of 1. And what do we get? So what's the limit as this, as a approaches infinity? You might say, wow, as a approaches infinity right here, this becomes a really big number. There's a minus sign in there, so it's going to be a really big negative number. But this is an exponent. a is an exponent right here."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You might say, wow, as a approaches infinity right here, this becomes a really big number. There's a minus sign in there, so it's going to be a really big negative number. But this is an exponent. a is an exponent right here. So e to the minus infinity is going to go to 0 much faster than this is going to go to infinity. This term right here is a much stronger function, I guess, the way you could see it. And you could try it out on your calculator, if you don't believe me."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "a is an exponent right here. So e to the minus infinity is going to go to 0 much faster than this is going to go to infinity. This term right here is a much stronger function, I guess, the way you could see it. And you could try it out on your calculator, if you don't believe me. This term is going to overpower this term. And so this whole thing is going to go to 0. Likewise, e to the minus, this e to the 0, this is 1, you're multiplying it times a 0."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And you could try it out on your calculator, if you don't believe me. This term is going to overpower this term. And so this whole thing is going to go to 0. Likewise, e to the minus, this e to the 0, this is 1, you're multiplying it times a 0. So this is also going to go to 0, which is convenient, because all of this stuff just disappears. And we're left with the Laplace transform of t is equal to 1 over s times the Laplace transform of 1. And we know what the Laplace transform of 1 is."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Likewise, e to the minus, this e to the 0, this is 1, you're multiplying it times a 0. So this is also going to go to 0, which is convenient, because all of this stuff just disappears. And we're left with the Laplace transform of t is equal to 1 over s times the Laplace transform of 1. And we know what the Laplace transform of 1 is. The Laplace transform of 1, we just did at the beginning of the video, was equal to 1 over s if we assume that s is greater than 0. In fact, we have to assume that s was greater than 0 here in order to assume that this goes to 0. Only if s is greater than 0, then when you get a minus infinity here, does this approach to 0."}, {"video_title": "Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we know what the Laplace transform of 1 is. The Laplace transform of 1, we just did at the beginning of the video, was equal to 1 over s if we assume that s is greater than 0. In fact, we have to assume that s was greater than 0 here in order to assume that this goes to 0. Only if s is greater than 0, then when you get a minus infinity here, does this approach to 0. So fair enough. So the Laplace transform of t is equal to 1 over s times 1 over s, which is equal to 1 over s squared, for s is greater than 0. So we have one more entry in our table, and then we can use this."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Let's do a couple of problems where the roots of the characteristic equation are complex. And just as a little bit of a review, and we'll put this up here in the corner so that it's useful for us, we learned that if the roots of our characteristic equation are r is equal to lambda plus or minus mu i, that the general solution for our differential equation is y is equal to e to the lambda x times c1, or some constant, cosine of mu x plus c2 times sine of mu x. With that said, let's do some problems. So let's see, this first one, our differential equation, I'll do this one in blue, our differential equation is the second derivative, y prime prime plus 4y prime plus 5y is equal to 0. And they actually give us some initial conditions. So they say y of 0 is equal to 1. And y prime of 0 is equal to 0."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So let's see, this first one, our differential equation, I'll do this one in blue, our differential equation is the second derivative, y prime prime plus 4y prime plus 5y is equal to 0. And they actually give us some initial conditions. So they say y of 0 is equal to 1. And y prime of 0 is equal to 0. So now we'll actually be able to figure out a particular solution, or the particular solution, for this differential equation. So let's write down the characteristic equation. So it's r squared plus 4r plus 5 is equal to 0."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "And y prime of 0 is equal to 0. So now we'll actually be able to figure out a particular solution, or the particular solution, for this differential equation. So let's write down the characteristic equation. So it's r squared plus 4r plus 5 is equal to 0. Let's break out our quadratic formula. The roots of this are going to be negative b, so minus 4, plus or minus the square root of b squared, so that's 16, minus 4 times a, well that's 1, times 1, times c, times 5. Goes out there."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So it's r squared plus 4r plus 5 is equal to 0. Let's break out our quadratic formula. The roots of this are going to be negative b, so minus 4, plus or minus the square root of b squared, so that's 16, minus 4 times a, well that's 1, times 1, times c, times 5. Goes out there. All of that over 2 times a. a is 1, so all of that over 2. And see, this simplifies. This equals minus 4 plus or minus, see, 16, this is 20, right, 4 times 1 times 5 is 20."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Goes out there. All of that over 2 times a. a is 1, so all of that over 2. And see, this simplifies. This equals minus 4 plus or minus, see, 16, this is 20, right, 4 times 1 times 5 is 20. So 16 minus 20 is minus 4 over 2. And that equals, let's see, this equals minus 4 plus or minus 2i, right, the square root of minus 4 is 2i. All of that over 2."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "This equals minus 4 plus or minus, see, 16, this is 20, right, 4 times 1 times 5 is 20. So 16 minus 20 is minus 4 over 2. And that equals, let's see, this equals minus 4 plus or minus 2i, right, the square root of minus 4 is 2i. All of that over 2. And so our roots to the characteristic equation are minus 2, just dividing both by 2, minus 2 plus or minus, we could say, i or 1i, right? So if we wanted to do some pattern match, or if we just wanted to do it really fast, what's our lambda? Well, our lambda is this minus 2."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "All of that over 2. And so our roots to the characteristic equation are minus 2, just dividing both by 2, minus 2 plus or minus, we could say, i or 1i, right? So if we wanted to do some pattern match, or if we just wanted to do it really fast, what's our lambda? Well, our lambda is this minus 2. Let me write that down. That's our lambda. And what's our mu?"}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Well, our lambda is this minus 2. Let me write that down. That's our lambda. And what's our mu? Well, mu is the coefficient on the i, so mu is 1. Mu is equal to 1. And now we're ready to write down our general solution."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "And what's our mu? Well, mu is the coefficient on the i, so mu is 1. Mu is equal to 1. And now we're ready to write down our general solution. So the general solution to this differential equation is y is equal to e to the lambda x, well, lambda is minus 2, minus 2x times c1 cosine of mu x, but mu is just 1, so c1 cosine of x, plus c2 sine of mu x, when mu is 1, so sine of x, fair enough. Now let's use our initial conditions to find the particular solution, or a particular solution. So when x is 0, y is equal to 1."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "And now we're ready to write down our general solution. So the general solution to this differential equation is y is equal to e to the lambda x, well, lambda is minus 2, minus 2x times c1 cosine of mu x, but mu is just 1, so c1 cosine of x, plus c2 sine of mu x, when mu is 1, so sine of x, fair enough. Now let's use our initial conditions to find the particular solution, or a particular solution. So when x is 0, y is equal to 1. So y is equal to 1 when x is 0. So 1 is equal to, let's substitute x as 0 here, so e to the minus 2 times 0, that's just 1. So this whole thing becomes 1, so we could just ignore it."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So when x is 0, y is equal to 1. So y is equal to 1 when x is 0. So 1 is equal to, let's substitute x as 0 here, so e to the minus 2 times 0, that's just 1. So this whole thing becomes 1, so we could just ignore it. It's just 1 times this thing. So I'll write that down. e to the 0 is 1, times c1 times cosine of 0, plus c2 times sine of 0."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So this whole thing becomes 1, so we could just ignore it. It's just 1 times this thing. So I'll write that down. e to the 0 is 1, times c1 times cosine of 0, plus c2 times sine of 0. Now what's sine of 0? Sine of 0 is 0, so this whole term is going to be 0. Cosine of 0 is 1."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "e to the 0 is 1, times c1 times cosine of 0, plus c2 times sine of 0. Now what's sine of 0? Sine of 0 is 0, so this whole term is going to be 0. Cosine of 0 is 1. So there we already solved for c1. We get this, this is 1, so 1 times c1 times 1 is equal to 1. So we get our first coefficient, c1 is equal to 1."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Cosine of 0 is 1. So there we already solved for c1. We get this, this is 1, so 1 times c1 times 1 is equal to 1. So we get our first coefficient, c1 is equal to 1. Now let's take the derivative of our general solution, and we can even substitute c1 in here, just so that we have to stop writing c1 all the time. And we can solve for c2. So right now we know that our general solution is y, we could call this our pseudo-general solution, because we already solved for c1."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So we get our first coefficient, c1 is equal to 1. Now let's take the derivative of our general solution, and we can even substitute c1 in here, just so that we have to stop writing c1 all the time. And we can solve for c2. So right now we know that our general solution is y, we could call this our pseudo-general solution, because we already solved for c1. y is equal to e to the minus 2x times c1, but we know that c1 is 1, so I'll write times cosine of x, plus c2 times sine of x. Now let's take the derivative of this so that we can use the second initial condition. So y prime is equal to, we're going to have to do a little bit of product rule here."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So right now we know that our general solution is y, we could call this our pseudo-general solution, because we already solved for c1. y is equal to e to the minus 2x times c1, but we know that c1 is 1, so I'll write times cosine of x, plus c2 times sine of x. Now let's take the derivative of this so that we can use the second initial condition. So y prime is equal to, we're going to have to do a little bit of product rule here. So what's the derivative of the first expression? It is minus 2 e to the minus 2x, and we multiply that times the second expression. Cosine of x plus c2 sine of x."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So y prime is equal to, we're going to have to do a little bit of product rule here. So what's the derivative of the first expression? It is minus 2 e to the minus 2x, and we multiply that times the second expression. Cosine of x plus c2 sine of x. And then we add that to just the regular first expression, so plus e to the minus 2x times the derivative of the second expression. So what's the derivative of cosine of x? It's minus sine of x, and then what's the derivative of c2 sine of x?"}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Cosine of x plus c2 sine of x. And then we add that to just the regular first expression, so plus e to the minus 2x times the derivative of the second expression. So what's the derivative of cosine of x? It's minus sine of x, and then what's the derivative of c2 sine of x? Well, it's plus c2 cosine of x. And let's see if we can do any kind of simplification here. Well, actually, the easiest way, instead of trying to simplify it algebraically and everything, let's just use our initial condition."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "It's minus sine of x, and then what's the derivative of c2 sine of x? Well, it's plus c2 cosine of x. And let's see if we can do any kind of simplification here. Well, actually, the easiest way, instead of trying to simplify it algebraically and everything, let's just use our initial condition. Our initial condition is y prime of 0 is equal to 0. So let me write that down. The second initial condition is y prime of 0 is equal to 0."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Well, actually, the easiest way, instead of trying to simplify it algebraically and everything, let's just use our initial condition. Our initial condition is y prime of 0 is equal to 0. So let me write that down. The second initial condition is y prime of 0 is equal to 0. So y prime, when x is equal to 0, is equal to 0. And let's substitute x is equal to 0 into this thing. We could have simplified this more, but let's not worry about that right now."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "The second initial condition is y prime of 0 is equal to 0. So y prime, when x is equal to 0, is equal to 0. And let's substitute x is equal to 0 into this thing. We could have simplified this more, but let's not worry about that right now. So if x is 0, this is going to be 1, right? e to the 0. e to the 0 is 1, so we're left with just minus 2. Minus 2 times e to the 0 times cosine of 0, that's 1, plus c2 times sine of 0."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "We could have simplified this more, but let's not worry about that right now. So if x is 0, this is going to be 1, right? e to the 0. e to the 0 is 1, so we're left with just minus 2. Minus 2 times e to the 0 times cosine of 0, that's 1, plus c2 times sine of 0. Sine of 0 is 0, so that's just 1 plus 0. Plus e to the minus 2 times 0, that's just 1, times minus sine of 0. Sine of 0 is just 0."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Minus 2 times e to the 0 times cosine of 0, that's 1, plus c2 times sine of 0. Sine of 0 is 0, so that's just 1 plus 0. Plus e to the minus 2 times 0, that's just 1, times minus sine of 0. Sine of 0 is just 0. Plus c2 times cosine of 0. Cosine of 0 is 1, so plus c2. That simplified things, didn't it?"}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Sine of 0 is just 0. Plus c2 times cosine of 0. Cosine of 0 is 1, so plus c2. That simplified things, didn't it? Let's see, we get 0 is equal to, this is just 1, minus 2 plus c2, or we get c2 is equal to 2. Add 2 to both sides, c2 is equal to 2. And then we have our particular solution."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "That simplified things, didn't it? Let's see, we get 0 is equal to, this is just 1, minus 2 plus c2, or we get c2 is equal to 2. Add 2 to both sides, c2 is equal to 2. And then we have our particular solution. I know it's c2 is equal to 2, c1 is equal to 1. Actually, let me erase some of this, just so that we can go from our general solution to our particular solution. Let me erase some of this."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "And then we have our particular solution. I know it's c2 is equal to 2, c1 is equal to 1. Actually, let me erase some of this, just so that we can go from our general solution to our particular solution. Let me erase some of this. So we had figured out, you can remember, c1 is 1 and c2 is 2. That's easy to memorize. So I'll just delete all of this."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Let me erase some of this. So we had figured out, you can remember, c1 is 1 and c2 is 2. That's easy to memorize. So I'll just delete all of this. I'll write it nice and big. So our particular solution, given these initial conditions, were, or are, or is, y of x is equal to, this was the general solution, e to the minus 2x times, we solved for c1, we got it's equal to 1, so we can just write cosine of x. And then we solved for c2, we figured out that that was 2, plus 2 sine of x."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "So I'll just delete all of this. I'll write it nice and big. So our particular solution, given these initial conditions, were, or are, or is, y of x is equal to, this was the general solution, e to the minus 2x times, we solved for c1, we got it's equal to 1, so we can just write cosine of x. And then we solved for c2, we figured out that that was 2, plus 2 sine of x. And there you go. We have our particular solution to this. Sorry, where did I write it?"}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "And then we solved for c2, we figured out that that was 2, plus 2 sine of x. And there you go. We have our particular solution to this. Sorry, where did I write it? To this differential equation with these initial conditions. And what's neat is, when we originally kind of proved this formula, when we originally showed this formula, we had all of these i's and we simplified. We said c2, it was a combination of some other constants times some i's."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "Sorry, where did I write it? To this differential equation with these initial conditions. And what's neat is, when we originally kind of proved this formula, when we originally showed this formula, we had all of these i's and we simplified. We said c2, it was a combination of some other constants times some i's. And we said, oh, we don't know whether they're imaginary or not, so let's just merge them into some constant. But what's interesting is, this particular solution has no i's anywhere in it. And so, well, that tells us a couple of neat things."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "We said c2, it was a combination of some other constants times some i's. And we said, oh, we don't know whether they're imaginary or not, so let's just merge them into some constant. But what's interesting is, this particular solution has no i's anywhere in it. And so, well, that tells us a couple of neat things. One, that if we had kept this c2 in terms of some multiple of i's, our constants actually would have had i's and they would have canceled out, et cetera. And it also tells us that this formula is useful beyond formulas that just involve imaginary numbers. For example, this differential equation, I don't see an i anywhere here."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "And so, well, that tells us a couple of neat things. One, that if we had kept this c2 in terms of some multiple of i's, our constants actually would have had i's and they would have canceled out, et cetera. And it also tells us that this formula is useful beyond formulas that just involve imaginary numbers. For example, this differential equation, I don't see an i anywhere here. I don't see an i anywhere here. And I don't see an i anywhere here. But given this differential equation, to get to this solution, we had to use imaginary numbers in between."}, {"video_title": "Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3", "Sentence": "For example, this differential equation, I don't see an i anywhere here. I don't see an i anywhere here. And I don't see an i anywhere here. But given this differential equation, to get to this solution, we had to use imaginary numbers in between. And I think this is the first time, if I'm remembering all my playlists correctly, this is the first time that we used imaginary numbers for something useful. We used it as an intermediary tool where we got a real, a non-imaginary solution to a real problem, a non-imaginary problem. But we used imaginary numbers as a tool to solve it."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "In this video, we're going to see if we can generalize this by trying to figure out the Laplace transform of t to the n, where n is any integer power greater than 0. So n is any positive integer greater than 0. So let's try it out. So we know from our definition of the Laplace transform that the Laplace transform of t to the n is equal to the integral from 0 to infinity of our function, well, let me write t to the n, times, and this is just the definition of the transform, e to the minus st dt. And similar to when we figured out this Laplace transform, your intuition might be that, hey, we should use integration by parts. And I showed it in the last video, I always forget it, but I just recorded it, so I do happen to remember it. So the integration by parts just tells us that the integral of uv prime is equal to uv minus the integral of, I view this as kind of the swap, so u prime u prime v. So this is just our integration by parts formula."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we know from our definition of the Laplace transform that the Laplace transform of t to the n is equal to the integral from 0 to infinity of our function, well, let me write t to the n, times, and this is just the definition of the transform, e to the minus st dt. And similar to when we figured out this Laplace transform, your intuition might be that, hey, we should use integration by parts. And I showed it in the last video, I always forget it, but I just recorded it, so I do happen to remember it. So the integration by parts just tells us that the integral of uv prime is equal to uv minus the integral of, I view this as kind of the swap, so u prime u prime v. So this is just our integration by parts formula. If you ever forget it, you can derive it in about 30 seconds from the product rule. And I did it in the last video because I hadn't used it for a while, so I had to re-derive it. So let's apply it here."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So the integration by parts just tells us that the integral of uv prime is equal to uv minus the integral of, I view this as kind of the swap, so u prime u prime v. So this is just our integration by parts formula. If you ever forget it, you can derive it in about 30 seconds from the product rule. And I did it in the last video because I hadn't used it for a while, so I had to re-derive it. So let's apply it here. So what do we want to make our v prime? It's always good to use the exponential function because that's easy to take the antiderivative of. So this is our v prime, in which case our v is just the antiderivative of that."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's apply it here. So what do we want to make our v prime? It's always good to use the exponential function because that's easy to take the antiderivative of. So this is our v prime, in which case our v is just the antiderivative of that. So it's e to the minus st over minus s. If we take the derivative of this, minus s divided by minus s cancels out, and you just get that. And then if we make our u, let me pick a good color here, if we make this equal to our u, what's our u prime? u prime is just going to be n times t to the n minus 1."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this is our v prime, in which case our v is just the antiderivative of that. So it's e to the minus st over minus s. If we take the derivative of this, minus s divided by minus s cancels out, and you just get that. And then if we make our u, let me pick a good color here, if we make this equal to our u, what's our u prime? u prime is just going to be n times t to the n minus 1. Fair enough. So let's apply the integration by parts. So this is going to be equal to uv, u already says t to the n, so u is t to the n, that's our u, times v, which is e, let me write this down, so it's minus, there's a minus sign there, so we put the minus, let me do it in that color, minus, I'm just rewriting this, e to the minus st over s. So that's the uv term right there, let me make that clear."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "u prime is just going to be n times t to the n minus 1. Fair enough. So let's apply the integration by parts. So this is going to be equal to uv, u already says t to the n, so u is t to the n, that's our u, times v, which is e, let me write this down, so it's minus, there's a minus sign there, so we put the minus, let me do it in that color, minus, I'm just rewriting this, e to the minus st over s. So that's the uv term right there, let me make that clear. Let me pick a good color here. So this term right here is this term right here. And of course, I'm going to have to evaluate this from 0 to infinity, so let me write that, 0 to infinity, I could put a little bracket there or something, but you know we're going to have to evaluate that."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this is going to be equal to uv, u already says t to the n, so u is t to the n, that's our u, times v, which is e, let me write this down, so it's minus, there's a minus sign there, so we put the minus, let me do it in that color, minus, I'm just rewriting this, e to the minus st over s. So that's the uv term right there, let me make that clear. Let me pick a good color here. So this term right here is this term right here. And of course, I'm going to have to evaluate this from 0 to infinity, so let me write that, 0 to infinity, I could put a little bracket there or something, but you know we're going to have to evaluate that. And from that, we're going to have to subtract the integral, and let me not forget our boundaries, 0 to infinity, of u prime, u prime is n times t to the n minus 1, that's our u prime, times v, times minus, let me put this minus out here, so minus e to the minus st over s, and then all of that, of course, we have our dt, and you have a minus, minus, these things become pluses, and let's see if we can simplify this a little bit. So we get our Laplace transform of t to the n, is equal to this evaluated at infinity and evaluated at 0. So when you evaluate, what's the limit of this as t approaches infinity?"}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And of course, I'm going to have to evaluate this from 0 to infinity, so let me write that, 0 to infinity, I could put a little bracket there or something, but you know we're going to have to evaluate that. And from that, we're going to have to subtract the integral, and let me not forget our boundaries, 0 to infinity, of u prime, u prime is n times t to the n minus 1, that's our u prime, times v, times minus, let me put this minus out here, so minus e to the minus st over s, and then all of that, of course, we have our dt, and you have a minus, minus, these things become pluses, and let's see if we can simplify this a little bit. So we get our Laplace transform of t to the n, is equal to this evaluated at infinity and evaluated at 0. So when you evaluate, what's the limit of this as t approaches infinity? As t approaches infinity, this term, you might say, oh, this becomes really big, and I went over this in the last video, but this term overpowers it, because you're going to have e to the minus infinity, if we assume that s is greater than 0. So if s is greater than 0, this term is going to win out and go to 0 much faster than this term is going to go to infinity. So when you evaluate it at infinity, when you evaluate this at infinity, you're going to get 0, and then you're going to subtract this evaluated at 0."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So when you evaluate, what's the limit of this as t approaches infinity? As t approaches infinity, this term, you might say, oh, this becomes really big, and I went over this in the last video, but this term overpowers it, because you're going to have e to the minus infinity, if we assume that s is greater than 0. So if s is greater than 0, this term is going to win out and go to 0 much faster than this term is going to go to infinity. So when you evaluate it at infinity, when you evaluate this at infinity, you're going to get 0, and then you're going to subtract this evaluated at 0. When it's evaluated at 0, it's just minus 0 to the n times e to the minus s times 0 over s. Well, this becomes 0 as well. So this whole term evaluated from 0 to infinity is all 0, which is a nice, convenient thing for us. And then we're going to have this next term right there."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So when you evaluate it at infinity, when you evaluate this at infinity, you're going to get 0, and then you're going to subtract this evaluated at 0. When it's evaluated at 0, it's just minus 0 to the n times e to the minus s times 0 over s. Well, this becomes 0 as well. So this whole term evaluated from 0 to infinity is all 0, which is a nice, convenient thing for us. And then we're going to have this next term right there. So let's take out the constant terms. This n and this s are constant. They're constant with respect to t. So we have plus n over s times the integral from 0 to infinity of t to the n minus 1 times e to the minus st dt."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then we're going to have this next term right there. So let's take out the constant terms. This n and this s are constant. They're constant with respect to t. So we have plus n over s times the integral from 0 to infinity of t to the n minus 1 times e to the minus st dt. Now, this should look reasonably familiar to you. This should look reasonably... What's the definition of the Laplace transform? The Laplace transform of any function is equal to the integral from 0 to infinity of that function times e to the minus st dt."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "They're constant with respect to t. So we have plus n over s times the integral from 0 to infinity of t to the n minus 1 times e to the minus st dt. Now, this should look reasonably familiar to you. This should look reasonably... What's the definition of the Laplace transform? The Laplace transform of any function is equal to the integral from 0 to infinity of that function times e to the minus st dt. Well, we have an e to the minus st dt. We're taking the integral from 0 to infinity. So this whole integral is equal to the Laplace transform of this, of t to the n minus 1."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "The Laplace transform of any function is equal to the integral from 0 to infinity of that function times e to the minus st dt. Well, we have an e to the minus st dt. We're taking the integral from 0 to infinity. So this whole integral is equal to the Laplace transform of this, of t to the n minus 1. So just that easily, because this term went to 0, we've simplified things. We get the Laplace transform of t to the n is equal to... This is all 0."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this whole integral is equal to the Laplace transform of this, of t to the n minus 1. So just that easily, because this term went to 0, we've simplified things. We get the Laplace transform of t to the n is equal to... This is all 0. It's equal to n over s. That's right there, times this integral right here, which we just figured out was the Laplace transform of t to the n minus 1. Well, this is a nice, neat simplification. We can now figure out the Laplace transform of a higher power in terms of the one power lower than that, but it still doesn't give me a generalized formula."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is all 0. It's equal to n over s. That's right there, times this integral right here, which we just figured out was the Laplace transform of t to the n minus 1. Well, this is a nice, neat simplification. We can now figure out the Laplace transform of a higher power in terms of the one power lower than that, but it still doesn't give me a generalized formula. So let's see if we can use this with this information to get a generalized formula. So the Laplace transform of just t... Let me write that down. I wrote that at the beginning of the problem."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We can now figure out the Laplace transform of a higher power in terms of the one power lower than that, but it still doesn't give me a generalized formula. So let's see if we can use this with this information to get a generalized formula. So the Laplace transform of just t... Let me write that down. I wrote that at the beginning of the problem. We get the Laplace transform... I could write this as t to the 1, which is just t, is equal to 1 over s squared, where s is greater than 0. Now, what happens if we take the Laplace transform of t squared?"}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I wrote that at the beginning of the problem. We get the Laplace transform... I could write this as t to the 1, which is just t, is equal to 1 over s squared, where s is greater than 0. Now, what happens if we take the Laplace transform of t squared? Well, we can just use this formula up here. The Laplace transform of t squared is equal to 2 over s times the Laplace transform of t, of just t to the 1, right? 2 minus 1."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, what happens if we take the Laplace transform of t squared? Well, we can just use this formula up here. The Laplace transform of t squared is equal to 2 over s times the Laplace transform of t, of just t to the 1, right? 2 minus 1. Times the Laplace transform of t to the 1. Well, we know what that is. This is equal to 2 over s times this, times 1 over s squared, which is equal to 2 over s to the 3rd."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "2 minus 1. Times the Laplace transform of t to the 1. Well, we know what that is. This is equal to 2 over s times this, times 1 over s squared, which is equal to 2 over s to the 3rd. Interesting. Let's see if we can do another one. What is..."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is equal to 2 over s times this, times 1 over s squared, which is equal to 2 over s to the 3rd. Interesting. Let's see if we can do another one. What is... I'll do it in the dark blue. The Laplace transform of t to the 3rd. Well, we just use this formula up here."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What is... I'll do it in the dark blue. The Laplace transform of t to the 3rd. Well, we just use this formula up here. It's n over s. In this case, n is 3. So it's 3 over s times the Laplace transform of t to the n minus 1, so t squared. We know what the Laplace transform of this one was."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, we just use this formula up here. It's n over s. In this case, n is 3. So it's 3 over s times the Laplace transform of t to the n minus 1, so t squared. We know what the Laplace transform of this one was. This is just this right there. So it's equal to 3 over s times this thing. I'm going to actually write it this way because I think it's interesting."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We know what the Laplace transform of this one was. This is just this right there. So it's equal to 3 over s times this thing. I'm going to actually write it this way because I think it's interesting. So I'll write the numerator times 2 times 1 over s squared, which is, we can write it as 3 factorial over... What is this? s to the 4th power. Let's do another one."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'm going to actually write it this way because I think it's interesting. So I'll write the numerator times 2 times 1 over s squared, which is, we can write it as 3 factorial over... What is this? s to the 4th power. Let's do another one. I think you already are getting the idea of what's going on. Laplace transform of t to the 4th power is what? It's equal to 4 over s times the Laplace transform of t to the 3rd power."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's do another one. I think you already are getting the idea of what's going on. Laplace transform of t to the 4th power is what? It's equal to 4 over s times the Laplace transform of t to the 3rd power. That's just 4 over s times this. It's 4 over s times 3 factorial over s to the 4th. Now, 4 times 3 factorial, that's just 4 factorial over s to the 5th."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It's equal to 4 over s times the Laplace transform of t to the 3rd power. That's just 4 over s times this. It's 4 over s times 3 factorial over s to the 4th. Now, 4 times 3 factorial, that's just 4 factorial over s to the 5th. You can just get the general principle, and we can prove this by induction. It's almost trivial based on what we've already done. The Laplace transform of t to the n is equal to n factorial over s to the n plus 1."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, 4 times 3 factorial, that's just 4 factorial over s to the 5th. You can just get the general principle, and we can prove this by induction. It's almost trivial based on what we've already done. The Laplace transform of t to the n is equal to n factorial over s to the n plus 1. We tried it out. We proved it directly for this base case right here. This is 1 factorial over s to the 1 plus 1."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "The Laplace transform of t to the n is equal to n factorial over s to the n plus 1. We tried it out. We proved it directly for this base case right here. This is 1 factorial over s to the 1 plus 1. If we know it's true for this, we know it's going to be true for the next increment. Induction proof is almost obvious, but you can even see it based on this. If you had to figure out the Laplace transform of t to the 10th, you could just keep doing this over and over again, but I think you see the pattern pretty clearly."}, {"video_title": "Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is 1 factorial over s to the 1 plus 1. If we know it's true for this, we know it's going to be true for the next increment. Induction proof is almost obvious, but you can even see it based on this. If you had to figure out the Laplace transform of t to the 10th, you could just keep doing this over and over again, but I think you see the pattern pretty clearly. Anyway, I thought that was a neat problem in of itself, outside of the fact it'll be useful when we figure out inverse and Laplace transforms. This is a pretty neat result. Laplace transform of t to the n, where n is some integer greater than 0, is equal to n factorial over s to the n plus 1, where s is also greater than 0."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Now that we are familiar with Euler's method, let's do an exercise that tests our mathematical understanding of it, or at least the process of using it. So it says, consider the differential equation. The derivative of y with respect to x is equal to three x minus two y. Let y is equal to g of x be a solution to the differential equation with the initial condition g of zero is equal to k, where k is constant. Euler's method, starting at x equals zero, with a step size of one, gives the approximation that g of two is approximately 4.5. Find the value of k. So once again, this is saying, hey look, we're gonna start with this initial condition. When x is equal to zero, y is equal to k, we're going to use Euler's method with a step size of one."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Let y is equal to g of x be a solution to the differential equation with the initial condition g of zero is equal to k, where k is constant. Euler's method, starting at x equals zero, with a step size of one, gives the approximation that g of two is approximately 4.5. Find the value of k. So once again, this is saying, hey look, we're gonna start with this initial condition. When x is equal to zero, y is equal to k, we're going to use Euler's method with a step size of one. So we're essentially going to use, we're gonna step once from zero to one, and then again from one to two, and then that approximation is going to give us 4.5. And so given that we started at k, we should be able to figure out what k was to get us to g of two being approximated as 4.5. So with that, I encourage you to pause the video and try to figure this out on your own."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "When x is equal to zero, y is equal to k, we're going to use Euler's method with a step size of one. So we're essentially going to use, we're gonna step once from zero to one, and then again from one to two, and then that approximation is going to give us 4.5. And so given that we started at k, we should be able to figure out what k was to get us to g of two being approximated as 4.5. So with that, I encourage you to pause the video and try to figure this out on your own. I am assuming you have tried to figure this out on your own. Now we can do it together. And I'll do the same thing that we did in the first video on Euler's method."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So with that, I encourage you to pause the video and try to figure this out on your own. I am assuming you have tried to figure this out on your own. Now we can do it together. And I'll do the same thing that we did in the first video on Euler's method. I'll make a little table here. So let me make a little table. I could draw a straighter line than that."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And I'll do the same thing that we did in the first video on Euler's method. I'll make a little table here. So let me make a little table. I could draw a straighter line than that. That's only marginally straighter, but it'll get the job done. So let's make this column x. I'm gonna give myself some space for y. I might do some calculation here. Y and then dy dx."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "I could draw a straighter line than that. That's only marginally straighter, but it'll get the job done. So let's make this column x. I'm gonna give myself some space for y. I might do some calculation here. Y and then dy dx. Now, we can start at our initial condition. When x is equal to zero, y is equal to k. When x is equal to zero, y is equal to k. And so what's our derivative going to be at that point? Well, dy dx is equal to three x minus two y."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Y and then dy dx. Now, we can start at our initial condition. When x is equal to zero, y is equal to k. When x is equal to zero, y is equal to k. And so what's our derivative going to be at that point? Well, dy dx is equal to three x minus two y. So in this case, it's three times zero minus two times k, which is just equal to negative two k. And so now we can increment one more step. We have a step size. I'll do this in a different color."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, dy dx is equal to three x minus two y. So in this case, it's three times zero minus two times k, which is just equal to negative two k. And so now we can increment one more step. We have a step size. I'll do this in a different color. We have a step size of one. So we're gonna, in each step, we're gonna increment, in each step, we're gonna increment x by one. And so we're now going to be at one."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "I'll do this in a different color. We have a step size of one. So we're gonna, in each step, we're gonna increment, in each step, we're gonna increment x by one. And so we're now going to be at one. Now, what's our new y going to be? Well, if we increment x by one and our slope is negative two k, that means we're going to increment y by negative two k times one, or just negative two k. So negative two k. So k plus negative two k is negative k. So our approximation using Euler's method gets us the point one, negative k. And then what is going to be our slope starting at that point? So one, negative k. Our slope is going to be three times our x, which is one, minus two times our y, which is negative k now."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so we're now going to be at one. Now, what's our new y going to be? Well, if we increment x by one and our slope is negative two k, that means we're going to increment y by negative two k times one, or just negative two k. So negative two k. So k plus negative two k is negative k. So our approximation using Euler's method gets us the point one, negative k. And then what is going to be our slope starting at that point? So one, negative k. Our slope is going to be three times our x, which is one, minus two times our y, which is negative k now. And this is equal to three plus two k. Three plus two k. And now we'll do another step of one, because that's our step size. Another, whoops, now I'm going to get to two. And this is the one that we care about, right?"}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So one, negative k. Our slope is going to be three times our x, which is one, minus two times our y, which is negative k now. And this is equal to three plus two k. Three plus two k. And now we'll do another step of one, because that's our step size. Another, whoops, now I'm going to get to two. And this is the one that we care about, right? Because we're trying to approximate g of two. So we have to say, oh, what does our approximation give us for y when x is equal to two? And we're going to have something expressed in k, but they're saying that's going to be 4.5, and then we could use that to solve for k. So what's this going to be?"}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And this is the one that we care about, right? Because we're trying to approximate g of two. So we have to say, oh, what does our approximation give us for y when x is equal to two? And we're going to have something expressed in k, but they're saying that's going to be 4.5, and then we could use that to solve for k. So what's this going to be? So if we increment by one in x, we should increment our y by one times three plus two k, which is just going to be, so we're going to increment by three plus two k. Three plus two k, or negative k plus three plus two k is just going to be three plus k. Three plus k. And they're telling us that our approximation gets that to be 4.5. So three plus k is equal to 4.5. So the k that we started with must have been, if we just subtract three from both sides, this is a decimal here, it must have been, k must be equal to 1.5."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And we're going to have something expressed in k, but they're saying that's going to be 4.5, and then we could use that to solve for k. So what's this going to be? So if we increment by one in x, we should increment our y by one times three plus two k, which is just going to be, so we're going to increment by three plus two k. Three plus two k, or negative k plus three plus two k is just going to be three plus k. Three plus k. And they're telling us that our approximation gets that to be 4.5. So three plus k is equal to 4.5. So the k that we started with must have been, if we just subtract three from both sides, this is a decimal here, it must have been, k must be equal to 1.5. And you can verify that. If this initial condition right over here, if g of zero is equal to 1.5, g of zero is equal to 1.5, and you put 1.5 over here, then over here you would get 4.5. And we're done."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "The second derivative of y minus 3 times the first derivative minus 4 times the y is equal to, and now instead of having an exponential function or a trigonometric function, we'll just have a simple, well, this just looks like an x squared term, but it's a polynomial. And you know how to solve the general solution of the homogenous equation if this were 0. So we're going to focus just now on the particular solution, then we can later add that to the general solution of the homogenous equation to get the solution. So what's a good guess for a particular solution? Well, when we had exponentials, we guessed that our solution would be an exponential. When we had trigonometric functions, we guessed that our solution would be trigonometric. So since we have a polynomial here that makes this differential equation non-homogenous, let's guess that a particular solution is a polynomial."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "So what's a good guess for a particular solution? Well, when we had exponentials, we guessed that our solution would be an exponential. When we had trigonometric functions, we guessed that our solution would be trigonometric. So since we have a polynomial here that makes this differential equation non-homogenous, let's guess that a particular solution is a polynomial. And that makes sense. If you take a polynomial, and actually a second degree polynomial, if you take a second degree polynomial, take its derivatives and add and subtract, you should hopefully get another second degree polynomial. So let's guess that it is ax squared plus bx plus c. Well, what would be its second derivative?"}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "So since we have a polynomial here that makes this differential equation non-homogenous, let's guess that a particular solution is a polynomial. And that makes sense. If you take a polynomial, and actually a second degree polynomial, if you take a second degree polynomial, take its derivatives and add and subtract, you should hopefully get another second degree polynomial. So let's guess that it is ax squared plus bx plus c. Well, what would be its second derivative? Well, its second derivative would be 2ax plus b. And then the third derivative, sorry, this is the first derivative, the second derivative would be 2a. And now we could substitute back into the original equation."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "So let's guess that it is ax squared plus bx plus c. Well, what would be its second derivative? Well, its second derivative would be 2ax plus b. And then the third derivative, sorry, this is the first derivative, the second derivative would be 2a. And now we could substitute back into the original equation. We get the second derivative, 2a, minus 3 times the first derivative, so minus 3 times this. So minus 6ax minus 3b minus 4 times the function itself, so minus 4ax squared minus 4bx minus 4c, that's just 4 times all of that, that's going to equal 4x squared. Now let's group our x squared, our x, and our constant terms, and then we could try to solve for the coefficients."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "And now we could substitute back into the original equation. We get the second derivative, 2a, minus 3 times the first derivative, so minus 3 times this. So minus 6ax minus 3b minus 4 times the function itself, so minus 4ax squared minus 4bx minus 4c, that's just 4 times all of that, that's going to equal 4x squared. Now let's group our x squared, our x, and our constant terms, and then we could try to solve for the coefficients. So let's see, I have 1x squared term here. So it's minus 4ax squared. And then what are my x terms?"}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "Now let's group our x squared, our x, and our constant terms, and then we could try to solve for the coefficients. So let's see, I have 1x squared term here. So it's minus 4ax squared. And then what are my x terms? I have minus 6ax minus 4bx. So let's say plus minus 6a minus 4b times x. I just added the coefficients. And then finally, we get our constant terms, 2a minus 3b minus 4c, so plus 2a minus 3b minus 4c."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "And then what are my x terms? I have minus 6ax minus 4bx. So let's say plus minus 6a minus 4b times x. I just added the coefficients. And then finally, we get our constant terms, 2a minus 3b minus 4c, so plus 2a minus 3b minus 4c. And all of that will equal 4x squared. Now how do we solve for a, b, and c? Well, whatever the x squared coefficients add up on this side, it should equal 4."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "And then finally, we get our constant terms, 2a minus 3b minus 4c, so plus 2a minus 3b minus 4c. And all of that will equal 4x squared. Now how do we solve for a, b, and c? Well, whatever the x squared coefficients add up on this side, it should equal 4. Whatever the x coefficients add up on this side, it should be equal to 0, right? Because you can view this as plus 0x, right? And then you could say plus 0 constant as well."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "Well, whatever the x squared coefficients add up on this side, it should equal 4. Whatever the x coefficients add up on this side, it should be equal to 0, right? Because you can view this as plus 0x, right? And then you could say plus 0 constant as well. So the constant should also add up to 0. So let's do that. So first, let's do the x squared term."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "And then you could say plus 0 constant as well. So the constant should also add up to 0. So let's do that. So first, let's do the x squared term. So minus 4a should be equal to 4. So minus 4a is equal to 4. 4 is equal to 4."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "So first, let's do the x squared term. So minus 4a should be equal to 4. So minus 4a is equal to 4. 4 is equal to 4. And then that tells us that a is equal to minus 1. Fair enough. Now, the x terms."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "4 is equal to 4. And then that tells us that a is equal to minus 1. Fair enough. Now, the x terms. These minus 6a minus 4b, that should be equal to 0, right? So let's write that down. We know what a is, so let's substitute."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "Now, the x terms. These minus 6a minus 4b, that should be equal to 0, right? So let's write that down. We know what a is, so let's substitute. So minus 6 times a. So minus 6 times minus 1. So that's 6 minus 4b is equal to 0."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "We know what a is, so let's substitute. So minus 6 times a. So minus 6 times minus 1. So that's 6 minus 4b is equal to 0. So we get 4b. I'm just putting 4b on this side and then switching. 4b is equal to 6."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "So that's 6 minus 4b is equal to 0. So we get 4b. I'm just putting 4b on this side and then switching. 4b is equal to 6. And b is equal to 6 divided by 4 is 3 over 2. And finally, the constant term should also equal 0. So let's add those."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "4b is equal to 6. And b is equal to 6 divided by 4 is 3 over 2. And finally, the constant term should also equal 0. So let's add those. Let's solve for those. Well, 2 times a, that's minus 2. Minus 3 times b, well, that's minus 3 times this."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "So let's add those. Let's solve for those. Well, 2 times a, that's minus 2. Minus 3 times b, well, that's minus 3 times this. So minus 9 halves minus 4c is equal to 0. So let's see. I don't want to make a careless mistake."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "Minus 3 times b, well, that's minus 3 times this. So minus 9 halves minus 4c is equal to 0. So let's see. I don't want to make a careless mistake. So this is minus 4 minus 9 over 2, right? Right. That's minus 4 over 2, minus 9 over 2."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "I don't want to make a careless mistake. So this is minus 4 minus 9 over 2, right? Right. That's minus 4 over 2, minus 9 over 2. And then we could take the 4c, put it on that side, is equal to 4c. And what's minus 4 minus 9? That's minus 13 over 2."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "That's minus 4 over 2, minus 9 over 2. And then we could take the 4c, put it on that side, is equal to 4c. And what's minus 4 minus 9? That's minus 13 over 2. So minus 13 over 2 is equal to 4c. Or c, divide both sides by 4, and then you get c is equal to minus 13 over 8. And I think I haven't made a careless mistake."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "That's minus 13 over 2. So minus 13 over 2 is equal to 4c. Or c, divide both sides by 4, and then you get c is equal to minus 13 over 8. And I think I haven't made a careless mistake. So if I haven't, then our particular solution we now know. And actually, let's just write it in common. Well, let me write the whole solution."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "And I think I haven't made a careless mistake. So if I haven't, then our particular solution we now know. And actually, let's just write it in common. Well, let me write the whole solution. So, and this is a nice stretch of horizontal real estate. So let's write our solution. Our solution is going to be equal to the particular solution, which is ax squared."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "Well, let me write the whole solution. So, and this is a nice stretch of horizontal real estate. So let's write our solution. Our solution is going to be equal to the particular solution, which is ax squared. So that's minus 1x squared. So minus 1x squared, right? ax squared plus bx plus 3 halves x plus c. Minus 13 over 8."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "Our solution is going to be equal to the particular solution, which is ax squared. So that's minus 1x squared. So minus 1x squared, right? ax squared plus bx plus 3 halves x plus c. Minus 13 over 8. So this is the particular solution. We solved for a, b, and c. We determined the undetermined coefficients. And now if we want the general solution, we add to that the general solution of the homogenous equation."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "ax squared plus bx plus 3 halves x plus c. Minus 13 over 8. So this is the particular solution. We solved for a, b, and c. We determined the undetermined coefficients. And now if we want the general solution, we add to that the general solution of the homogenous equation. What was that? y prime minus 3y prime minus 4y is equal to 0. And we've solved this multiple times."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "And now if we want the general solution, we add to that the general solution of the homogenous equation. What was that? y prime minus 3y prime minus 4y is equal to 0. And we've solved this multiple times. We know that the general solution of the homogenous equation is c1 e to the 4x plus c2 e to the minus x. Right? You just take the characteristic equation, r squared minus 3r minus 4."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "And we've solved this multiple times. We know that the general solution of the homogenous equation is c1 e to the 4x plus c2 e to the minus x. Right? You just take the characteristic equation, r squared minus 3r minus 4. You get, what did you get? You get r minus 4 times r plus 1, and then that's how you get minus 1 and 4. Anyway, so this is the general solution to the homogenous equation."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "You just take the characteristic equation, r squared minus 3r minus 4. You get, what did you get? You get r minus 4 times r plus 1, and then that's how you get minus 1 and 4. Anyway, so this is the general solution to the homogenous equation. This is a particular solution to the non-homogenous equation. The general solution to the homogenous equation is going to be the sum of the two. So let's add that."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "Anyway, so this is the general solution to the homogenous equation. This is a particular solution to the non-homogenous equation. The general solution to the homogenous equation is going to be the sum of the two. So let's add that. So plus c1 e to the 4x plus c2 e to the minus x. So there you go. I don't think that was too painful."}, {"video_title": "Undetermined coefficients 3 Second order differential equations Khan Academy.mp3", "Sentence": "So let's add that. So plus c1 e to the 4x plus c2 e to the minus x. So there you go. I don't think that was too painful. The most painful part was just making sure that you don't make a careless mistake with the algebra. But using a fairly straightforward, really algebraic technique, we were able to get a fairly fancy solution to this second order linear non-homogenous differential equation with constant coefficients. See you in the next video."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "But the application here, at least I don't see the connection. Homogenous differential equation. And even within differential equations, we'll learn later, there's a different type of homogenous differential equation, and those are called homogenous linear differential equations, but they mean something actually quite different. But anyway, for this purposes, I'm going to show you homogenous differential equations, and what we're dealing with are going to be first order equations. And what does a homogenous differential equation mean? Well, let's say I had just a regular first order differential equation that could be written like this. So dy dx is equal to some function of x and y."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "But anyway, for this purposes, I'm going to show you homogenous differential equations, and what we're dealing with are going to be first order equations. And what does a homogenous differential equation mean? Well, let's say I had just a regular first order differential equation that could be written like this. So dy dx is equal to some function of x and y. And let's say we try to do this, and it's not separable, and it's not exact. What we learn is that if it can be homogenous, if this is a homogenous differential equation, that we can make a variable substitution, and that variable substitution allows this equation to turn into a separable one. But before I need to show you that, I need to tell you, well what does it mean to be homogenous?"}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "So dy dx is equal to some function of x and y. And let's say we try to do this, and it's not separable, and it's not exact. What we learn is that if it can be homogenous, if this is a homogenous differential equation, that we can make a variable substitution, and that variable substitution allows this equation to turn into a separable one. But before I need to show you that, I need to tell you, well what does it mean to be homogenous? Well, if I can algebraically manipulate this right side of this equation so that I can actually rewrite it, instead of a function of x and y, if I could rewrite this differential equation so that dy dx is equal to some function, I don't know, let's call that g, or we'll call it capital F. If I can rewrite it algebraically so it's a function of y divided by x, then I can make a variable substitution that makes it separable. So right now it seems all confusing. Let me show you an example."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "But before I need to show you that, I need to tell you, well what does it mean to be homogenous? Well, if I can algebraically manipulate this right side of this equation so that I can actually rewrite it, instead of a function of x and y, if I could rewrite this differential equation so that dy dx is equal to some function, I don't know, let's call that g, or we'll call it capital F. If I can rewrite it algebraically so it's a function of y divided by x, then I can make a variable substitution that makes it separable. So right now it seems all confusing. Let me show you an example. So let's say that, and I'll just show you the example, show you it's homogenous, and then we'll just do the substitution. So let's say that my differential equation is d, the derivative of y with respect to x, is equal to x plus y over x. And you can, if you'd like, you can try to separate, make this a separable, but it's not that trivial to solve, or at least I'm looking at an inspection, it doesn't seem that trivial to solve."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "Let me show you an example. So let's say that, and I'll just show you the example, show you it's homogenous, and then we'll just do the substitution. So let's say that my differential equation is d, the derivative of y with respect to x, is equal to x plus y over x. And you can, if you'd like, you can try to separate, make this a separable, but it's not that trivial to solve, or at least I'm looking at an inspection, it doesn't seem that trivial to solve. And as we see right here, we have the derivative, it's equal to some function of x and y. And my question is to you, can I just algebraically rewrite this so it becomes a function of y over x? Well sure, if we just divide both of these top terms by x, right, this is the same thing as x over x plus y over x."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "And you can, if you'd like, you can try to separate, make this a separable, but it's not that trivial to solve, or at least I'm looking at an inspection, it doesn't seem that trivial to solve. And as we see right here, we have the derivative, it's equal to some function of x and y. And my question is to you, can I just algebraically rewrite this so it becomes a function of y over x? Well sure, if we just divide both of these top terms by x, right, this is the same thing as x over x plus y over x. This equation is the same thing as dy over dx is equal to this, which is the same thing as rewriting this whole equation, I'm going to switch colors arbitrarily, as this. dy over dx is equal to x divided by x is equal to 1, if we assume x doesn't equal 0, plus y over x. So you're probably wondering, what did I mean by a function of y over x?"}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "Well sure, if we just divide both of these top terms by x, right, this is the same thing as x over x plus y over x. This equation is the same thing as dy over dx is equal to this, which is the same thing as rewriting this whole equation, I'm going to switch colors arbitrarily, as this. dy over dx is equal to x divided by x is equal to 1, if we assume x doesn't equal 0, plus y over x. So you're probably wondering, what did I mean by a function of y over x? Well you can see it here. When I just algebraically manipulated this equation, I got 1 plus y over x. So if I said that y over x is equal to some third variable, this is just a function of that third variable."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "So you're probably wondering, what did I mean by a function of y over x? Well you can see it here. When I just algebraically manipulated this equation, I got 1 plus y over x. So if I said that y over x is equal to some third variable, this is just a function of that third variable. And actually I'm going to do that right now. So let's say, let's make a substitution for y over x. So let's say that v, and I'll do v in a different color, let's say that v is equal to y over x, or another way, if you just multiply both sides by x, you could write that y is equal to xv."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "So if I said that y over x is equal to some third variable, this is just a function of that third variable. And actually I'm going to do that right now. So let's say, let's make a substitution for y over x. So let's say that v, and I'll do v in a different color, let's say that v is equal to y over x, or another way, if you just multiply both sides by x, you could write that y is equal to xv. And we're going to substitute v for y over x, but we're also going to have to substitute dy over dx. So let's figure out what that is in terms of the derivatives of v. So the derivative of y with respect to x is equal to, what's the derivative of this with respect to x? Well if we assume that v is also a function of x, then we're just going to use the product rule."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "So let's say that v, and I'll do v in a different color, let's say that v is equal to y over x, or another way, if you just multiply both sides by x, you could write that y is equal to xv. And we're going to substitute v for y over x, but we're also going to have to substitute dy over dx. So let's figure out what that is in terms of the derivatives of v. So the derivative of y with respect to x is equal to, what's the derivative of this with respect to x? Well if we assume that v is also a function of x, then we're just going to use the product rule. So the derivative of x is 1 times v, plus x times the derivative of v with respect to x. And now we can substitute this and this back into this equation, and we get. So dy over dx, that is equal to this."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "Well if we assume that v is also a function of x, then we're just going to use the product rule. So the derivative of x is 1 times v, plus x times the derivative of v with respect to x. And now we can substitute this and this back into this equation, and we get. So dy over dx, that is equal to this. So we get v plus x dv dx, derivative of v with respect to x, is equal to, that's just the left hand side, it's equal to 1 plus y over x, but we're making this substitution that v is equal to y over x, is equal to 1 plus v. And now this should be pretty straightforward. So let's see, we can subtract v from both sides of this equation. And then what do we have left?"}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "So dy over dx, that is equal to this. So we get v plus x dv dx, derivative of v with respect to x, is equal to, that's just the left hand side, it's equal to 1 plus y over x, but we're making this substitution that v is equal to y over x, is equal to 1 plus v. And now this should be pretty straightforward. So let's see, we can subtract v from both sides of this equation. And then what do we have left? We have x dv dx is equal to 1. Let's divide both sides by x, and we get the derivative of v with respect to x is equal to 1 over x. It should maybe start becoming a little bit clear what the solution here is, but let's just keep going forward."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "And then what do we have left? We have x dv dx is equal to 1. Let's divide both sides by x, and we get the derivative of v with respect to x is equal to 1 over x. It should maybe start becoming a little bit clear what the solution here is, but let's just keep going forward. So if we multiply both sides by dx, we get dv is equal to 1 over x times dx. Now we can take the antiderivative of both sides, integrate both sides, and we're left with v is equal to the natural log of the absolute value of x plus c. And we are kind of done, but it would be nice to get this solution in terms of just y and x, and not have this third variable v here, because our original problem was just in terms of y and x. So let's do that."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "It should maybe start becoming a little bit clear what the solution here is, but let's just keep going forward. So if we multiply both sides by dx, we get dv is equal to 1 over x times dx. Now we can take the antiderivative of both sides, integrate both sides, and we're left with v is equal to the natural log of the absolute value of x plus c. And we are kind of done, but it would be nice to get this solution in terms of just y and x, and not have this third variable v here, because our original problem was just in terms of y and x. So let's do that. What was v? We made the substitution that v is equal to y over x, so let's reverse substitute it now, or unsubstitute it. So we get y over x is equal to the natural log of x plus c, some constant."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "So let's do that. What was v? We made the substitution that v is equal to y over x, so let's reverse substitute it now, or unsubstitute it. So we get y over x is equal to the natural log of x plus c, some constant. Multiply both sides times x, and you get y is equal to x times the natural log of x plus c. And we're done. We solved that seemingly inseparable differential equation by recognizing that it was homogenous, and making that variable substitution v is equal to y over x. That turned it into a separable equation in terms of v, and then we solved it, and then we unsubstituted it back, and we got the solution to the differential equation."}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "So we get y over x is equal to the natural log of x plus c, some constant. Multiply both sides times x, and you get y is equal to x times the natural log of x plus c. And we're done. We solved that seemingly inseparable differential equation by recognizing that it was homogenous, and making that variable substitution v is equal to y over x. That turned it into a separable equation in terms of v, and then we solved it, and then we unsubstituted it back, and we got the solution to the differential equation. You can verify it for yourself that y is equal to the x natural log of the absolute value of x plus c. Oh, actually I made a mistake. y over x is equal to the natural log of x plus c. If I multiply both sides of this equation times x, what's the solution? It's not just x natural log of x. I have to multiply this times x too, right?"}, {"video_title": "First order homogenous equations First order differential equations Khan Academy.mp3", "Sentence": "That turned it into a separable equation in terms of v, and then we solved it, and then we unsubstituted it back, and we got the solution to the differential equation. You can verify it for yourself that y is equal to the x natural log of the absolute value of x plus c. Oh, actually I made a mistake. y over x is equal to the natural log of x plus c. If I multiply both sides of this equation times x, what's the solution? It's not just x natural log of x. I have to multiply this times x too, right? I'm multiplying a distributive property. That was an amateur mistake. So the correct solution is y is equal to x natural log of the absolute value of x plus x times c. And if you wanted to figure out c, I would have to give you some initial conditions, and then you could solve for c, and that would be the particular solution then for this differential equation."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "Now that you've had a little bit of exposure to what a convolution is, I can introduce you to the convolution theorem, or at least the convolution theorem, or at least in the context of, there may be other convolution theorems, but we're talking about differential equations in Laplace transforms. So this is the convolution theorem as applies to Laplace transforms. And it tells us that if I have a function, f of t, and I can define its Laplace transform as, let's see, the Laplace transform of f of t is capital F of s, we've done that before. And if I have another function, g of t, and I take its Laplace transform, that of course is capital G of s, that if we were to convolute these two functions, so if I were to take f and I were to convolute it with g, which is going to be another function of t, and we already saw this, we saw that in the last video, I convoluted sine and cosine. So this is going to be a function of t. That the Laplace transform of this thing, and this is the crux of the theorem, the Laplace transform of the convolution of these two functions is equal to the products of their Laplace transforms. It equals F of s, big capital F of s, times big capital G of s. Now this might seem very abstract and very hard to handle for you right now, so let's do an actual example, and actually even better, let's do an inverse Laplace transform with an example. Actually, let me write one more thing."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "And if I have another function, g of t, and I take its Laplace transform, that of course is capital G of s, that if we were to convolute these two functions, so if I were to take f and I were to convolute it with g, which is going to be another function of t, and we already saw this, we saw that in the last video, I convoluted sine and cosine. So this is going to be a function of t. That the Laplace transform of this thing, and this is the crux of the theorem, the Laplace transform of the convolution of these two functions is equal to the products of their Laplace transforms. It equals F of s, big capital F of s, times big capital G of s. Now this might seem very abstract and very hard to handle for you right now, so let's do an actual example, and actually even better, let's do an inverse Laplace transform with an example. Actually, let me write one more thing. If this is true, then we could also do it the other way. We could also say that F, and I'll just do it all in yellow, it takes me too much time to keep switching colors, that the convolution of F and G, which is just a function of t, I could just say it's the inverse Laplace transform of F of s times G of s. I couldn't resist it, let me switch colors. F of s times G of s. There you go."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "Actually, let me write one more thing. If this is true, then we could also do it the other way. We could also say that F, and I'll just do it all in yellow, it takes me too much time to keep switching colors, that the convolution of F and G, which is just a function of t, I could just say it's the inverse Laplace transform of F of s times G of s. I couldn't resist it, let me switch colors. F of s times G of s. There you go. Now, what good does all of this do? Well, we can take inverse Laplace transforms. Let's just say that I had, let me write it down here."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "F of s times G of s. There you go. Now, what good does all of this do? Well, we can take inverse Laplace transforms. Let's just say that I had, let me write it down here. Let's say I gave you, I told you that the following expression or function, let's say H of s is equal to 2s over s squared plus 1. We did this long differential equation at the end, we end up with this thing, and we have to take the inverse Laplace transform of it. We want to figure out the inverse Laplace transform of H of s, or the inverse Laplace transform of this thing right there."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "Let's just say that I had, let me write it down here. Let's say I gave you, I told you that the following expression or function, let's say H of s is equal to 2s over s squared plus 1. We did this long differential equation at the end, we end up with this thing, and we have to take the inverse Laplace transform of it. We want to figure out the inverse Laplace transform of H of s, or the inverse Laplace transform of this thing right there. We want to figure out the inverse Laplace transform of this expression right here, 2s over s squared plus 1 squared. Don't want to lose that right there. Now, can we write this as a product of two Laplace transforms that we do know?"}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "We want to figure out the inverse Laplace transform of H of s, or the inverse Laplace transform of this thing right there. We want to figure out the inverse Laplace transform of this expression right here, 2s over s squared plus 1 squared. Don't want to lose that right there. Now, can we write this as a product of two Laplace transforms that we do know? Let's try to do it. We can rewrite this, so this is the inverse Laplace transform. Let me rewrite this expression down here."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "Now, can we write this as a product of two Laplace transforms that we do know? Let's try to do it. We can rewrite this, so this is the inverse Laplace transform. Let me rewrite this expression down here. I can rewrite 2s over s squared plus 1 squared. This is the same thing as, let me write it this way, 2 times 1 over s squared plus 1, times s over s squared plus 1. I just kind of broke it up."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "Let me rewrite this expression down here. I can rewrite 2s over s squared plus 1 squared. This is the same thing as, let me write it this way, 2 times 1 over s squared plus 1, times s over s squared plus 1. I just kind of broke it up. If you multiply the numerators here, you get 2 times 1 times s, or 2s. If you multiply the denominators here, s squared plus 1 times s squared plus 1, that's just s squared plus 1 squared. This is the same thing."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "I just kind of broke it up. If you multiply the numerators here, you get 2 times 1 times s, or 2s. If you multiply the denominators here, s squared plus 1 times s squared plus 1, that's just s squared plus 1 squared. This is the same thing. If we want to take the inverse Laplace transform of this, it's the same thing as taking the inverse Laplace transform of this right here. Now, this something should hopefully start popping out at you. If these were separate transforms, if they were on their own, we know what this is."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "This is the same thing. If we want to take the inverse Laplace transform of this, it's the same thing as taking the inverse Laplace transform of this right here. Now, this something should hopefully start popping out at you. If these were separate transforms, if they were on their own, we know what this is. If we call this f of s, if we say that this is a Laplace transform of some function, we know what that function is. This piece right here, I'm just doing a little dotted line around it, this is the Laplace transform of sine of t. Then, if we draw a little box around this one right here, this is the Laplace transform of cosine of t, g of s. This is the Laplace transform of sine of t, or we could write that this implies that f of t is equal to sine of t. You should recognize that one by now. This implies that g of t, if we define this as the Laplace transform of g, this means that g of t is equal to cosine of t. Of course, when you take the inverse Laplace transform, you could take the 2s out."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "If these were separate transforms, if they were on their own, we know what this is. If we call this f of s, if we say that this is a Laplace transform of some function, we know what that function is. This piece right here, I'm just doing a little dotted line around it, this is the Laplace transform of sine of t. Then, if we draw a little box around this one right here, this is the Laplace transform of cosine of t, g of s. This is the Laplace transform of sine of t, or we could write that this implies that f of t is equal to sine of t. You should recognize that one by now. This implies that g of t, if we define this as the Laplace transform of g, this means that g of t is equal to cosine of t. Of course, when you take the inverse Laplace transform, you could take the 2s out. You could say, now what can we say? We can now say that the inverse, or we could actually have a better thing to do, instead of taking the 2 out so I can leave it nice and clean, if we were to draw a box around this whole thing and define this whole thing as f of s, then f of s is the Laplace transform of 2 sine of t. I just wanted to include that 2. I didn't want to leave that out and confuse the issue."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "This implies that g of t, if we define this as the Laplace transform of g, this means that g of t is equal to cosine of t. Of course, when you take the inverse Laplace transform, you could take the 2s out. You could say, now what can we say? We can now say that the inverse, or we could actually have a better thing to do, instead of taking the 2 out so I can leave it nice and clean, if we were to draw a box around this whole thing and define this whole thing as f of s, then f of s is the Laplace transform of 2 sine of t. I just wanted to include that 2. I didn't want to leave that out and confuse the issue. I wanted a very pure f of s times g of s. This expression right here is the product of the Laplace transform of 2 sine of t and the Laplace transform of cosine of t. Now, our convolution theorem told us this right here, that if we want to take the inverse Laplace transform of the Laplace transforms of 2 functions, I know that sounds very confusing, but you just kind of pattern match. This thing that I had here, I could rewrite it as a product of 2 Laplace transforms I can recognize. This right here is the Laplace transform of 2 sine of t. This is the Laplace transform of cosine of t. We just wrote that as g of s and f of s. If I have an expression written like this, I can take the inverse Laplace transform and it will be equal to the convolution of the original functions."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "I didn't want to leave that out and confuse the issue. I wanted a very pure f of s times g of s. This expression right here is the product of the Laplace transform of 2 sine of t and the Laplace transform of cosine of t. Now, our convolution theorem told us this right here, that if we want to take the inverse Laplace transform of the Laplace transforms of 2 functions, I know that sounds very confusing, but you just kind of pattern match. This thing that I had here, I could rewrite it as a product of 2 Laplace transforms I can recognize. This right here is the Laplace transform of 2 sine of t. This is the Laplace transform of cosine of t. We just wrote that as g of s and f of s. If I have an expression written like this, I can take the inverse Laplace transform and it will be equal to the convolution of the original functions. It will be equal to the convolution of the inverse of g and the inverse of f. Let me write it this way. I could write it like this. We know that f of t is equal to the inverse Laplace transform of f of s. We know that g of t is equal to the inverse Laplace transform of g of s. We could rewrite the convolution theorem as the inverse Laplace transform of, I'm going to try to stay true to the colors, of f of s times g of s is equal to, I'm just restating this convolution theorem right here, this is equal to the convolution of the inverse Laplace transform of f of s with the inverse Laplace transform of g of s. I'm not sure if that helps you or not, but if you go back to this example, it might."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "This right here is the Laplace transform of 2 sine of t. This is the Laplace transform of cosine of t. We just wrote that as g of s and f of s. If I have an expression written like this, I can take the inverse Laplace transform and it will be equal to the convolution of the original functions. It will be equal to the convolution of the inverse of g and the inverse of f. Let me write it this way. I could write it like this. We know that f of t is equal to the inverse Laplace transform of f of s. We know that g of t is equal to the inverse Laplace transform of g of s. We could rewrite the convolution theorem as the inverse Laplace transform of, I'm going to try to stay true to the colors, of f of s times g of s is equal to, I'm just restating this convolution theorem right here, this is equal to the convolution of the inverse Laplace transform of f of s with the inverse Laplace transform of g of s. I'm not sure if that helps you or not, but if you go back to this example, it might. This is f of s right here. 2 over s squared plus 1. That's f of s in our example."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "We know that f of t is equal to the inverse Laplace transform of f of s. We know that g of t is equal to the inverse Laplace transform of g of s. We could rewrite the convolution theorem as the inverse Laplace transform of, I'm going to try to stay true to the colors, of f of s times g of s is equal to, I'm just restating this convolution theorem right here, this is equal to the convolution of the inverse Laplace transform of f of s with the inverse Laplace transform of g of s. I'm not sure if that helps you or not, but if you go back to this example, it might. This is f of s right here. 2 over s squared plus 1. That's f of s in our example. The g of s was s over s squared plus 1. All I got that from is I just broke this up into two things that I recognize. If I multiply this together, I get back to my original thing that I was trying to take the inverse Laplace transform of."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "That's f of s in our example. The g of s was s over s squared plus 1. All I got that from is I just broke this up into two things that I recognize. If I multiply this together, I get back to my original thing that I was trying to take the inverse Laplace transform of. The convolution theorem just says that the inverse Laplace transform of this is equal to the inverse Laplace transform of 2 over s squared plus 1 convoluted with the inverse Laplace transform of our g of s, of s over s squared plus 1. We know what these things are. I already told them to you, but they should be somewhat second nature now."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "If I multiply this together, I get back to my original thing that I was trying to take the inverse Laplace transform of. The convolution theorem just says that the inverse Laplace transform of this is equal to the inverse Laplace transform of 2 over s squared plus 1 convoluted with the inverse Laplace transform of our g of s, of s over s squared plus 1. We know what these things are. I already told them to you, but they should be somewhat second nature now. This is 2 times sine of t. Take the Laplace transform of sine of t, you get 1 over s squared plus 1, and then you multiply it by 2, you get the 2 up there. You're going to have to convolute that with the inverse Laplace transform of this thing here. We already went over this."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "I already told them to you, but they should be somewhat second nature now. This is 2 times sine of t. Take the Laplace transform of sine of t, you get 1 over s squared plus 1, and then you multiply it by 2, you get the 2 up there. You're going to have to convolute that with the inverse Laplace transform of this thing here. We already went over this. This is cosine of t. Our result so far, let me be very clear, it's always good to take a step back and just think about what we're doing, much less why we're doing it. Let's see, the Laplace transform, the inverse Laplace transform of this thing up in this top left corner of 2s over s squared plus 1 squared, which before we did what we're doing now was very hard to figure out. Actually, this should be a curly bracket right here, but you get the idea."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "We already went over this. This is cosine of t. Our result so far, let me be very clear, it's always good to take a step back and just think about what we're doing, much less why we're doing it. Let's see, the Laplace transform, the inverse Laplace transform of this thing up in this top left corner of 2s over s squared plus 1 squared, which before we did what we're doing now was very hard to figure out. Actually, this should be a curly bracket right here, but you get the idea. It's equal to this. It's equal to 2 sine of t convoluted with cosine of t. You're like, Sal, throughout this whole process, I've already forgotten what it means to convolute two functions, so let's convolute them. I'll just write the definition, or the definition we're using of the convolution, that f convoluted with g, it's going to be a function of t, I'll just write this shorthand, is equal to the integral from 0 to t of f of t minus tau times g of tau d tau."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "Actually, this should be a curly bracket right here, but you get the idea. It's equal to this. It's equal to 2 sine of t convoluted with cosine of t. You're like, Sal, throughout this whole process, I've already forgotten what it means to convolute two functions, so let's convolute them. I'll just write the definition, or the definition we're using of the convolution, that f convoluted with g, it's going to be a function of t, I'll just write this shorthand, is equal to the integral from 0 to t of f of t minus tau times g of tau d tau. So 2 sine of t convoluted with cosine of t is equal to, so this is equal to, let me do a neutral color, the integral from 0 to t of 2 sine of t minus tau times the cosine of tau d tau. Now, if you watched the very last video I made, I actually solved this, or I solved a very similar thing to this. If we take the 2 out, we get 2 times the integral from 0 to t of sine of t minus tau times the cosine of tau."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "I'll just write the definition, or the definition we're using of the convolution, that f convoluted with g, it's going to be a function of t, I'll just write this shorthand, is equal to the integral from 0 to t of f of t minus tau times g of tau d tau. So 2 sine of t convoluted with cosine of t is equal to, so this is equal to, let me do a neutral color, the integral from 0 to t of 2 sine of t minus tau times the cosine of tau d tau. Now, if you watched the very last video I made, I actually solved this, or I solved a very similar thing to this. If we take the 2 out, we get 2 times the integral from 0 to t of sine of t minus tau times the cosine of tau. I actually solved this in the previous video. This right here, this is a convolution of sine of t and cosine of t. It's sine of t convoluted with cosine of t. I show you in the previous video, just watch that video where I introduce a convolution, that this thing right here is equal to 1 half t sine of t. Now, if this thing is equal to 1 half t sine of t, and I have to multiply it by 2, then we get our big result that the inverse Laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t, which is just 2 times this thing here, which is 2 times 1 half. Those cancel out, so it equals t sine of t. So all of this mess, and once you get the hang of it, you won't have to go through all of these steps, but the key is to recognize that this could be broken down as a product of 2 Laplace transforms that you recognize."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "If we take the 2 out, we get 2 times the integral from 0 to t of sine of t minus tau times the cosine of tau. I actually solved this in the previous video. This right here, this is a convolution of sine of t and cosine of t. It's sine of t convoluted with cosine of t. I show you in the previous video, just watch that video where I introduce a convolution, that this thing right here is equal to 1 half t sine of t. Now, if this thing is equal to 1 half t sine of t, and I have to multiply it by 2, then we get our big result that the inverse Laplace transform of 2s over s squared plus 1 squared is equal to the convolution of 2 sine of t with cosine of t, which is just 2 times this thing here, which is 2 times 1 half. Those cancel out, so it equals t sine of t. So all of this mess, and once you get the hang of it, you won't have to go through all of these steps, but the key is to recognize that this could be broken down as a product of 2 Laplace transforms that you recognize. This could be broken down as a product of 2 Laplace transforms we recognized. This was a Laplace transform of 2 sine of t. This was a Laplace transform of cosine of t. So the inverse Laplace transform of our original thing, our original expression, is just the convolution of that with that. And if you watched the previous video, you'll realize that actually calculating that convolution was no simple task, but it can be done."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "Those cancel out, so it equals t sine of t. So all of this mess, and once you get the hang of it, you won't have to go through all of these steps, but the key is to recognize that this could be broken down as a product of 2 Laplace transforms that you recognize. This could be broken down as a product of 2 Laplace transforms we recognized. This was a Laplace transform of 2 sine of t. This was a Laplace transform of cosine of t. So the inverse Laplace transform of our original thing, our original expression, is just the convolution of that with that. And if you watched the previous video, you'll realize that actually calculating that convolution was no simple task, but it can be done. So you actually can get an integral form. Even if it can't be done, you can get your answer at least in terms of some integral. I haven't proven the convolution theorem to you just yet."}, {"video_title": "The convolution and the laplace transform Laplace transform Khan Academy.mp3", "Sentence": "And if you watched the previous video, you'll realize that actually calculating that convolution was no simple task, but it can be done. So you actually can get an integral form. Even if it can't be done, you can get your answer at least in terms of some integral. I haven't proven the convolution theorem to you just yet. I'll do that in a future video. But hopefully this gave you a little bit of a sense of how you can use it to actually take inverse Laplace transforms. And remember, the reason why we're learning to take inverse Laplace transforms and we have all of these tools to do it is because that's always that last step when you're solving these differential equations using your Laplace transforms."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "It was different than n with respect to x. We said, oh boy, it's not exact. But we said, what if we could multiply both sides of this equation by some function that would make it exact? And we called that mu. And in the last video, we actually solved for mu. We said, well, if we multiply both sides of this equation by mu of x is equal to x, it should make this into an exact differential equation. It's important to note, there might have been a function of y that if I multiplied by both sides, it would also make it exact."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "And we called that mu. And in the last video, we actually solved for mu. We said, well, if we multiply both sides of this equation by mu of x is equal to x, it should make this into an exact differential equation. It's important to note, there might have been a function of y that if I multiplied by both sides, it would also make it exact. There might have been a function of x and y that would have done the trick. But really, our whole goal is just to make this exact. So it doesn't matter which one we pick, which integrating factor, and this is called the integrating factor."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "It's important to note, there might have been a function of y that if I multiplied by both sides, it would also make it exact. There might have been a function of x and y that would have done the trick. But really, our whole goal is just to make this exact. So it doesn't matter which one we pick, which integrating factor, and this is called the integrating factor. Which integrating factor we pick. So anyway, let's do it now. Let's solve the problem."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So it doesn't matter which one we pick, which integrating factor, and this is called the integrating factor. Which integrating factor we pick. So anyway, let's do it now. Let's solve the problem. So let's multiply both sides of this equation by mu, and mu of x is just x. So we multiply both sides by x. So we get, see, if you multiply this term by x, you get 3x squared y plus xy squared plus, we're multiplying these terms by x now, plus x to the third plus x squared y, y prime is equal to 0."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "Let's solve the problem. So let's multiply both sides of this equation by mu, and mu of x is just x. So we multiply both sides by x. So we get, see, if you multiply this term by x, you get 3x squared y plus xy squared plus, we're multiplying these terms by x now, plus x to the third plus x squared y, y prime is equal to 0. Well now, first of all, just as a reality check, let's make sure that this is now an exact equation. So what's the partial of this expression, or this kind of sub-function, with respect to y? Well, it's 3x squared, that's just a kind of constant coefficient of y, plus 2xy, that's the partial with respect to y of that expression."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So we get, see, if you multiply this term by x, you get 3x squared y plus xy squared plus, we're multiplying these terms by x now, plus x to the third plus x squared y, y prime is equal to 0. Well now, first of all, just as a reality check, let's make sure that this is now an exact equation. So what's the partial of this expression, or this kind of sub-function, with respect to y? Well, it's 3x squared, that's just a kind of constant coefficient of y, plus 2xy, that's the partial with respect to y of that expression. Now let's take the partial of this with respect to x. So we get 3x squared plus 2xy. And there we have it."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "Well, it's 3x squared, that's just a kind of constant coefficient of y, plus 2xy, that's the partial with respect to y of that expression. Now let's take the partial of this with respect to x. So we get 3x squared plus 2xy. And there we have it. The partial of this with respect to y is equal to the partial of this with respect to n. So we now have an exact equation whose solution should be the same as this. All we did is we multiplied both sides of this equation by x. So it really shouldn't change the solution of that equation or that differential equation."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "And there we have it. The partial of this with respect to y is equal to the partial of this with respect to n. So we now have an exact equation whose solution should be the same as this. All we did is we multiplied both sides of this equation by x. So it really shouldn't change the solution of that equation or that differential equation. So it's exact. Let's solve it. So how do we do that?"}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So it really shouldn't change the solution of that equation or that differential equation. So it's exact. Let's solve it. So how do we do that? Well, what we say is, since we've shown this exact, we know that there's some function xi where the partial derivative of xi with respect to x is equal to this expression right here. So it's equal to 3x squared y plus xy squared. Let's take the antiderivative of both sides with respect to x, and we'll get xi is equal to what?"}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So how do we do that? Well, what we say is, since we've shown this exact, we know that there's some function xi where the partial derivative of xi with respect to x is equal to this expression right here. So it's equal to 3x squared y plus xy squared. Let's take the antiderivative of both sides with respect to x, and we'll get xi is equal to what? Let's see, it's x to the third y plus, let's see, x, we could write 1 half x squared y squared. And of course, this xi is a function of x and y, so when you take the partial with respect to x, when you go that way, you might have lost some function that's only a function of y. So instead of a plus c here, there could have been a whole function of y that we lost."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "Let's take the antiderivative of both sides with respect to x, and we'll get xi is equal to what? Let's see, it's x to the third y plus, let's see, x, we could write 1 half x squared y squared. And of course, this xi is a function of x and y, so when you take the partial with respect to x, when you go that way, you might have lost some function that's only a function of y. So instead of a plus c here, there could have been a whole function of y that we lost. So we'll add that back when we take the antiderivative. So this is our xi, but we're not completely done yet because we have to somehow figure out what this function of y is. And the way we figure that out is we use the information that the partial of this with respect to y should be equal to this."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So instead of a plus c here, there could have been a whole function of y that we lost. So we'll add that back when we take the antiderivative. So this is our xi, but we're not completely done yet because we have to somehow figure out what this function of y is. And the way we figure that out is we use the information that the partial of this with respect to y should be equal to this. So let's set that up. So what's the partial of this expression with respect to y? So it would be, so I could write xi, the partial of xi with respect to y is equal to, let's see, x to the third plus, let's see, 2 times 1 half, so it's just x squared y plus h prime of y."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "And the way we figure that out is we use the information that the partial of this with respect to y should be equal to this. So let's set that up. So what's the partial of this expression with respect to y? So it would be, so I could write xi, the partial of xi with respect to y is equal to, let's see, x to the third plus, let's see, 2 times 1 half, so it's just x squared y plus h prime of y. That's the partial of a function purely of y with respect to y. And then that has to equal our new n, or the new expression we got after we multiplied by the integrating factor. So that's going to be equal to this right here."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So it would be, so I could write xi, the partial of xi with respect to y is equal to, let's see, x to the third plus, let's see, 2 times 1 half, so it's just x squared y plus h prime of y. That's the partial of a function purely of y with respect to y. And then that has to equal our new n, or the new expression we got after we multiplied by the integrating factor. So that's going to be equal to this right here. This will hopefully be making sense to you at this point. So that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side, so let's subtract both of those terms from both sides."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So that's going to be equal to this right here. This will hopefully be making sense to you at this point. So that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side, so let's subtract both of those terms from both sides. So x to the third, x to the third, x squared y, x squared y, and we're left with h prime of y is equal to 0. Or you could say that h of y is equal to some constant. And so there's really no y, I guess, extra function of y."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "And interesting enough, both of these terms are on this side, so let's subtract both of those terms from both sides. So x to the third, x to the third, x squared y, x squared y, and we're left with h prime of y is equal to 0. Or you could say that h of y is equal to some constant. And so there's really no y, I guess, extra function of y. There's just some constant left over. So for our purposes, we can just say that xi is equal to this, because this is just a constant. We're going to take the antiderivative anyway and get a constant on the right-hand side."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "And so there's really no y, I guess, extra function of y. There's just some constant left over. So for our purposes, we can just say that xi is equal to this, because this is just a constant. We're going to take the antiderivative anyway and get a constant on the right-hand side. And in the previous videos, the constants all merged together. So we'll just assume that that is our xi. And we know that this differential equation up here can be rewritten as the derivative of xi with respect to x."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "We're going to take the antiderivative anyway and get a constant on the right-hand side. And in the previous videos, the constants all merged together. So we'll just assume that that is our xi. And we know that this differential equation up here can be rewritten as the derivative of xi with respect to x. And that just falls out of the partial derivative chain rule. The derivative of xi with respect to x is equal to 0. If you took the derivative of xi with respect to x, it should be equal to this whole thing, just using the partial derivative chain rule."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "And we know that this differential equation up here can be rewritten as the derivative of xi with respect to x. And that just falls out of the partial derivative chain rule. The derivative of xi with respect to x is equal to 0. If you took the derivative of xi with respect to x, it should be equal to this whole thing, just using the partial derivative chain rule. Well, we know what xi is, so we can write. Or actually, we could use this fact to say, well, if we integrate both sides, that the solution of this differential equation is that xi is equal to c. I just took the antiderivative of both sides. So a solution to the differential equation is xi is equal to c. So xi is equal to x to the third y plus 1 half x squared y squared."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "If you took the derivative of xi with respect to x, it should be equal to this whole thing, just using the partial derivative chain rule. Well, we know what xi is, so we can write. Or actually, we could use this fact to say, well, if we integrate both sides, that the solution of this differential equation is that xi is equal to c. I just took the antiderivative of both sides. So a solution to the differential equation is xi is equal to c. So xi is equal to x to the third y plus 1 half x squared y squared. And we could have said plus c here, but we know the solution is that xi is equal to c, so we'll just write that there. I could have written a plus c here, but then you have a plus c here, and you have another constant there, and you could just subtract them from both sides, and they just merge into another arbitrary constant. But anyway, there we have it."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So a solution to the differential equation is xi is equal to c. So xi is equal to x to the third y plus 1 half x squared y squared. And we could have said plus c here, but we know the solution is that xi is equal to c, so we'll just write that there. I could have written a plus c here, but then you have a plus c here, and you have another constant there, and you could just subtract them from both sides, and they just merge into another arbitrary constant. But anyway, there we have it. We had a differential equation that at least superficially looked exact. But then when we tested the exactness of it, it was not exact, but we multiplied it by an integrating factor. In the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "But anyway, there we have it. We had a differential equation that at least superficially looked exact. But then when we tested the exactness of it, it was not exact, but we multiplied it by an integrating factor. In the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x. And when we did that, we tested it, and true enough, it was exact. And so given that it was exact, we knew that a xi would exist where the derivative of xi with respect to x would be equal to this entire expression. So we could rewrite our differential equation like this, and we'd know that a solution is xi is equal to c. And to solve for xi, we just say, OK, the partial derivative of xi with respect to x is going to be this thing, antiderivative of both sides."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "In the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x. And when we did that, we tested it, and true enough, it was exact. And so given that it was exact, we knew that a xi would exist where the derivative of xi with respect to x would be equal to this entire expression. So we could rewrite our differential equation like this, and we'd know that a solution is xi is equal to c. And to solve for xi, we just say, OK, the partial derivative of xi with respect to x is going to be this thing, antiderivative of both sides. There's some constant h of y, not constant, there's some function of y, h of y that we might have lost when we took the partial with respect to x. So to figure that out, we take this expression, take the partial with respect to y, and set that equal to our n expression. And by doing that, we figured out that that function of y is really just some constant."}, {"video_title": "Integrating factors 2 First order differential equations Khan Academy.mp3", "Sentence": "So we could rewrite our differential equation like this, and we'd know that a solution is xi is equal to c. And to solve for xi, we just say, OK, the partial derivative of xi with respect to x is going to be this thing, antiderivative of both sides. There's some constant h of y, not constant, there's some function of y, h of y that we might have lost when we took the partial with respect to x. So to figure that out, we take this expression, take the partial with respect to y, and set that equal to our n expression. And by doing that, we figured out that that function of y is really just some constant. And we could have written that here. We could have written that plus c. But we know, we could call that c1 or something, but we know that the solution of our original differential equation is xi is equal to c. So the solution of our differential equation is xi, x third y plus 1 half x squared y squared is equal to c. We could have had this plus c1 here and then subtracted both sides. But I think I've said it so many times that you understand why if h of y is just a c, you can kind of ignore it."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "which of the differential equations are separable. And I encourage you to pause this video and see which of these are actually separable. Now, the way that I approach this is I try to solve for the derivative, and if when I solve for the derivative, if I get dy dx is equal to some function of y times some other function of x, then I say, okay, this is separable, because I could rewrite this as, I could divide both sides by g of y, and I get one over g of y, which is itself a function of y, times dy is equal to h of x dx. You would go from this first equation to the second equation just by dividing both sides by g of y and multiplying both sides by dx, and then it's clear you have a separable equation, you can integrate both sides. But the key is let's solve for the derivative and see if we can put this in a form where we have the product of a function of y times a function of x. So let's do it with this first one here. So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "You would go from this first equation to the second equation just by dividing both sides by g of y and multiplying both sides by dx, and then it's clear you have a separable equation, you can integrate both sides. But the key is let's solve for the derivative and see if we can put this in a form where we have the product of a function of y times a function of x. So let's do it with this first one here. So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y. So I subtracted y from both sides. Now, let's see, if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually, I'm gonna write it this way, I'm gonna write it three minus y times one over x. And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y. So I subtracted y from both sides. Now, let's see, if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually, I'm gonna write it this way, I'm gonna write it three minus y times one over x. And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x. So this indeed is separable. And I could show you, I can multiply both sides by dx, and I can divide both sides by three minus y now, and I would get one over three minus y dy is equal to one over x dx. So clearly, this one right over here is separable."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x. So this indeed is separable. And I could show you, I can multiply both sides by dx, and I can divide both sides by three minus y now, and I would get one over three minus y dy is equal to one over x dx. So clearly, this one right over here is separable. Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a different color so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So clearly, this one right over here is separable. Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a different color so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once. I'm gonna subtract two x from both sides. I am going to subtract two y from both sides. So I'm gonna subtract two y from both sides."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once. I'm gonna subtract two x from both sides. I am going to subtract two y from both sides. So I'm gonna subtract two y from both sides. I'm gonna add one to both sides. So I'm gonna add one to both sides. And then what am I going to get if I do that?"}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm gonna subtract two y from both sides. I'm gonna add one to both sides. So I'm gonna add one to both sides. And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x minus two y plus one. And now let's see, I can divide everything by two."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x minus two y plus one. And now let's see, I can divide everything by two. I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two. So I'm gonna get negative x minus y and then I'm going to get plus 1 1 2. So it's not obvious to me how I can write this as a product of a function of x and a function of y."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "And now let's see, I can divide everything by two. I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two. So I'm gonna get negative x minus y and then I'm going to get plus 1 1 2. So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y. So this one I'm gonna say is not separable."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y. So this one I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating, I can rewrite this as, well, this is dy dx. If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So this one I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating, I can rewrite this as, well, this is dy dx. If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx. So clearly separable. Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx. So clearly separable. Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y is equal to x."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "They've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x over x plus y. And here, my algebraic toolkit of how do I separate x and y so I can write this as a function of x times a function of y, not obvious to me here. So this one is not separable."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "I know everything we've done so far has really just been a toolkit of being able to solve them. But the whole reason is that because differential equations can describe a lot of systems and that we can actually model them that way. And we know that in the real world, everything isn't these nice, continuous functions. So over the next couple of videos, we're going to talk about functions that are a little bit more discontinuous than what you might be used to even in kind of your traditional calculus or traditional pre-calculus class. And the first one is the unit step function. Let's write it as u, and I'll put a little subscript c here of t. And it's defined as when t is 0, when t is less than, whatever subscript I put here, when t is less than c. And it's defined as 1, that's why we call it the unit step function, when t is greater than or equal to c. And if I had to graph this, you could graph it as well, but it's not too difficult to graph. Let me draw my x-axis right here."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So over the next couple of videos, we're going to talk about functions that are a little bit more discontinuous than what you might be used to even in kind of your traditional calculus or traditional pre-calculus class. And the first one is the unit step function. Let's write it as u, and I'll put a little subscript c here of t. And it's defined as when t is 0, when t is less than, whatever subscript I put here, when t is less than c. And it's defined as 1, that's why we call it the unit step function, when t is greater than or equal to c. And if I had to graph this, you could graph it as well, but it's not too difficult to graph. Let me draw my x-axis right here. I'll do a little thicker line. That's my x-axis right there. This is my y-axis right there."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Let me draw my x-axis right here. I'll do a little thicker line. That's my x-axis right there. This is my y-axis right there. And when we talk about Laplace transforms, which we'll talk about shortly, we only care about t is greater than 0, because we saw our definition of a Laplace transform, we're always taking the integral from 0 to infinity. So we're only dealing with the positive x-axis. But anyway, by this definition, it would be 0 all the way until you get to some value c. So you'd be 0 until you get to c. And then at c, you jump, and the point c is included."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "This is my y-axis right there. And when we talk about Laplace transforms, which we'll talk about shortly, we only care about t is greater than 0, because we saw our definition of a Laplace transform, we're always taking the integral from 0 to infinity. So we're only dealing with the positive x-axis. But anyway, by this definition, it would be 0 all the way until you get to some value c. So you'd be 0 until you get to c. And then at c, you jump, and the point c is included. x is equal to c here. So it's included, so I'll put a dot there, because it's greater than or equal to c. You're at 1. So this is 1 right here."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "But anyway, by this definition, it would be 0 all the way until you get to some value c. So you'd be 0 until you get to c. And then at c, you jump, and the point c is included. x is equal to c here. So it's included, so I'll put a dot there, because it's greater than or equal to c. You're at 1. So this is 1 right here. And then you go forward for all of time. And you're like, Sal, you just said that differential equations, we're going to model things. Why is this type of a function useful?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So this is 1 right here. And then you go forward for all of time. And you're like, Sal, you just said that differential equations, we're going to model things. Why is this type of a function useful? Well, in the real world, sometimes you do have something that essentially jolts something, that moves it from this position to that position. And obviously, nothing can move it immediately like this, but you might have some system. It could be an electrical system or a mechanical system, where maybe the behavior looks something like this, where maybe it moves it like that or something."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Why is this type of a function useful? Well, in the real world, sometimes you do have something that essentially jolts something, that moves it from this position to that position. And obviously, nothing can move it immediately like this, but you might have some system. It could be an electrical system or a mechanical system, where maybe the behavior looks something like this, where maybe it moves it like that or something. And this function is a pretty good analytic approximation for some type of real world behavior like this, when something just gets moved. Whenever we solve these differential equations analytically, we're really just trying to get a pure model of something. Eventually, we'll see that it doesn't perfectly describe things, but it helps describe it enough for us to get a sense of what's going to happen."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "It could be an electrical system or a mechanical system, where maybe the behavior looks something like this, where maybe it moves it like that or something. And this function is a pretty good analytic approximation for some type of real world behavior like this, when something just gets moved. Whenever we solve these differential equations analytically, we're really just trying to get a pure model of something. Eventually, we'll see that it doesn't perfectly describe things, but it helps describe it enough for us to get a sense of what's going to happen. Sometimes it will completely describe things, but anyway, we can ignore that for now. So let me get rid of these things right there. So the first question is, well, what if something doesn't jar just like that?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Eventually, we'll see that it doesn't perfectly describe things, but it helps describe it enough for us to get a sense of what's going to happen. Sometimes it will completely describe things, but anyway, we can ignore that for now. So let me get rid of these things right there. So the first question is, well, what if something doesn't jar just like that? What if I want to construct more fancy unit functions or more fancy step functions? Let's say I wanted to construct something that looked like this. Let me say this is my y-axis, this is my x-axis."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So the first question is, well, what if something doesn't jar just like that? What if I want to construct more fancy unit functions or more fancy step functions? Let's say I wanted to construct something that looked like this. Let me say this is my y-axis, this is my x-axis. And let's say I wanted to construct something that is at let me do it a different color, let's say it's at 2 until I get to pi. And then from pi until forever, it just stays at 0. So how could I construct this function right here using my unit step function?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Let me say this is my y-axis, this is my x-axis. And let's say I wanted to construct something that is at let me do it a different color, let's say it's at 2 until I get to pi. And then from pi until forever, it just stays at 0. So how could I construct this function right here using my unit step function? So what if I had written it as, so my unit step function is 0 initially. So what if I make it 2 minus a unit step function that starts at pi? So if I define my function here, will this work?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So how could I construct this function right here using my unit step function? So what if I had written it as, so my unit step function is 0 initially. So what if I make it 2 minus a unit step function that starts at pi? So if I define my function here, will this work? Well, this unit step function, when we pass pi, is only going to be equal to 1. But we want this thing to equal to 0, so it has to be 2 minus 2. So I'll have to put a 2 here."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So if I define my function here, will this work? Well, this unit step function, when we pass pi, is only going to be equal to 1. But we want this thing to equal to 0, so it has to be 2 minus 2. So I'll have to put a 2 here. And this should work. When we go from any value below pi, when t is less than pi here, this becomes 0. So our function will just evaluate to 2, which is right there."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So I'll have to put a 2 here. And this should work. When we go from any value below pi, when t is less than pi here, this becomes 0. So our function will just evaluate to 2, which is right there. But as soon as we hit t is equal to pi, pi is a c in this example, as soon as we hit that, the unit step function becomes 1. We multiply that by 2, and we have 2 minus 2, and then we end up here with 0. Now, that might be nice and everything, but let's say you wanted for it to go back up."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So our function will just evaluate to 2, which is right there. But as soon as we hit t is equal to pi, pi is a c in this example, as soon as we hit that, the unit step function becomes 1. We multiply that by 2, and we have 2 minus 2, and then we end up here with 0. Now, that might be nice and everything, but let's say you wanted for it to go back up. Let's say that instead of it going like this, let me kind of erase that by overdrawing the x-axis again. We want the function to jump up again. Let's say it's some value, let's say it's at 2 pi."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Now, that might be nice and everything, but let's say you wanted for it to go back up. Let's say that instead of it going like this, let me kind of erase that by overdrawing the x-axis again. We want the function to jump up again. Let's say it's some value, let's say it's at 2 pi. We want the function to jump up again. How could we construct this? We could make it jump to anything."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Let's say it's some value, let's say it's at 2 pi. We want the function to jump up again. How could we construct this? We could make it jump to anything. Let's say we want it to jump back to 2. Well, we could just add another unit step function here, something that would have been 0 all along, all the way up until this point. But then at 2 pi, it jumps."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "We could make it jump to anything. Let's say we want it to jump back to 2. Well, we could just add another unit step function here, something that would have been 0 all along, all the way up until this point. But then at 2 pi, it jumps. So it would be, in this case, our c would be 2 pi. So unit step function. We want it to jump to 2."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "But then at 2 pi, it jumps. So it would be, in this case, our c would be 2 pi. So unit step function. We want it to jump to 2. This would just jump to 1 by itself, so let's multiply it by 2. And now we have this function. So you could imagine, you could make an arbitrarily complicated function of things jumping up and down to different levels based on different, essentially linear combinations of these unit step functions."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "We want it to jump to 2. This would just jump to 1 by itself, so let's multiply it by 2. And now we have this function. So you could imagine, you could make an arbitrarily complicated function of things jumping up and down to different levels based on different, essentially linear combinations of these unit step functions. Now, what if I wanted to do something a little bit fancier? What if I wanted to do something that, let's say I have some function that looks like this. Let me draw some function."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So you could imagine, you could make an arbitrarily complicated function of things jumping up and down to different levels based on different, essentially linear combinations of these unit step functions. Now, what if I wanted to do something a little bit fancier? What if I wanted to do something that, let's say I have some function that looks like this. Let me draw some function. I should draw straighter than that. I should have some standards. So let's say that just my regular f of t, this is x."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Let me draw some function. I should draw straighter than that. I should have some standards. So let's say that just my regular f of t, this is x. Actually, why am I doing x? This should be the t-axis. We're doing the time domain."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So let's say that just my regular f of t, this is x. Actually, why am I doing x? This should be the t-axis. We're doing the time domain. It could have been x. And then this is, we could call this f of t. So let me draw some arbitrary f of t. Let's say my function looks something crazy like that. Now, what if I'm modeling a physical system?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "We're doing the time domain. It could have been x. And then this is, we could call this f of t. So let me draw some arbitrary f of t. Let's say my function looks something crazy like that. Now, what if I'm modeling a physical system? This is my f of t. What if I'm modeling a physical system that doesn't do this, that actually, at some point, well, actually, let's say it stays at 0 until some value. Let's say it goes to 0 until, I'll call that c again. And then at c, f of t kind of starts up."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Now, what if I'm modeling a physical system? This is my f of t. What if I'm modeling a physical system that doesn't do this, that actually, at some point, well, actually, let's say it stays at 0 until some value. Let's say it goes to 0 until, I'll call that c again. And then at c, f of t kind of starts up. So right at c, f of t should start up. So it just kind of goes like this. So essentially, what we have here is a combination of it's 0 all the way, and then we have a shifted f of t. So at c, we have a shifted f of t. So it shifts that way."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And then at c, f of t kind of starts up. So right at c, f of t should start up. So it just kind of goes like this. So essentially, what we have here is a combination of it's 0 all the way, and then we have a shifted f of t. So at c, we have a shifted f of t. So it shifts that way. So how can we construct this yellow function, where it's essentially a shifted version of this green function, but it's 0 below c? This green function might have continued. It might have gone something like this."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So essentially, what we have here is a combination of it's 0 all the way, and then we have a shifted f of t. So at c, we have a shifted f of t. So it shifts that way. So how can we construct this yellow function, where it's essentially a shifted version of this green function, but it's 0 below c? This green function might have continued. It might have gone something like this. It might have continued and done something crazy. But what we did is we shifted it from here to there, and then we zeroed out everything before c. So how could we do that? Well, just shifting this function you've learned in your Algebra 2 or your Precalculus classes, to shift a function by c to the right, you just replace your t with a t minus c. So this function right here is f of t minus c. And to make sure I get it right, what I always do is I imagine, OK, what's going to happen when t is equal to c?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "It might have gone something like this. It might have continued and done something crazy. But what we did is we shifted it from here to there, and then we zeroed out everything before c. So how could we do that? Well, just shifting this function you've learned in your Algebra 2 or your Precalculus classes, to shift a function by c to the right, you just replace your t with a t minus c. So this function right here is f of t minus c. And to make sure I get it right, what I always do is I imagine, OK, what's going to happen when t is equal to c? When t is equal to c, you're going to have a c minus a c, and you're going to have f of 0. So f of 0 should be the same. So when t is equal to c, this value, the value of the function should be equivalent to the value of the original green function at 0."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Well, just shifting this function you've learned in your Algebra 2 or your Precalculus classes, to shift a function by c to the right, you just replace your t with a t minus c. So this function right here is f of t minus c. And to make sure I get it right, what I always do is I imagine, OK, what's going to happen when t is equal to c? When t is equal to c, you're going to have a c minus a c, and you're going to have f of 0. So f of 0 should be the same. So when t is equal to c, this value, the value of the function should be equivalent to the value of the original green function at 0. So it's equivalent to that value, which makes sense. If we go up one more above c, so let's say this is one more above c, so we get to this point. When you, if t is c plus 1, then when you put c plus 1 minus c, you just have f of 1."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So when t is equal to c, this value, the value of the function should be equivalent to the value of the original green function at 0. So it's equivalent to that value, which makes sense. If we go up one more above c, so let's say this is one more above c, so we get to this point. When you, if t is c plus 1, then when you put c plus 1 minus c, you just have f of 1. And f of 1 is really just this point right here. And so it'll be that f of 1. So it makes sense."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "When you, if t is c plus 1, then when you put c plus 1 minus c, you just have f of 1. And f of 1 is really just this point right here. And so it'll be that f of 1. So it makes sense. So as we move one forward here, we're essentially at the same function value as we were there. So the shift works. But I said that we had to also, if I just shifted this function, you would have all this other stuff."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So it makes sense. So as we move one forward here, we're essentially at the same function value as we were there. So the shift works. But I said that we had to also, if I just shifted this function, you would have all this other stuff. Because you would have had all this other stuff when the function was back here, still going on. The function, I'll draw it lightly, would still continue. But I said I wanted to zero out this function before we reach c. So how can I zero out that function?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "But I said that we had to also, if I just shifted this function, you would have all this other stuff. Because you would have had all this other stuff when the function was back here, still going on. The function, I'll draw it lightly, would still continue. But I said I wanted to zero out this function before we reach c. So how can I zero out that function? Well, I think it's pretty obvious to you. I started this video talking about the unit step function. So what if I multiply the unit step function times this thing, what's going to happen?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "But I said I wanted to zero out this function before we reach c. So how can I zero out that function? Well, I think it's pretty obvious to you. I started this video talking about the unit step function. So what if I multiply the unit step function times this thing, what's going to happen? So what if I, my new function, I call it the unit step function up till c of t, times f of t minus c. So what's going to happen? Until we get to c, the unit step function is 0 when it's less than c. So you're going to have 0 times, I don't care what this is, 0 times anything is 0. So this function is going to be 0."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So what if I multiply the unit step function times this thing, what's going to happen? So what if I, my new function, I call it the unit step function up till c of t, times f of t minus c. So what's going to happen? Until we get to c, the unit step function is 0 when it's less than c. So you're going to have 0 times, I don't care what this is, 0 times anything is 0. So this function is going to be 0. Once you hit c, the unit step function becomes 1. So once you pass c, this thing becomes a 1. And you're just left with 1 times your function."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So this function is going to be 0. Once you hit c, the unit step function becomes 1. So once you pass c, this thing becomes a 1. And you're just left with 1 times your function. So then your function can behave as it would like to behave. And you actually shifted it. This t minus c is what actually shifted this green function over to the right."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And you're just left with 1 times your function. So then your function can behave as it would like to behave. And you actually shifted it. This t minus c is what actually shifted this green function over to the right. And this is actually going to be a very useful way to look at, or a very usefully constructed function. And in a second, we're going to figure out the Laplace transform of this. And you're going to appreciate, I think, why this is a useful function to look at."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "This t minus c is what actually shifted this green function over to the right. And this is actually going to be a very useful way to look at, or a very usefully constructed function. And in a second, we're going to figure out the Laplace transform of this. And you're going to appreciate, I think, why this is a useful function to look at. But now you understand at least what it is and why it essentially shifts a function and zeros out everything before that point. Well, I told you that this is a useful function. So we should add its Laplace transform to our library of Laplace transforms."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And you're going to appreciate, I think, why this is a useful function to look at. But now you understand at least what it is and why it essentially shifts a function and zeros out everything before that point. Well, I told you that this is a useful function. So we should add its Laplace transform to our library of Laplace transforms. So let's do that. So let's take the Laplace transform of this, of the unit step function up to c. I'm doing it in fairly general terms. In the next video, we'll do a bunch of examples where we can apply this."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So we should add its Laplace transform to our library of Laplace transforms. So let's do that. So let's take the Laplace transform of this, of the unit step function up to c. I'm doing it in fairly general terms. In the next video, we'll do a bunch of examples where we can apply this. But we should at least prove to ourselves what the Laplace transform of this thing is. Well, the Laplace transform of anything, or our definition of it so far, is the integral from 0 to infinity of e to the minus st times our function. So our function in this case is the unit step function, u sub c of t times f of t minus c dt."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "In the next video, we'll do a bunch of examples where we can apply this. But we should at least prove to ourselves what the Laplace transform of this thing is. Well, the Laplace transform of anything, or our definition of it so far, is the integral from 0 to infinity of e to the minus st times our function. So our function in this case is the unit step function, u sub c of t times f of t minus c dt. And it seems very general. It seems very hard to evaluate this integral at first. But maybe we can make some form of a substitution to get it into a term that we can appreciate."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So our function in this case is the unit step function, u sub c of t times f of t minus c dt. And it seems very general. It seems very hard to evaluate this integral at first. But maybe we can make some form of a substitution to get it into a term that we can appreciate. So let's make a substitution here. Let me pick a nice variable to work with. I don't know."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "But maybe we can make some form of a substitution to get it into a term that we can appreciate. So let's make a substitution here. Let me pick a nice variable to work with. I don't know. We're not using an x anywhere. We might as well use an x. That's the most fun variable to work with."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "I don't know. We're not using an x anywhere. We might as well use an x. That's the most fun variable to work with. x. Sometimes you'll see in a lot of math classes they introduce these crazy Latin alphabets. And that by itself makes it hard to understand."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "That's the most fun variable to work with. x. Sometimes you'll see in a lot of math classes they introduce these crazy Latin alphabets. And that by itself makes it hard to understand. So I like to stay away from those crazy Latin alphabets. So we'll just use a regular x. Let's make a substitution."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And that by itself makes it hard to understand. So I like to stay away from those crazy Latin alphabets. So we'll just use a regular x. Let's make a substitution. Let's say that x is equal to t minus c. Or if we added t to both sides, we could say that t is equal to x plus c. Let's see what happens to our substitution. And also if we took the derivative of both sides of this, or I guess the differential, you would get dx is equal to dt. Or I mean if you took dx with respect to dt, you would get that to be equal to 1. c is just a constant."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Let's make a substitution. Let's say that x is equal to t minus c. Or if we added t to both sides, we could say that t is equal to x plus c. Let's see what happens to our substitution. And also if we took the derivative of both sides of this, or I guess the differential, you would get dx is equal to dt. Or I mean if you took dx with respect to dt, you would get that to be equal to 1. c is just a constant. And if you multiply both sides by dt, you get dx is equal to dt, and that's a nice substitution. So what is our integral going to become with this substitution? So our integral, this was t equals 0 to t is equal to infinity."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Or I mean if you took dx with respect to dt, you would get that to be equal to 1. c is just a constant. And if you multiply both sides by dt, you get dx is equal to dt, and that's a nice substitution. So what is our integral going to become with this substitution? So our integral, this was t equals 0 to t is equal to infinity. When t is equal to 0, what is x going to be equal to? Well, x is going to be equal to minus c. Actually, before I go there, let me actually take a step back, because we could go in this direction. We could actually simplify it more before we do that."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So our integral, this was t equals 0 to t is equal to infinity. When t is equal to 0, what is x going to be equal to? Well, x is going to be equal to minus c. Actually, before I go there, let me actually take a step back, because we could go in this direction. We could actually simplify it more before we do that. Let's go back to our original integral before we even made our substitution. If we're taking the integral from 0 to infinity of this thing, we already said, what does this integral look like? Or what does this function look like?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "We could actually simplify it more before we do that. Let's go back to our original integral before we even made our substitution. If we're taking the integral from 0 to infinity of this thing, we already said, what does this integral look like? Or what does this function look like? It's 0. We have this unit step function sitting right here. We have the unit step function sitting right there."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Or what does this function look like? It's 0. We have this unit step function sitting right here. We have the unit step function sitting right there. So this whole expression is going to be 0 until we get to c. This whole thing, by definition, this unit step function is 0 until we get to c. So everything's going to be zeroed out until we get to c. So we could essentially say, we don't have to take the integral from t equals 0 to t equals infinity. We could take the integral. I'll just use that old integral sign."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "We have the unit step function sitting right there. So this whole expression is going to be 0 until we get to c. This whole thing, by definition, this unit step function is 0 until we get to c. So everything's going to be zeroed out until we get to c. So we could essentially say, we don't have to take the integral from t equals 0 to t equals infinity. We could take the integral. I'll just use that old integral sign. We could just take the integral from t is equal to c to t is equal to infinity of e to the minus st, the unit step function, uc of t times f of t minus c dt. In fact, at this point, this unit step function has no use anymore because before t is equal to c, it's 0. And now that we're only worried about values above c, it's equal to 1."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "I'll just use that old integral sign. We could just take the integral from t is equal to c to t is equal to infinity of e to the minus st, the unit step function, uc of t times f of t minus c dt. In fact, at this point, this unit step function has no use anymore because before t is equal to c, it's 0. And now that we're only worried about values above c, it's equal to 1. So it equals 1 in this context. I want to make that very clear to you. What did I do just here?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And now that we're only worried about values above c, it's equal to 1. So it equals 1 in this context. I want to make that very clear to you. What did I do just here? I changed our bottom boundary from 0 to c. And I think you might realize why I did it when I was working with the substitution. Because this will simplify things if we do this ahead of time. So if we have this unit step function, this thing is going to zero out this entire integral before we get to c. Remember, this definite integral is really just the area under this curve, right?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "What did I do just here? I changed our bottom boundary from 0 to c. And I think you might realize why I did it when I was working with the substitution. Because this will simplify things if we do this ahead of time. So if we have this unit step function, this thing is going to zero out this entire integral before we get to c. Remember, this definite integral is really just the area under this curve, right? Of this whole function, of the unit step function times all of this stuff. All of this stuff, when we multiply it, it's going to be 0 until we get to some value c. And then above c, it's going to be e to the minus st times f of t minus c. So it's going to start doing all this crazy stuff. So if we want to essentially find the area under this curve, we can ignore all the stuff that happens before c. So instead of going from t equals 0 to infinity, we can go from t is equal to c to infinity."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So if we have this unit step function, this thing is going to zero out this entire integral before we get to c. Remember, this definite integral is really just the area under this curve, right? Of this whole function, of the unit step function times all of this stuff. All of this stuff, when we multiply it, it's going to be 0 until we get to some value c. And then above c, it's going to be e to the minus st times f of t minus c. So it's going to start doing all this crazy stuff. So if we want to essentially find the area under this curve, we can ignore all the stuff that happens before c. So instead of going from t equals 0 to infinity, we can go from t is equal to c to infinity. Because there was no area before t was equal to c. So that's all I did here. And then the other thing I said is that the unit step function, it's going to be 1 over this entire range of potential t values. So we can just kind of ignore it, right?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So if we want to essentially find the area under this curve, we can ignore all the stuff that happens before c. So instead of going from t equals 0 to infinity, we can go from t is equal to c to infinity. Because there was no area before t was equal to c. So that's all I did here. And then the other thing I said is that the unit step function, it's going to be 1 over this entire range of potential t values. So we can just kind of ignore it, right? It's just going to be 1 this entire time. So our integral simplifies to the definite integral from t is equal to c to t is equal to infinity of e to the minus st times f of t minus c dt. And this will simplify it a good bit."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So we can just kind of ignore it, right? It's just going to be 1 this entire time. So our integral simplifies to the definite integral from t is equal to c to t is equal to infinity of e to the minus st times f of t minus c dt. And this will simplify it a good bit. I was going down the other road when I did the substitution first, which would have worked. But I think the argument as to why I could have changed the boundaries would have been a harder argument to make. So now that we had this, let's go back and make that substitution that x is equal to t minus c. So our integral becomes, I'll do it in green, when t is equal to c, what is x?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And this will simplify it a good bit. I was going down the other road when I did the substitution first, which would have worked. But I think the argument as to why I could have changed the boundaries would have been a harder argument to make. So now that we had this, let's go back and make that substitution that x is equal to t minus c. So our integral becomes, I'll do it in green, when t is equal to c, what is x? Then x is 0, right? c minus c is 0. When t is equal to infinity, what is x?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So now that we had this, let's go back and make that substitution that x is equal to t minus c. So our integral becomes, I'll do it in green, when t is equal to c, what is x? Then x is 0, right? c minus c is 0. When t is equal to infinity, what is x? Well, x is infinity minus any constant. It's still going to be infinity. Or the limit as t approaches infinity, x is still going to be infinity here."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "When t is equal to infinity, what is x? Well, x is infinity minus any constant. It's still going to be infinity. Or the limit as t approaches infinity, x is still going to be infinity here. And it's the integral of e to the minus s. But now instead of a t, we have the substitution. If we said x is equal to t minus c, then we can just add c to both sides. You get t is equal to x plus c. So you get x plus c there."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Or the limit as t approaches infinity, x is still going to be infinity here. And it's the integral of e to the minus s. But now instead of a t, we have the substitution. If we said x is equal to t minus c, then we can just add c to both sides. You get t is equal to x plus c. So you get x plus c there. And then times a function, f of t minus c. But we said t minus c is the same thing as x. And dt is the same thing as dx. Showed you that right there."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "You get t is equal to x plus c. So you get x plus c there. And then times a function, f of t minus c. But we said t minus c is the same thing as x. And dt is the same thing as dx. Showed you that right there. So we can write this as dx. Now this is starting to look a little bit interesting. So what is this equal to?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Showed you that right there. So we can write this as dx. Now this is starting to look a little bit interesting. So what is this equal to? This is equal to the integral from 0 to infinity. Let me expand this out of e to the minus sx minus sc times f of x dx. Now what is this equal to?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So what is this equal to? This is equal to the integral from 0 to infinity. Let me expand this out of e to the minus sx minus sc times f of x dx. Now what is this equal to? Well, we could factor out an e to the minus sc and bring it outside of the integral. Because this has nothing to do with what we're taking the integral with respect to. So let's do that."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Now what is this equal to? Well, we could factor out an e to the minus sc and bring it outside of the integral. Because this has nothing to do with what we're taking the integral with respect to. So let's do that. Let me take this guy out. And just to not confuse you, let me rewrite the whole thing, 0 to infinity. I could rewrite this e term as e to the minus sx times e to the minus sc, right?"}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So let's do that. Let me take this guy out. And just to not confuse you, let me rewrite the whole thing, 0 to infinity. I could rewrite this e term as e to the minus sx times e to the minus sc, right? Common base. So if I were to multiply these two, I could just add the exponents, which you would get that up there, times f of x dx. This is a constant term with respect to x."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "I could rewrite this e term as e to the minus sx times e to the minus sc, right? Common base. So if I were to multiply these two, I could just add the exponents, which you would get that up there, times f of x dx. This is a constant term with respect to x. So we can just factor it out. We can just factor this thing out right there. So then you get e to the minus sc times the integral from 0 to infinity of e to the minus sx times f of x dx."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "This is a constant term with respect to x. So we can just factor it out. We can just factor this thing out right there. So then you get e to the minus sc times the integral from 0 to infinity of e to the minus sx times f of x dx. Now what were we doing here the whole time? We were taking the Laplace transform of the unit step function that goes up to c, and then it's 0 up to c, and it's 1 after that, of t times some shifted function, f of t minus c. And now we got that as being equal to this thing. And we made a substitution."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "So then you get e to the minus sc times the integral from 0 to infinity of e to the minus sx times f of x dx. Now what were we doing here the whole time? We were taking the Laplace transform of the unit step function that goes up to c, and then it's 0 up to c, and it's 1 after that, of t times some shifted function, f of t minus c. And now we got that as being equal to this thing. And we made a substitution. We simplified it a little bit. e to the minus sc times the integral from 0 to infinity of e to the minus sx f of x dx. Now something about the tablet doesn't work properly right around this period."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And we made a substitution. We simplified it a little bit. e to the minus sc times the integral from 0 to infinity of e to the minus sx f of x dx. Now something about the tablet doesn't work properly right around this period. But this should look interesting to you. What is this? This is the Laplace transform of f of x."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Now something about the tablet doesn't work properly right around this period. But this should look interesting to you. What is this? This is the Laplace transform of f of x. Let me write that down. What's the Laplace transform of, well, I could write it as f of t or f of x. The Laplace transform of f of t is equal to the integral from 0 to infinity of e to the minus st times f of t dt."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "This is the Laplace transform of f of x. Let me write that down. What's the Laplace transform of, well, I could write it as f of t or f of x. The Laplace transform of f of t is equal to the integral from 0 to infinity of e to the minus st times f of t dt. I mean, this and this are the exact same thing. We're just using a t here. We're using an x here."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "The Laplace transform of f of t is equal to the integral from 0 to infinity of e to the minus st times f of t dt. I mean, this and this are the exact same thing. We're just using a t here. We're using an x here. No difference. They're just letters. But they're, oh, let me, this is f of t. e to the minus st times f of t dt."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "We're using an x here. No difference. They're just letters. But they're, oh, let me, this is f of t. e to the minus st times f of t dt. I could have also rewritten it as the Laplace transform of f of t. I could write this as the integral from 0 to infinity of e to the minus sy times f of y dy. I could do it by anything because this is a definite integral. The y's are going to disappear."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "But they're, oh, let me, this is f of t. e to the minus st times f of t dt. I could have also rewritten it as the Laplace transform of f of t. I could write this as the integral from 0 to infinity of e to the minus sy times f of y dy. I could do it by anything because this is a definite integral. The y's are going to disappear. And we've seen that. All you're left with is a function of s. This ends up being some capital, well, we could write some capital function of s. So this is interesting. This is a Laplace transform of f of t times some scaling factor, and that's what we set out to show."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "The y's are going to disappear. And we've seen that. All you're left with is a function of s. This ends up being some capital, well, we could write some capital function of s. So this is interesting. This is a Laplace transform of f of t times some scaling factor, and that's what we set out to show. So we can now show that the Laplace transform of the unit step function times some function t minus c is equal to this function right here, e to the minus sc, where this c is the same as this c right here, times the Laplace transform of f of t times the Laplace transform of f of t. This is equal to, because it's looking funny there, e to the minus sc times the Laplace transform of f of t. So this is our result. Now what does this mean? Let's see if we backfilled it somehow."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "This is a Laplace transform of f of t times some scaling factor, and that's what we set out to show. So we can now show that the Laplace transform of the unit step function times some function t minus c is equal to this function right here, e to the minus sc, where this c is the same as this c right here, times the Laplace transform of f of t times the Laplace transform of f of t. This is equal to, because it's looking funny there, e to the minus sc times the Laplace transform of f of t. So this is our result. Now what does this mean? Let's see if we backfilled it somehow. What does this mean? What can we do with this? Well, let's say we had the Laplace transform."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Let's see if we backfilled it somehow. What does this mean? What can we do with this? Well, let's say we had the Laplace transform. Let's say we wanted to figure out the Laplace transform of the unit step function that starts off at pi of t. And let's say we're taking, I don't know, something that we know well, sine of t minus pi. So we shifted it, right? This thing is really malfunctioning at this point right here."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "Well, let's say we had the Laplace transform. Let's say we wanted to figure out the Laplace transform of the unit step function that starts off at pi of t. And let's say we're taking, I don't know, something that we know well, sine of t minus pi. So we shifted it, right? This thing is really malfunctioning at this point right here. Let me pause it. I just paused, sorry if that was a little disconcerting. I just paused the video because it was having trouble recording at some point on my little board."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "This thing is really malfunctioning at this point right here. Let me pause it. I just paused, sorry if that was a little disconcerting. I just paused the video because it was having trouble recording at some point on my little board. So let me rewrite the result that we proved just now. We showed that the Laplace transform of the unit step function, t, and it goes to 1 at some value c, times some function that's shifted by c to the right. It's equal to e to the minus cs times the Laplace transform of just the unshifted function."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "I just paused the video because it was having trouble recording at some point on my little board. So let me rewrite the result that we proved just now. We showed that the Laplace transform of the unit step function, t, and it goes to 1 at some value c, times some function that's shifted by c to the right. It's equal to e to the minus cs times the Laplace transform of just the unshifted function. That was our result. That was a big takeaway from this video. And if this seems like some Byzantine, hard to understand result, we can apply it."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "It's equal to e to the minus cs times the Laplace transform of just the unshifted function. That was our result. That was a big takeaway from this video. And if this seems like some Byzantine, hard to understand result, we can apply it. So let's say the Laplace transform, and this is what I was doing right before the actual pen tablet started malfunctioning. If we want to take the Laplace transform of the unit step function that goes to 1 at pi, t, times the sine function shifted by pi to the right, we know that this is going to be equal to e to the minus cs. c is pi in this case, so minus pi s times the Laplace transform of the unshifted function."}, {"video_title": "Laplace transform of the unit step function Laplace transform Khan Academy.mp3", "Sentence": "And if this seems like some Byzantine, hard to understand result, we can apply it. So let's say the Laplace transform, and this is what I was doing right before the actual pen tablet started malfunctioning. If we want to take the Laplace transform of the unit step function that goes to 1 at pi, t, times the sine function shifted by pi to the right, we know that this is going to be equal to e to the minus cs. c is pi in this case, so minus pi s times the Laplace transform of the unshifted function. So in this case, it's the Laplace transform of sine of t. And we know what the Laplace transform of sine of t is. It's just 1 over s squared plus 1. So the Laplace transform of this thing here, which before this video seemed like something crazy, we now know is this times this."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "Let's do some more examples with exact differential equations. And I'm getting these problems from page 80 of my old college differential equations book. This is the fifth edition of Elementary Differential Equations by William Boyce and Richard DePrima. I want to make sure they get credit that I'm not making up these problems. I'm getting it from their book. Anyway, so I'm just going to give a bunch of equations. We have to figure out if they're exact."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "I want to make sure they get credit that I'm not making up these problems. I'm getting it from their book. Anyway, so I'm just going to give a bunch of equations. We have to figure out if they're exact. And if they are exact, we'll use what we know about exact differential equations to figure out their solutions. So the first one they have is 2x plus 3 plus 2y minus 2 times y prime is equal to 0. So this is our m of x and y, although this is only a function of x."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "We have to figure out if they're exact. And if they are exact, we'll use what we know about exact differential equations to figure out their solutions. So the first one they have is 2x plus 3 plus 2y minus 2 times y prime is equal to 0. So this is our m of x and y, although this is only a function of x. And this is our n. You could say that's m or that's n. You could also say that if this is exact, well, first let's just say it's exact before we start talking about xi. So what's the partial of this with respect to y? The partial of m with respect to y."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So this is our m of x and y, although this is only a function of x. And this is our n. You could say that's m or that's n. You could also say that if this is exact, well, first let's just say it's exact before we start talking about xi. So what's the partial of this with respect to y? The partial of m with respect to y. Well, there's no y here, so it's 0. The rate of change that this changes with respect to y is 0. And what's the rate of change that this changes with respect to x?"}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "The partial of m with respect to y. Well, there's no y here, so it's 0. The rate of change that this changes with respect to y is 0. And what's the rate of change that this changes with respect to x? The partial of n with respect to x is equal to, well, there's no x here, right? So these are just constants from an x point of view, so this is all going to be 0. But we do see that they're both 0."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "And what's the rate of change that this changes with respect to x? The partial of n with respect to x is equal to, well, there's no x here, right? So these are just constants from an x point of view, so this is all going to be 0. But we do see that they're both 0. So m sub y, or the partial with respect to y, is equal to the partial with respect to x. So this is exact. And actually, we don't have to use exact equations here."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "But we do see that they're both 0. So m sub y, or the partial with respect to y, is equal to the partial with respect to x. So this is exact. And actually, we don't have to use exact equations here. We'll do it just so that we get used to it. But if you looked here, you actually could have figured out that this is actually a separable equation. But anyway, this is exact."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "And actually, we don't have to use exact equations here. We'll do it just so that we get used to it. But if you looked here, you actually could have figured out that this is actually a separable equation. But anyway, this is exact. So knowing that it's exact, it tells us there's some function xi, where xi is a function of x and y, where xi sub x is equal to this function, is equal to 2x plus 3. And xi, I shouldn't say sub x, I should say the partial of xi with respect to x. And the partial of xi with respect to y is equal to this, 2y minus 2."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "But anyway, this is exact. So knowing that it's exact, it tells us there's some function xi, where xi is a function of x and y, where xi sub x is equal to this function, is equal to 2x plus 3. And xi, I shouldn't say sub x, I should say the partial of xi with respect to x. And the partial of xi with respect to y is equal to this, 2y minus 2. And if we can find our xi, we know that this is just the derivative of xi, right? Because we know that the derivative with respect to x of xi is equal to the partial of xi with respect to x plus the partial of xi with respect to y times y prime. So this is just the same form as that."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "And the partial of xi with respect to y is equal to this, 2y minus 2. And if we can find our xi, we know that this is just the derivative of xi, right? Because we know that the derivative with respect to x of xi is equal to the partial of xi with respect to x plus the partial of xi with respect to y times y prime. So this is just the same form as that. So if we can figure out y, then we could rewrite this equation as dx, the derivative of xi with respect to x, is equal to 0, right? This, let me switch colors just because it gets monotonous. This right here, if we can find a xi where the partial with respect to x is this, the partial with respect to y is this, then this can be rewritten as this, right?"}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So this is just the same form as that. So if we can figure out y, then we could rewrite this equation as dx, the derivative of xi with respect to x, is equal to 0, right? This, let me switch colors just because it gets monotonous. This right here, if we can find a xi where the partial with respect to x is this, the partial with respect to y is this, then this can be rewritten as this, right? And how do we know that? Because the derivative of xi with respect to x, using the partial derivative chain rules, is this. And this is the same."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "This right here, if we can find a xi where the partial with respect to x is this, the partial with respect to y is this, then this can be rewritten as this, right? And how do we know that? Because the derivative of xi with respect to x, using the partial derivative chain rules, is this. And this is the same. This partial with respect to x, that's this. Partial with respect to y is this times y prime. So this is a whole point of exact equations."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "And this is the same. This partial with respect to x, that's this. Partial with respect to y is this times y prime. So this is a whole point of exact equations. But anyway, so let's figure out what our xi is. Actually, before we figure out, if the derivative of xi with respect to x is 0, then if you integrate both sides, the solution of this equation is xi is equal to c. So using this information, if we can solve for xi, then we know that the solution of this differential equation is xi is equal to c. And if we have some initial conditions, we could solve for c. So let's solve for xi. So let's integrate both sides of this equation with respect to x."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So this is a whole point of exact equations. But anyway, so let's figure out what our xi is. Actually, before we figure out, if the derivative of xi with respect to x is 0, then if you integrate both sides, the solution of this equation is xi is equal to c. So using this information, if we can solve for xi, then we know that the solution of this differential equation is xi is equal to c. And if we have some initial conditions, we could solve for c. So let's solve for xi. So let's integrate both sides of this equation with respect to x. And then we get xi is equal to x squared plus 3x plus some function of y. Let's call it h of y. And remember, normally when you take an antiderivative, you have just a plus c here, right?"}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So let's integrate both sides of this equation with respect to x. And then we get xi is equal to x squared plus 3x plus some function of y. Let's call it h of y. And remember, normally when you take an antiderivative, you have just a plus c here, right? But you can kind of say we took an antipartial derivative. So when you took a partial derivative with respect to x, not only do you lose constants, that's why we have a plus c normally, but you also lose anything that's a function of just y and not x. So for example, take the partial derivative of this with respect to x."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "And remember, normally when you take an antiderivative, you have just a plus c here, right? But you can kind of say we took an antipartial derivative. So when you took a partial derivative with respect to x, not only do you lose constants, that's why we have a plus c normally, but you also lose anything that's a function of just y and not x. So for example, take the partial derivative of this with respect to x. You're going to get this, right? Because the partial derivative of a function purely of y with respect to x is going to be 0, so it will disappear. So anyway, we take the antiderivative of this, we get this."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So for example, take the partial derivative of this with respect to x. You're going to get this, right? Because the partial derivative of a function purely of y with respect to x is going to be 0, so it will disappear. So anyway, we take the antiderivative of this, we get this. Now we use this information. We take the partial of this expression, and we say, well, the partial of this expression with respect to y has to equal this, and then we can solve for h of y, and we'll be done. So let's do that."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So anyway, we take the antiderivative of this, we get this. Now we use this information. We take the partial of this expression, and we say, well, the partial of this expression with respect to y has to equal this, and then we can solve for h of y, and we'll be done. So let's do that. So the partial of xi with respect to y is equal to, well, that's going to be 0, 0, 0. It's going to be, right? This part is a function of x."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So let's do that. So the partial of xi with respect to y is equal to, well, that's going to be 0, 0, 0. It's going to be, right? This part is a function of x. You take the partial with respect to y, it's 0, because these are constants from a y point of view. So you're left with h prime of y. So we know that h prime of y, which is the partial of xi with respect to y, is equal to this."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "This part is a function of x. You take the partial with respect to y, it's 0, because these are constants from a y point of view. So you're left with h prime of y. So we know that h prime of y, which is the partial of xi with respect to y, is equal to this. So h prime of y is equal to 2y minus 2. And then if we wanted to figure out what h of y is, we get h of y, just integrate both sides with respect to y, is equal to y squared minus 2y. Now you could have a plus c there, but if you watch the previous example, you'll see that that c kind of merges with the other c, so you don't have to worry about it right now."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So we know that h prime of y, which is the partial of xi with respect to y, is equal to this. So h prime of y is equal to 2y minus 2. And then if we wanted to figure out what h of y is, we get h of y, just integrate both sides with respect to y, is equal to y squared minus 2y. Now you could have a plus c there, but if you watch the previous example, you'll see that that c kind of merges with the other c, so you don't have to worry about it right now. So what is our xi function, as we know it now, not worrying about the plus c? It is xi of x and y is equal to x squared plus 3x plus h of y, which we figured out is this, plus y squared minus 2y. And we know a solution of our original differential equation is xi is equal to c. So the solution of our differential equation is this is equal to c. It's x squared plus 3x plus y squared minus 2y is equal to c. And if you had some initial conditions, you could test it."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "Now you could have a plus c there, but if you watch the previous example, you'll see that that c kind of merges with the other c, so you don't have to worry about it right now. So what is our xi function, as we know it now, not worrying about the plus c? It is xi of x and y is equal to x squared plus 3x plus h of y, which we figured out is this, plus y squared minus 2y. And we know a solution of our original differential equation is xi is equal to c. So the solution of our differential equation is this is equal to c. It's x squared plus 3x plus y squared minus 2y is equal to c. And if you had some initial conditions, you could test it. And I encourage you to test this out on this original equation, or I encourage you to take the derivative of xi and prove to yourself that if you took the derivative of xi with respect to x here, implicitly, that you would get this differential equation. Anyway, let's do another one. The more examples you see, the better."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "And we know a solution of our original differential equation is xi is equal to c. So the solution of our differential equation is this is equal to c. It's x squared plus 3x plus y squared minus 2y is equal to c. And if you had some initial conditions, you could test it. And I encourage you to test this out on this original equation, or I encourage you to take the derivative of xi and prove to yourself that if you took the derivative of xi with respect to x here, implicitly, that you would get this differential equation. Anyway, let's do another one. The more examples you see, the better. So let's see, this one says 2x plus 4y plus 2x minus 2y, y prime is equal to 0. So what's the partial of this with respect to y? So m, the partial of m with respect to y, this is 0, so it's equal to 4."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "The more examples you see, the better. So let's see, this one says 2x plus 4y plus 2x minus 2y, y prime is equal to 0. So what's the partial of this with respect to y? So m, the partial of m with respect to y, this is 0, so it's equal to 4. What's the partial of this with respect to x, just this part right here? The partial of n with respect to x is 2, this is 0. So the partial of this with respect to y is different than the partial of n with respect to x."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So m, the partial of m with respect to y, this is 0, so it's equal to 4. What's the partial of this with respect to x, just this part right here? The partial of n with respect to x is 2, this is 0. So the partial of this with respect to y is different than the partial of n with respect to x. So this is not exact. So we can't solve this using our exact methodology. So that was a fairly straightforward problem."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So the partial of this with respect to y is different than the partial of n with respect to x. So this is not exact. So we can't solve this using our exact methodology. So that was a fairly straightforward problem. Let's do another one. Let's see, I'm running out of time, so I want to do one that's not too complicated. Let's see, we have 3x squared minus 2xy."}, {"video_title": "Exact equations example 2 First order differential equations Khan Academy.mp3", "Sentence": "So that was a fairly straightforward problem. Let's do another one. Let's see, I'm running out of time, so I want to do one that's not too complicated. Let's see, we have 3x squared minus 2xy. Actually, let me do this in the next problem. I don't want to rush these things. I will continue this in the next video."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "And the first is to show that it is a linear operator. What does that mean? Well, let's say I wanted to take the Laplace transform of the sum of the, we call it the weighted sum of two functions. So say some constant, c1 times my first function, f of t, plus some constant, c2, times my second function, g of t. Well, by the definition of the Laplace transform, this would be equal to the improper integral from 0 to infinity of e to the minus st times whatever our function that we're taking the Laplace transform of, so times c1 f of t plus c2 g of t. I think you know where this is going. All of that dt. And then that is equal to, integral from 0 to infinity, let's just distribute the e to the minus st. That is equal to what? That is equal to c1 e to the minus st f of t plus c2 e to the minus st g of t. And all of that times dt."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "So say some constant, c1 times my first function, f of t, plus some constant, c2, times my second function, g of t. Well, by the definition of the Laplace transform, this would be equal to the improper integral from 0 to infinity of e to the minus st times whatever our function that we're taking the Laplace transform of, so times c1 f of t plus c2 g of t. I think you know where this is going. All of that dt. And then that is equal to, integral from 0 to infinity, let's just distribute the e to the minus st. That is equal to what? That is equal to c1 e to the minus st f of t plus c2 e to the minus st g of t. And all of that times dt. And just by the definition of how, or the properties of integrals work, we know that we can split this up into two integrals, right? If the integral of the sum of two functions is equal to the sum of their integrals. And these are just constants."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "That is equal to c1 e to the minus st f of t plus c2 e to the minus st g of t. And all of that times dt. And just by the definition of how, or the properties of integrals work, we know that we can split this up into two integrals, right? If the integral of the sum of two functions is equal to the sum of their integrals. And these are just constants. So this is going to be equal to c1 times the integral from 0 to infinity of e to the minus st times f of t, d of t, plus c2 times the integral from 0 to infinity of e to the minus st g of t, dt. And this was just a very long-winded way of saying, well, what is this? This is the Laplace transform of f of t. This is the Laplace transform of g of t. So this is equal to c1 times the Laplace transform of f of t plus c2 times, this is the Laplace transform, the Laplace transform of g of t. And so we have just shown that the Laplace transform is a linear operator, right?"}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "And these are just constants. So this is going to be equal to c1 times the integral from 0 to infinity of e to the minus st times f of t, d of t, plus c2 times the integral from 0 to infinity of e to the minus st g of t, dt. And this was just a very long-winded way of saying, well, what is this? This is the Laplace transform of f of t. This is the Laplace transform of g of t. So this is equal to c1 times the Laplace transform of f of t plus c2 times, this is the Laplace transform, the Laplace transform of g of t. And so we have just shown that the Laplace transform is a linear operator, right? The Laplace transform of this is equal to this. So essentially, you can kind of break up the sum and go and take out the constants and just take the Laplace transform. That's something useful to know, and you might have guessed that that was the case anyway, but now you know for sure."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "This is the Laplace transform of f of t. This is the Laplace transform of g of t. So this is equal to c1 times the Laplace transform of f of t plus c2 times, this is the Laplace transform, the Laplace transform of g of t. And so we have just shown that the Laplace transform is a linear operator, right? The Laplace transform of this is equal to this. So essentially, you can kind of break up the sum and go and take out the constants and just take the Laplace transform. That's something useful to know, and you might have guessed that that was the case anyway, but now you know for sure. Now we'll do something which I consider even more interesting. And this is actually going to be a big clue as to why Laplace transforms are extremely useful for solving differential equations. So let's say I wanted to find the Laplace transform of f prime of t. So I have some f of t. I take its derivative, and then I want the Laplace transform of that."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "That's something useful to know, and you might have guessed that that was the case anyway, but now you know for sure. Now we'll do something which I consider even more interesting. And this is actually going to be a big clue as to why Laplace transforms are extremely useful for solving differential equations. So let's say I wanted to find the Laplace transform of f prime of t. So I have some f of t. I take its derivative, and then I want the Laplace transform of that. Let's see if we can find a relationship between the Laplace transform of the derivative of a function and the Laplace transform of the function. So we're going to use some integration by parts here. So let's just put the integration."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "So let's say I wanted to find the Laplace transform of f prime of t. So I have some f of t. I take its derivative, and then I want the Laplace transform of that. Let's see if we can find a relationship between the Laplace transform of the derivative of a function and the Laplace transform of the function. So we're going to use some integration by parts here. So let's just put the integration. Well, actually, let me just say what this is. First of all, this is equal to the integral from 0 to infinity of e to the minus st times f prime of t dt. And to solve this, we're going to use integration by parts."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "So let's just put the integration. Well, actually, let me just say what this is. First of all, this is equal to the integral from 0 to infinity of e to the minus st times f prime of t dt. And to solve this, we're going to use integration by parts. Let me write it in the corner just so you remember what it is. So I think I memorized it because I recorded that last video not too long ago. That the integral of u v prime, because that will match what we have up here better, is equal to both functions without the derivatives, u v, minus the integral of the opposite."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "And to solve this, we're going to use integration by parts. Let me write it in the corner just so you remember what it is. So I think I memorized it because I recorded that last video not too long ago. That the integral of u v prime, because that will match what we have up here better, is equal to both functions without the derivatives, u v, minus the integral of the opposite. So the opposite is u prime v. So here, the substitution is pretty clear, right? Because we want to end up with f of x, right? So let's make v prime is f prime."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "That the integral of u v prime, because that will match what we have up here better, is equal to both functions without the derivatives, u v, minus the integral of the opposite. So the opposite is u prime v. So here, the substitution is pretty clear, right? Because we want to end up with f of x, right? So let's make v prime is f prime. And let's make u e to the minus st. So let's do that. u is going to be e to the minus st. And v is going to equal what?"}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "So let's make v prime is f prime. And let's make u e to the minus st. So let's do that. u is going to be e to the minus st. And v is going to equal what? v is going to equal f prime of t. And then u prime would be minus s e to the minus st. And then v prime is f prime of t. So v is just going to be equal to f of t. Hope I didn't say that wrong the first time. But you see what I'm saying. This is u."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "u is going to be e to the minus st. And v is going to equal what? v is going to equal f prime of t. And then u prime would be minus s e to the minus st. And then v prime is f prime of t. So v is just going to be equal to f of t. Hope I didn't say that wrong the first time. But you see what I'm saying. This is u. That's u. And this is v prime. And if this is v prime, then if you take the antiderivative of both sides, then v is equal to f of t. So let's apply integration by parts."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "This is u. That's u. And this is v prime. And if this is v prime, then if you take the antiderivative of both sides, then v is equal to f of t. So let's apply integration by parts. So this Laplace transform, which is this, is equal to u v, which is equal to e to the minus st. This is equal to e to the minus st times v. f of t minus the integral. And of course, we're going to have to evaluate this from 0 to infinity."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "And if this is v prime, then if you take the antiderivative of both sides, then v is equal to f of t. So let's apply integration by parts. So this Laplace transform, which is this, is equal to u v, which is equal to e to the minus st. This is equal to e to the minus st times v. f of t minus the integral. And of course, we're going to have to evaluate this from 0 to infinity. I'll keep the improper integral with us the whole time, I won't switch back and forth between the definite and indefinite integral. So minus this part. So the integral from 0 to infinity of u prime."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "And of course, we're going to have to evaluate this from 0 to infinity. I'll keep the improper integral with us the whole time, I won't switch back and forth between the definite and indefinite integral. So minus this part. So the integral from 0 to infinity of u prime. u prime is minus se to the minus st times v. v is f of t dt. Let's see, we have a minus and a minus. Let's make both of these pluses."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "So the integral from 0 to infinity of u prime. u prime is minus se to the minus st times v. v is f of t dt. Let's see, we have a minus and a minus. Let's make both of these pluses. This s is just a constant, so we can bring it out. So that is equal to e to the minus st f of t, evaluated from 0 to infinity, or as we approach infinity, plus s times the integral from 0 to infinity of e to the minus st f of t dt. And here we see, what is this?"}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "Let's make both of these pluses. This s is just a constant, so we can bring it out. So that is equal to e to the minus st f of t, evaluated from 0 to infinity, or as we approach infinity, plus s times the integral from 0 to infinity of e to the minus st f of t dt. And here we see, what is this? This is the Laplace transform of f of t, right? So this is equal to, let's evaluate this part. So when we evaluate at infinity, as we approach infinity, e to the minus infinity approaches 0. f of infinity, now this is an interesting question."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "And here we see, what is this? This is the Laplace transform of f of t, right? So this is equal to, let's evaluate this part. So when we evaluate at infinity, as we approach infinity, e to the minus infinity approaches 0. f of infinity, now this is an interesting question. f of infinity, I don't know, that could be large, that could be small, that could approach some, that approaches some value, right? This approaches 0, so we're not sure. If this increases faster than this approaches 0, then this will diverge."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "So when we evaluate at infinity, as we approach infinity, e to the minus infinity approaches 0. f of infinity, now this is an interesting question. f of infinity, I don't know, that could be large, that could be small, that could approach some, that approaches some value, right? This approaches 0, so we're not sure. If this increases faster than this approaches 0, then this will diverge. I won't go into the mathematics of whether this converges or diverges, but let's just say, in very rough terms, that this will converge to 0 if f of t grows slower than e to the minus st shrinks. And maybe later on we'll do some more rigorous definitions of under what conditions will f of t actually, will this expression actually converge. But let's assume that f of t grows slower than e to the st, or it converges, or it grows slower than this, or it diverges slower than this converges is another way to view it, or this grows slower than this shrinks."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "If this increases faster than this approaches 0, then this will diverge. I won't go into the mathematics of whether this converges or diverges, but let's just say, in very rough terms, that this will converge to 0 if f of t grows slower than e to the minus st shrinks. And maybe later on we'll do some more rigorous definitions of under what conditions will f of t actually, will this expression actually converge. But let's assume that f of t grows slower than e to the st, or it converges, or it grows slower than this, or it diverges slower than this converges is another way to view it, or this grows slower than this shrinks. So if this grows slower than this shrinks, then this whole expression will approach 0, and then you want to subtract this whole expression evaluated at 0. So e to the 0 is 1 times f of 0. So that's just f of 0 plus s times, we said this is the Laplace transform of f of t. That's our definition."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "But let's assume that f of t grows slower than e to the st, or it converges, or it grows slower than this, or it diverges slower than this converges is another way to view it, or this grows slower than this shrinks. So if this grows slower than this shrinks, then this whole expression will approach 0, and then you want to subtract this whole expression evaluated at 0. So e to the 0 is 1 times f of 0. So that's just f of 0 plus s times, we said this is the Laplace transform of f of t. That's our definition. So the Laplace transform of f of t. And now we have an interesting property. What was the left-hand side of everything we were doing? The Laplace transform of f prime of t. So let me just write it all over again."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "So that's just f of 0 plus s times, we said this is the Laplace transform of f of t. That's our definition. So the Laplace transform of f of t. And now we have an interesting property. What was the left-hand side of everything we were doing? The Laplace transform of f prime of t. So let me just write it all over again. So we now know that, and I'll switch colors, the Laplace transform of f prime of t is equal to s times the Laplace transform of f of t minus f of 0. And now let's just extend this further. What is the Laplace transform?"}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "The Laplace transform of f prime of t. So let me just write it all over again. So we now know that, and I'll switch colors, the Laplace transform of f prime of t is equal to s times the Laplace transform of f of t minus f of 0. And now let's just extend this further. What is the Laplace transform? And this is a really useful thing to know. What is the Laplace transform of f prime of t? Well, we could do a little pattern matching here, right?"}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "What is the Laplace transform? And this is a really useful thing to know. What is the Laplace transform of f prime of t? Well, we could do a little pattern matching here, right? That's going to be s times the Laplace transform of its antiderivative times the Laplace transform of f prime of t, right? This goes to this. That's an antiderivative."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "Well, we could do a little pattern matching here, right? That's going to be s times the Laplace transform of its antiderivative times the Laplace transform of f prime of t, right? This goes to this. That's an antiderivative. This goes to this. That's one antiderivative. Minus f prime of 0, right?"}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "That's an antiderivative. This goes to this. That's one antiderivative. Minus f prime of 0, right? But then what's the Laplace transform of this? This is going to be equal to s times the Laplace transform of f prime of t, but what's that? That's this, right?"}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "Minus f prime of 0, right? But then what's the Laplace transform of this? This is going to be equal to s times the Laplace transform of f prime of t, but what's that? That's this, right? That's s times the Laplace transform of f of t minus f of 0, right? I just substituted this with this. Minus f prime of 0."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "That's this, right? That's s times the Laplace transform of f of t minus f of 0, right? I just substituted this with this. Minus f prime of 0. And we get the Laplace transform of the second derivative is equal to s squared times the Laplace transform of our function, f of t, minus s times f of 0 minus f prime of 0. And I think you're starting to see a pattern here. This is the Laplace transform of f prime prime of t. And I think you're starting to see why the Laplace transform is useful."}, {"video_title": "Laplace as linear operator and Laplace of derivatives Laplace transform Khan Academy.mp3", "Sentence": "Minus f prime of 0. And we get the Laplace transform of the second derivative is equal to s squared times the Laplace transform of our function, f of t, minus s times f of 0 minus f prime of 0. And I think you're starting to see a pattern here. This is the Laplace transform of f prime prime of t. And I think you're starting to see why the Laplace transform is useful. It turns derivatives into multiplication, into multiplications by s. And actually, as you'll see later, it turns integration to divisions by s. And you can take arbitrary derivatives and just keep multiplying by s. And you see this pattern. And I'm running out of time, but I'll leave it up to you to figure out what the Laplace transform of the third derivative of f is. See you in the next video."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "One, trying to figure out whether the equations are exact. And then if you know they're exact, how do you figure out the xi and figure out the solution of the differential equation? So the next one in my book is 3x squared minus 2xy plus 2 times dx plus 6y squared minus x squared plus 3 times dy is equal to 0. So just the way it was written, this isn't superficially in that form that we want, right? What's the form that we want? We want some function of x and y plus another function of x and y times y prime or dy dx is equal to 0. We're close."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So just the way it was written, this isn't superficially in that form that we want, right? What's the form that we want? We want some function of x and y plus another function of x and y times y prime or dy dx is equal to 0. We're close. How could we get this equation into this form? We just divide both sides of this equation by dx, right? And then we get 3x squared minus 2xy plus 2."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "We're close. How could we get this equation into this form? We just divide both sides of this equation by dx, right? And then we get 3x squared minus 2xy plus 2. We're dividing by dx so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx so that becomes dy dx is equal to what's 0 divided by dx?"}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And then we get 3x squared minus 2xy plus 2. We're dividing by dx so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx so that becomes dy dx is equal to what's 0 divided by dx? Well, it's just 0. And there we have it. We have written this in the form that we need, in this form, and now we need to prove to ourselves that this is an exact equation."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And then we're dividing by dx so that becomes dy dx is equal to what's 0 divided by dx? Well, it's just 0. And there we have it. We have written this in the form that we need, in this form, and now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of m? This is the m function, right?"}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "We have written this in the form that we need, in this form, and now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of m? This is the m function, right? This is the m function. What's the, this was a plus here. What's the partial of this with respect to y?"}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "This is the m function, right? This is the m function. What's the, this was a plus here. What's the partial of this with respect to y? This would be 0. This would be minus 2x and then just a 2. So that's the partial of this with respect to y is minus 2x."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "What's the partial of this with respect to y? This would be 0. This would be minus 2x and then just a 2. So that's the partial of this with respect to y is minus 2x. What's the partial of n with respect to x? This would be 0. This would be minus 2x."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So that's the partial of this with respect to y is minus 2x. What's the partial of n with respect to x? This would be 0. This would be minus 2x. So there you have it. The partial of m with respect to y is equal to the partial of n with respect to x. My is equal to nx."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "This would be minus 2x. So there you have it. The partial of m with respect to y is equal to the partial of n with respect to x. My is equal to nx. So we are dealing with an exact equation. So now we have to find xi. The partial of xi with respect to x is equal to m, which is equal to 3x squared minus 2xy plus 2."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "My is equal to nx. So we are dealing with an exact equation. So now we have to find xi. The partial of xi with respect to x is equal to m, which is equal to 3x squared minus 2xy plus 2. Take the antiderivative with respect to x on both sides and you get xi is equal to x to the third minus x squared y, because y is just a constant, plus 2x plus some function of y, right? Because we know xi is a function of x and y. So when you take a derivative, when you take a partial with respect to just x, a pure function of just y would get lost."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "The partial of xi with respect to x is equal to m, which is equal to 3x squared minus 2xy plus 2. Take the antiderivative with respect to x on both sides and you get xi is equal to x to the third minus x squared y, because y is just a constant, plus 2x plus some function of y, right? Because we know xi is a function of x and y. So when you take a derivative, when you take a partial with respect to just x, a pure function of just y would get lost. So it's like the constant when we first learned taking antiderivatives. And now to figure out xi, we just have to solve for h of y. And how do we do that?"}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So when you take a derivative, when you take a partial with respect to just x, a pure function of just y would get lost. So it's like the constant when we first learned taking antiderivatives. And now to figure out xi, we just have to solve for h of y. And how do we do that? Well, let's take the partial of xi with respect to y. And that's going to be equal to this right here. So the partial of xi with respect to y, this is 0, this is minus x squared, so it's minus x squared, this is 0, plus h prime of y is going to be equal to what?"}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And how do we do that? Well, let's take the partial of xi with respect to y. And that's going to be equal to this right here. So the partial of xi with respect to y, this is 0, this is minus x squared, so it's minus x squared, this is 0, plus h prime of y is going to be equal to what? That's going to be equal to our n of xy. It's going to be equal to this. And then we can solve for this."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So the partial of xi with respect to y, this is 0, this is minus x squared, so it's minus x squared, this is 0, plus h prime of y is going to be equal to what? That's going to be equal to our n of xy. It's going to be equal to this. And then we can solve for this. So that's going to be equal to 6y squared minus x squared plus 3. Can add x squared to both sides to get rid of this and this. And then we're left with h prime of y is equal to 6y squared plus 3."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And then we can solve for this. So that's going to be equal to 6y squared minus x squared plus 3. Can add x squared to both sides to get rid of this and this. And then we're left with h prime of y is equal to 6y squared plus 3. Antiderivative. So h of y is equal to 2y cubed plus 3y. And you could put a plus c there, but the plus c merges later on when we solve the differential equation, so you don't have to worry about it too much."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And then we're left with h prime of y is equal to 6y squared plus 3. Antiderivative. So h of y is equal to 2y cubed plus 3y. And you could put a plus c there, but the plus c merges later on when we solve the differential equation, so you don't have to worry about it too much. So what is our function xi? I'll write it in a new color. Our function xi as a function of x and y is equal to x to the third minus x squared y plus 2x plus h of y, which we just solved for."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And you could put a plus c there, but the plus c merges later on when we solve the differential equation, so you don't have to worry about it too much. So what is our function xi? I'll write it in a new color. Our function xi as a function of x and y is equal to x to the third minus x squared y plus 2x plus h of y, which we just solved for. So h of y is plus 2y to the third plus 3y. And then there could be a plus c there, but you'll see that it doesn't matter much. Actually, I want to do something a little bit different."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "Our function xi as a function of x and y is equal to x to the third minus x squared y plus 2x plus h of y, which we just solved for. So h of y is plus 2y to the third plus 3y. And then there could be a plus c there, but you'll see that it doesn't matter much. Actually, I want to do something a little bit different. I'm not just going to chug through the problem. I want to kind of go back to the intuition, because I don't want this to be completely mechanical. Let me just show you what the derivative, using what we knew before you even learned anything about the partial derivative chain rule."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "Actually, I want to do something a little bit different. I'm not just going to chug through the problem. I want to kind of go back to the intuition, because I don't want this to be completely mechanical. Let me just show you what the derivative, using what we knew before you even learned anything about the partial derivative chain rule. What is the derivative of xi with respect to x? Here we just use our implicit differentiation skills. So the derivative of this, I'll do it in a new color, is 3x squared minus, now we're going to have to use the chain rule here, so the derivative of the first expression with respect to x is, well, let me just put the minus sign and I can put like that, so it's 2x times y plus the first function, x squared times the derivative of the second function with respect to x."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "Let me just show you what the derivative, using what we knew before you even learned anything about the partial derivative chain rule. What is the derivative of xi with respect to x? Here we just use our implicit differentiation skills. So the derivative of this, I'll do it in a new color, is 3x squared minus, now we're going to have to use the chain rule here, so the derivative of the first expression with respect to x is, well, let me just put the minus sign and I can put like that, so it's 2x times y plus the first function, x squared times the derivative of the second function with respect to x. Well, that's just y prime. The derivative of y with respect to y is 1 times the derivative of y with respect to x, which is just y prime. Fair enough."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So the derivative of this, I'll do it in a new color, is 3x squared minus, now we're going to have to use the chain rule here, so the derivative of the first expression with respect to x is, well, let me just put the minus sign and I can put like that, so it's 2x times y plus the first function, x squared times the derivative of the second function with respect to x. Well, that's just y prime. The derivative of y with respect to y is 1 times the derivative of y with respect to x, which is just y prime. Fair enough. Plus, the derivative of this with respect to x is easy, 2. Plus, the derivative of this with respect to x, well, let's take the derivative of this with respect to y first. We're just doing implicit differentiation in the chain rule."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "Fair enough. Plus, the derivative of this with respect to x is easy, 2. Plus, the derivative of this with respect to x, well, let's take the derivative of this with respect to y first. We're just doing implicit differentiation in the chain rule. So this is plus 6y squared, and then we're using the chain rule, so we took the derivative with respect to y, then you have to multiply that times the derivative of y with respect to x, which is just y prime. Plus, the derivative of this with respect to y is 3 times, just doing the chain rule, the derivative of y with respect to x, so that's y prime. And let's try to see if we can simplify this."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "We're just doing implicit differentiation in the chain rule. So this is plus 6y squared, and then we're using the chain rule, so we took the derivative with respect to y, then you have to multiply that times the derivative of y with respect to x, which is just y prime. Plus, the derivative of this with respect to y is 3 times, just doing the chain rule, the derivative of y with respect to x, so that's y prime. And let's try to see if we can simplify this. So we get this is equal to 3x squared minus 2xy plus 2, so that's this term, this term, and this term, plus, let's say, let's just put the y prime outside, y prime times, see, I have a negative sign out here, minus x squared plus 6y squared plus 3. So this is the derivative of our xi as we solved it. And notice, look at this closely, and notice that that is the same, hopefully it's the same, as our original problem."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And let's try to see if we can simplify this. So we get this is equal to 3x squared minus 2xy plus 2, so that's this term, this term, and this term, plus, let's say, let's just put the y prime outside, y prime times, see, I have a negative sign out here, minus x squared plus 6y squared plus 3. So this is the derivative of our xi as we solved it. And notice, look at this closely, and notice that that is the same, hopefully it's the same, as our original problem. What was our original problem that we started working with? The original problem was 3x squared minus 2xy plus 2 plus 6y squared minus x squared plus 3 times y prime is equal to 0. So this was our original problem."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "And notice, look at this closely, and notice that that is the same, hopefully it's the same, as our original problem. What was our original problem that we started working with? The original problem was 3x squared minus 2xy plus 2 plus 6y squared minus x squared plus 3 times y prime is equal to 0. So this was our original problem. And notice that the derivative of xi with respect to x, just using implicit differentiation, is exactly this. So hopefully this gives you a little intuition of why we can just rewrite this equation as the derivative with respect to x of xi, which is a function of x and y, is equal to 0. Because this is the derivative of xi with respect to x. I wrote it out here."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So this was our original problem. And notice that the derivative of xi with respect to x, just using implicit differentiation, is exactly this. So hopefully this gives you a little intuition of why we can just rewrite this equation as the derivative with respect to x of xi, which is a function of x and y, is equal to 0. Because this is the derivative of xi with respect to x. I wrote it out here. It's the same thing. It's right here. So that equals 0."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "Because this is the derivative of xi with respect to x. I wrote it out here. It's the same thing. It's right here. So that equals 0. So if we take the antiderivative of both sides, we know that the solution of this differential equation is that xi of x and y is equal to c is the solution. And we know what xi is, so we just set that equal to c, and we have the implicit, we have a solution to the differential equation, all this defined implicitly. So the solution, you don't have to do this every time."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So that equals 0. So if we take the antiderivative of both sides, we know that the solution of this differential equation is that xi of x and y is equal to c is the solution. And we know what xi is, so we just set that equal to c, and we have the implicit, we have a solution to the differential equation, all this defined implicitly. So the solution, you don't have to do this every time. This step right here, you wouldn't have to do if you were taking a test, unless the teacher explicitly asked for it. I just wanted to kind of make sure that you know what you're doing, that you're not just doing things completely mechanically. That you really see that the derivative of xi really does give you, like we solved for xi, and I just want to show that the derivative of xi with respect to x, just using implicit differentiation and our standard chain rule, actually gives you the left-hand side of the differential equation, which was our original problem."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "So the solution, you don't have to do this every time. This step right here, you wouldn't have to do if you were taking a test, unless the teacher explicitly asked for it. I just wanted to kind of make sure that you know what you're doing, that you're not just doing things completely mechanically. That you really see that the derivative of xi really does give you, like we solved for xi, and I just want to show that the derivative of xi with respect to x, just using implicit differentiation and our standard chain rule, actually gives you the left-hand side of the differential equation, which was our original problem. And then that's how we know that the derivative of xi with respect to x is equal to 0, because our original differential equation was equal to 0. You take the antiderivative of both sides of this, you get xi is equal to c, is the solution of the differential equation. Or if you wanted to write it out, xi is this thing."}, {"video_title": "Exact equations example 3 First order differential equations Khan Academy.mp3", "Sentence": "That you really see that the derivative of xi really does give you, like we solved for xi, and I just want to show that the derivative of xi with respect to x, just using implicit differentiation and our standard chain rule, actually gives you the left-hand side of the differential equation, which was our original problem. And then that's how we know that the derivative of xi with respect to x is equal to 0, because our original differential equation was equal to 0. You take the antiderivative of both sides of this, you get xi is equal to c, is the solution of the differential equation. Or if you wanted to write it out, xi is this thing. Our solution to the differential equation is x to the third minus x squared y plus 2x plus 2y to the third plus 3y is equal to c, is the implicitly defined solution of our original differential equation. Anyway, I've run out of time again. I will see you in the next video."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "That's what's written here. 0.25 y is equal to 0. And they've actually given us some initial conditions. They said that y of 0 is equal to 2, and y prime of 0 is equal to 1 third. So like we've done in every one of these constant coefficient linear second order homogenous differential equations, let's get the characteristic equation. So that's r squared minus r plus 0.25, or we could even call it plus 1 fourth. So let's see."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "They said that y of 0 is equal to 2, and y prime of 0 is equal to 1 third. So like we've done in every one of these constant coefficient linear second order homogenous differential equations, let's get the characteristic equation. So that's r squared minus r plus 0.25, or we could even call it plus 1 fourth. So let's see. When I'm just inspecting this, it always confuses me when I have fractions, so it becomes very hard to factor. So let's just do the quadratic formula. So the roots of this are going to be r is equal to negative b."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "So let's see. When I'm just inspecting this, it always confuses me when I have fractions, so it becomes very hard to factor. So let's just do the quadratic formula. So the roots of this are going to be r is equal to negative b. Well, b is negative 1. So negative b is going to be 1 plus or minus the square root of b squared. b is negative 1, so that squared is 1."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "So the roots of this are going to be r is equal to negative b. Well, b is negative 1. So negative b is going to be 1 plus or minus the square root of b squared. b is negative 1, so that squared is 1. Minus 4 times a, which is 1, times c. Well, 4 times 1 times 0.25, that's 1. So notice that when you have a repeated root, this under the square root becomes 0. That makes sense, because this plus or minus in the quadratic formula gives you two roots, whether they be real or complex, but if the square root is 0, you're adding plus or minus 0, and you're only left with one root."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "b is negative 1, so that squared is 1. Minus 4 times a, which is 1, times c. Well, 4 times 1 times 0.25, that's 1. So notice that when you have a repeated root, this under the square root becomes 0. That makes sense, because this plus or minus in the quadratic formula gives you two roots, whether they be real or complex, but if the square root is 0, you're adding plus or minus 0, and you're only left with one root. Anyway, we're not done yet. What's the denominator of the quadratic equation? 2a."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "That makes sense, because this plus or minus in the quadratic formula gives you two roots, whether they be real or complex, but if the square root is 0, you're adding plus or minus 0, and you're only left with one root. Anyway, we're not done yet. What's the denominator of the quadratic equation? 2a. So a is 1 over 2. So our one repeated root is 1 plus or minus 0 over 2, or it equals 1 half. And like we learned in the last video, you might just say, oh, well, maybe the solution is just y is equal to c e to the 1 half x."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "2a. So a is 1 over 2. So our one repeated root is 1 plus or minus 0 over 2, or it equals 1 half. And like we learned in the last video, you might just say, oh, well, maybe the solution is just y is equal to c e to the 1 half x. But like we pointed out last time, you have two initial conditions, and this solution is not general enough for two initial conditions. And then last time we said, OK, if this isn't general enough, maybe some solution that was some function of x times e to the 1 half x, maybe that would be our solution. We said, it turns out it is."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "And like we learned in the last video, you might just say, oh, well, maybe the solution is just y is equal to c e to the 1 half x. But like we pointed out last time, you have two initial conditions, and this solution is not general enough for two initial conditions. And then last time we said, OK, if this isn't general enough, maybe some solution that was some function of x times e to the 1 half x, maybe that would be our solution. We said, it turns out it is. And so that more general solution that we found, that we figured out that v of x is actually equal to some constant plus x times some other constant. So our more general solution is y is equal to c1 times e to the 1 half x plus c2 times x e to the 1 half x. I forgot the x here. Let me draw a line here so you don't get confused."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "We said, it turns out it is. And so that more general solution that we found, that we figured out that v of x is actually equal to some constant plus x times some other constant. So our more general solution is y is equal to c1 times e to the 1 half x plus c2 times x e to the 1 half x. I forgot the x here. Let me draw a line here so you don't get confused. Anyway, that's the reasoning. That's how we came up with this thing. And it is good to know, because later on when you want to know more theory of differential equations, and that's really the whole point about learning this, if your whole goal isn't just to pass an exam."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "Let me draw a line here so you don't get confused. Anyway, that's the reasoning. That's how we came up with this thing. And it is good to know, because later on when you want to know more theory of differential equations, and that's really the whole point about learning this, if your whole goal isn't just to pass an exam. It's good to know. But when you're actually solving these, you could just kind of know the template. If I have a repeated root, well, I just put that repeated root twice, and one of them gets an x in front of it, and they have two constants."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "And it is good to know, because later on when you want to know more theory of differential equations, and that's really the whole point about learning this, if your whole goal isn't just to pass an exam. It's good to know. But when you're actually solving these, you could just kind of know the template. If I have a repeated root, well, I just put that repeated root twice, and one of them gets an x in front of it, and they have two constants. But anyway, this is our general solution, and now we can use our initial conditions to solve for c1 and c2. So let's just figure out the derivative of this first so it becomes easy to substitute in for c2. So y prime is equal to 1 half c1 e to the 1 half x plus."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "If I have a repeated root, well, I just put that repeated root twice, and one of them gets an x in front of it, and they have two constants. But anyway, this is our general solution, and now we can use our initial conditions to solve for c1 and c2. So let's just figure out the derivative of this first so it becomes easy to substitute in for c2. So y prime is equal to 1 half c1 e to the 1 half x plus. Now this becomes a little bit more complicated. We're going to have to use the product rule here. So let's see."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "So y prime is equal to 1 half c1 e to the 1 half x plus. Now this becomes a little bit more complicated. We're going to have to use the product rule here. So let's see. Plus c2 times derivative of x is 1 times e to the 1 half x. That's the product rule. Plus the derivative of e to the 1 half x times x."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "So let's see. Plus c2 times derivative of x is 1 times e to the 1 half x. That's the product rule. Plus the derivative of e to the 1 half x times x. So that's 1 half x e to the 1 half x. Or we can write, I don't want to lose this stuff up here, we can write that it equals, let's see. I have 1 half, so I have c2 times e to the 1 half x, and I have 1 half times c1 e to the 1 half x."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "Plus the derivative of e to the 1 half x times x. So that's 1 half x e to the 1 half x. Or we can write, I don't want to lose this stuff up here, we can write that it equals, let's see. I have 1 half, so I have c2 times e to the 1 half x, and I have 1 half times c1 e to the 1 half x. So I could say it's equal to e to the 1 half x times c1 over 2, that's that, plus c2, that takes care of these two terms, plus c2 over 2 x e to the 1 half x. And now let's use our initial conditions. Let me actually clear up some space, because I think it's nice to have our initial conditions up here, so we can see them."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "I have 1 half, so I have c2 times e to the 1 half x, and I have 1 half times c1 e to the 1 half x. So I could say it's equal to e to the 1 half x times c1 over 2, that's that, plus c2, that takes care of these two terms, plus c2 over 2 x e to the 1 half x. And now let's use our initial conditions. Let me actually clear up some space, because I think it's nice to have our initial conditions up here, so we can see them. So let me delete all this stuff here that hopefully makes sense to you by now. We know the characteristic equation, we figured out the general solution. I don't want to erase our initial conditions."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "Let me actually clear up some space, because I think it's nice to have our initial conditions up here, so we can see them. So let me delete all this stuff here that hopefully makes sense to you by now. We know the characteristic equation, we figured out the general solution. I don't want to erase our initial conditions. We figured out the general solution was this. I'll keep our general solution there. And so now we just substitute our initial conditions into our general solution and the derivative of the general solution, and hopefully we can get meaningful answers."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "I don't want to erase our initial conditions. We figured out the general solution was this. I'll keep our general solution there. And so now we just substitute our initial conditions into our general solution and the derivative of the general solution, and hopefully we can get meaningful answers. So substituting into our general solution, y of 0 is equal to 2. So y is equal to 2 when x is equal to 0. So c1, when x is equal to 0, all the e terms become 1, right?"}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "And so now we just substitute our initial conditions into our general solution and the derivative of the general solution, and hopefully we can get meaningful answers. So substituting into our general solution, y of 0 is equal to 2. So y is equal to 2 when x is equal to 0. So c1, when x is equal to 0, all the e terms become 1, right? This one will become 1. And then notice, we have an x e to the 0. So now this x is 0, so this whole term is going to be equal to 0."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "So c1, when x is equal to 0, all the e terms become 1, right? This one will become 1. And then notice, we have an x e to the 0. So now this x is 0, so this whole term is going to be equal to 0. So we're done. c1 is equal to 2. That was pretty straightforward."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "So now this x is 0, so this whole term is going to be equal to 0. So we're done. c1 is equal to 2. That was pretty straightforward. This x actually made it a lot easier. So c1 is equal to 2. And now we can use the derivative."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "That was pretty straightforward. This x actually made it a lot easier. So c1 is equal to 2. And now we can use the derivative. So let's see, this is the first derivative. So let's see. And I'll substitute c1 in there so we can just solve for c2."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "And now we can use the derivative. So let's see, this is the first derivative. So let's see. And I'll substitute c1 in there so we can just solve for c2. So our first derivative is y prime is equal to, let's see, c1 1 half plus c2. So it's, well, I'll write this first. It's equal to 2 over 2, so it's 1 plus c2 times e to the 1 half x plus c2 over 2 times x e to the 1 half x."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "And I'll substitute c1 in there so we can just solve for c2. So our first derivative is y prime is equal to, let's see, c1 1 half plus c2. So it's, well, I'll write this first. It's equal to 2 over 2, so it's 1 plus c2 times e to the 1 half x plus c2 over 2 times x e to the 1 half x. There was an x here. So when x is equal to 0, y prime is equal to 1 third. So 1 third is equal to, well, x is equal to 0, this will be 1, so it's equal to 1 plus c2."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "It's equal to 2 over 2, so it's 1 plus c2 times e to the 1 half x plus c2 over 2 times x e to the 1 half x. There was an x here. So when x is equal to 0, y prime is equal to 1 third. So 1 third is equal to, well, x is equal to 0, this will be 1, so it's equal to 1 plus c2. And then this term, when x is equal to 0, this whole thing becomes 0, right? Because this x just cancels out the whole thing. You multiply it by 0, you get 0."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "So 1 third is equal to, well, x is equal to 0, this will be 1, so it's equal to 1 plus c2. And then this term, when x is equal to 0, this whole thing becomes 0, right? Because this x just cancels out the whole thing. You multiply it by 0, you get 0. So then we get 1 third is equal to 1 plus c2, or that c2 is equal to 1 third minus 1 is equal to minus 2 thirds. And now we have our particular solution. Let me write it down and put a box around it."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "You multiply it by 0, you get 0. So then we get 1 third is equal to 1 plus c2, or that c2 is equal to 1 third minus 1 is equal to minus 2 thirds. And now we have our particular solution. Let me write it down and put a box around it. So this was our general solution, our particular solution, given these initial conditions, for this repeated root problem is y is equal to c1. We figured that out to be 2 fairly quickly. 2 e to the 1 half x plus c2."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "Let me write it down and put a box around it. So this was our general solution, our particular solution, given these initial conditions, for this repeated root problem is y is equal to c1. We figured that out to be 2 fairly quickly. 2 e to the 1 half x plus c2. c2 is minus 2 thirds. So minus 2 thirds x e to the 1 half x. And we are done."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "2 e to the 1 half x plus c2. c2 is minus 2 thirds. So minus 2 thirds x e to the 1 half x. And we are done. There is our particular solution. So once again, kind of the proof of how do you get to this, why is there this x in there? And it wasn't a proof."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "And we are done. There is our particular solution. So once again, kind of the proof of how do you get to this, why is there this x in there? And it wasn't a proof. It was really more of just to show you the intuition of where that came from. And it did introduce you to a method called reduction of order to figure out what that function v was, which ended up just being c1 plus c2 times x. But all of that can be pretty complicated."}, {"video_title": "Repeated roots of the characteristic equations part 2 Khan Academy.mp3", "Sentence": "And it wasn't a proof. It was really more of just to show you the intuition of where that came from. And it did introduce you to a method called reduction of order to figure out what that function v was, which ended up just being c1 plus c2 times x. But all of that can be pretty complicated. But you see that once you know the pattern, or once you know that this is going to be the general solution, they're pretty easy to solve. Characteristic equation, get your general solution, figure out the derivative of the general solution, and then substitute your initial conditions to solve for your constants, and you're done. Anyway, I'll see you in the next video."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "Let's solve another second order linear homogenous differential equation. And this one, well, I won't give you the details before I actually write it down. So the differential equation is 4 times the second derivative of y with respect to x minus 8 times the first derivative plus 3 times the function times y is equal to 0. And we have our initial conditions. y of 0 is equal to 2. And we have y prime of 0 is equal to 1 half. Now, I could go into the whole thing, y is equal to e to the rx as a solution, substitute it in, then factor out e to the rx and have the characteristic equation."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And we have our initial conditions. y of 0 is equal to 2. And we have y prime of 0 is equal to 1 half. Now, I could go into the whole thing, y is equal to e to the rx as a solution, substitute it in, then factor out e to the rx and have the characteristic equation. And if you want to see all of that over again, you might want to watch the previous video just to see where that characteristic equation comes from. But in this video, I'm just going to show you literally how quickly you can do these type of problems mechanically. So if this is our original differential equation, the characteristic equation is going to be, and I'll do this in a different color, 4r squared minus 8r plus 3r is equal to 0."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "Now, I could go into the whole thing, y is equal to e to the rx as a solution, substitute it in, then factor out e to the rx and have the characteristic equation. And if you want to see all of that over again, you might want to watch the previous video just to see where that characteristic equation comes from. But in this video, I'm just going to show you literally how quickly you can do these type of problems mechanically. So if this is our original differential equation, the characteristic equation is going to be, and I'll do this in a different color, 4r squared minus 8r plus 3r is equal to 0. And watch the previous video if you don't know where this characteristic equation comes from. But if you want to do these problems really quick, you just substitute the second derivatives with an r squared, the first derivatives with an r, and then the function with oh, sorry, no, this is supposed to be a constant. And then the coefficient on the original function is just the constant, right?"}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "So if this is our original differential equation, the characteristic equation is going to be, and I'll do this in a different color, 4r squared minus 8r plus 3r is equal to 0. And watch the previous video if you don't know where this characteristic equation comes from. But if you want to do these problems really quick, you just substitute the second derivatives with an r squared, the first derivatives with an r, and then the function with oh, sorry, no, this is supposed to be a constant. And then the coefficient on the original function is just the constant, right? I think you see what I did. Second derivative, r squared, first derivative, r, no derivative, you could say that's r to the 0 or just 1, but this is our characteristic equation. And now we can just figure out its roots."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And then the coefficient on the original function is just the constant, right? I think you see what I did. Second derivative, r squared, first derivative, r, no derivative, you could say that's r to the 0 or just 1, but this is our characteristic equation. And now we can just figure out its roots. This is not a trivial one for me to factor, so if it's not trivial, you can use the quadratic equation. So we could say the solution of this is r is equal to negative b, well, b is negative 8, so it's positive 8, 8 plus or minus the square root of b squared, so that's 64, minus 4 times a, which is 4, times c, which is 3, all of that over 2a. 2 times 4 is 8."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And now we can just figure out its roots. This is not a trivial one for me to factor, so if it's not trivial, you can use the quadratic equation. So we could say the solution of this is r is equal to negative b, well, b is negative 8, so it's positive 8, 8 plus or minus the square root of b squared, so that's 64, minus 4 times a, which is 4, times c, which is 3, all of that over 2a. 2 times 4 is 8. That equals 8 plus or minus square root of 64 minus, what's 16 times 3 minus 48, all of that over 8. What's 64 minus 48? Let's see."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "2 times 4 is 8. That equals 8 plus or minus square root of 64 minus, what's 16 times 3 minus 48, all of that over 8. What's 64 minus 48? Let's see. 12, it's 16, right? 10 plus 48 is 58, then another, so it's 16. So we have r is equal to 8 plus or minus the square root of 16 over 8 is equal to 8 plus or minus 4 over 8."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "Let's see. 12, it's 16, right? 10 plus 48 is 58, then another, so it's 16. So we have r is equal to 8 plus or minus the square root of 16 over 8 is equal to 8 plus or minus 4 over 8. That equals 1 plus or minus 1 half. So the two solutions of this characteristic equation, and ignore that, let me scratch that out in black so that you know that that's not like a 30 or something. The two solutions of this characteristic equation are r is equal to, well, 1 plus 1 half is equal to 3 halves, and r is equal to 1 minus 1 half is equal to 1 half."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "So we have r is equal to 8 plus or minus the square root of 16 over 8 is equal to 8 plus or minus 4 over 8. That equals 1 plus or minus 1 half. So the two solutions of this characteristic equation, and ignore that, let me scratch that out in black so that you know that that's not like a 30 or something. The two solutions of this characteristic equation are r is equal to, well, 1 plus 1 half is equal to 3 halves, and r is equal to 1 minus 1 half is equal to 1 half. So we know our two r's, and we know that from previous experience in the last video that y is equal to c times e to the rx is a solution. So the general solution of this differential equation is y is equal to c1 times e, let's use our first r, e to the 3 halves x plus c2 times e to the 1 half x. This differential equations problem was literally just a problem in using the quadratic equation, and once you figure out the r's, you have your general solution."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "The two solutions of this characteristic equation are r is equal to, well, 1 plus 1 half is equal to 3 halves, and r is equal to 1 minus 1 half is equal to 1 half. So we know our two r's, and we know that from previous experience in the last video that y is equal to c times e to the rx is a solution. So the general solution of this differential equation is y is equal to c1 times e, let's use our first r, e to the 3 halves x plus c2 times e to the 1 half x. This differential equations problem was literally just a problem in using the quadratic equation, and once you figure out the r's, you have your general solution. Now we just have to use our initial conditions. So to know the initial conditions, we need to know y of x, and we need to know y prime of x. Let's just do that right now."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "This differential equations problem was literally just a problem in using the quadratic equation, and once you figure out the r's, you have your general solution. Now we just have to use our initial conditions. So to know the initial conditions, we need to know y of x, and we need to know y prime of x. Let's just do that right now. So what's y prime? y prime of our general solution is equal to 3 halves times c1 e to the 3 halves x plus, derivative of the inside, 1 half times c2 e to the 1 half x. And now let's use our actual initial conditions."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "Let's just do that right now. So what's y prime? y prime of our general solution is equal to 3 halves times c1 e to the 3 halves x plus, derivative of the inside, 1 half times c2 e to the 1 half x. And now let's use our actual initial conditions. I don't want to lose them. Let me rewrite them down here so I can scroll down. So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1 half."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And now let's use our actual initial conditions. I don't want to lose them. Let me rewrite them down here so I can scroll down. So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1 half. Those are our initial conditions. So let's use that information. So y of 0, what happens when you substitute x is equal to 0 here?"}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1 half. Those are our initial conditions. So let's use that information. So y of 0, what happens when you substitute x is equal to 0 here? You get c1 times e to the 0, essentially, so that's just 1, plus c2, well that's just e to the 0 again because x is 0, is equal to, so this is when x is equal to 0, what is y? y is equal to 2. y of 0 is equal to 2. And then let's use the second equation."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "So y of 0, what happens when you substitute x is equal to 0 here? You get c1 times e to the 0, essentially, so that's just 1, plus c2, well that's just e to the 0 again because x is 0, is equal to, so this is when x is equal to 0, what is y? y is equal to 2. y of 0 is equal to 2. And then let's use the second equation. So when we substitute x is equal to 0 in the derivative, so when x is 0, we get 3 halves c1, this goes to 1 again, plus 1 half c2, this is 1 again, e to the 1 half times 0 is e to the 0, which is 1, is equal to, so when x is 0 for the derivative, y is equal to 1 half, or the derivative is 1 half at that point, or the slope is 1 half at that point. And now we have two equations in two unknowns, and we could do a couple, we could solve it a ton of ways, I think you know how to solve them. Let's multiply the top equation, I don't know, let's multiply it by 3 halves, and what do we get?"}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And then let's use the second equation. So when we substitute x is equal to 0 in the derivative, so when x is 0, we get 3 halves c1, this goes to 1 again, plus 1 half c2, this is 1 again, e to the 1 half times 0 is e to the 0, which is 1, is equal to, so when x is 0 for the derivative, y is equal to 1 half, or the derivative is 1 half at that point, or the slope is 1 half at that point. And now we have two equations in two unknowns, and we could do a couple, we could solve it a ton of ways, I think you know how to solve them. Let's multiply the top equation, I don't know, let's multiply it by 3 halves, and what do we get? We get, I'll do it in a different color, we get 3 halves c1 plus 3 halves c2 is equal to, what's 3 halves times 2? It's equal to 3. And now, let's subtract, well, I don't want to confuse you, so let's just subtract the bottom from the top, so this cancels out."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "Let's multiply the top equation, I don't know, let's multiply it by 3 halves, and what do we get? We get, I'll do it in a different color, we get 3 halves c1 plus 3 halves c2 is equal to, what's 3 halves times 2? It's equal to 3. And now, let's subtract, well, I don't want to confuse you, so let's just subtract the bottom from the top, so this cancels out. What's 1 half minus 3 halves? 1 half minus 1 and 1 half. Well, that's just minus 1, right?"}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And now, let's subtract, well, I don't want to confuse you, so let's just subtract the bottom from the top, so this cancels out. What's 1 half minus 3 halves? 1 half minus 1 and 1 half. Well, that's just minus 1, right? So minus c2 is equal to, what's 1 half minus 3? That's minus 2 and 1 half, or minus 5 halves. And so we get c2 is equal to 5 halves, and we could substitute back to this top equation, c1 plus 5 halves is equal to 2, or c1 is equal to 2, which is the same thing as 4 halves minus 5 halves, which is equal to minus 1 half."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "Well, that's just minus 1, right? So minus c2 is equal to, what's 1 half minus 3? That's minus 2 and 1 half, or minus 5 halves. And so we get c2 is equal to 5 halves, and we could substitute back to this top equation, c1 plus 5 halves is equal to 2, or c1 is equal to 2, which is the same thing as 4 halves minus 5 halves, which is equal to minus 1 half. And now we can just substitute c1 and c2 back into our general solution, and we have found the particular solution of this differential equation, which is y is equal to c1, c1 is minus 1 half, e to the 3 halves x plus c2, which is 5 halves, e to the 1 half x. And we are done. And it might seem really fancy."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And so we get c2 is equal to 5 halves, and we could substitute back to this top equation, c1 plus 5 halves is equal to 2, or c1 is equal to 2, which is the same thing as 4 halves minus 5 halves, which is equal to minus 1 half. And now we can just substitute c1 and c2 back into our general solution, and we have found the particular solution of this differential equation, which is y is equal to c1, c1 is minus 1 half, e to the 3 halves x plus c2, which is 5 halves, e to the 1 half x. And we are done. And it might seem really fancy. We're solving a differential equation, our solution has e in it, and we're taking derivatives, and we're doing all sorts of things. But really, the meat of this problem was solving a quadratic, which was our characteristic equation. And watch the previous video just to see why this characteristic equation works."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And it might seem really fancy. We're solving a differential equation, our solution has e in it, and we're taking derivatives, and we're doing all sorts of things. But really, the meat of this problem was solving a quadratic, which was our characteristic equation. And watch the previous video just to see why this characteristic equation works. But it's very easy to come up with the characteristic equation, right? I think you obviously see that y prime turns into r squared, y prime turns into r, and then y just turns into 1, essentially. So you solve a quadratic, and then after doing that, you just have to take one derivative."}, {"video_title": "2nd Order Linear Homogeneous Differential Equations 4 Khan Academy.mp3", "Sentence": "And watch the previous video just to see why this characteristic equation works. But it's very easy to come up with the characteristic equation, right? I think you obviously see that y prime turns into r squared, y prime turns into r, and then y just turns into 1, essentially. So you solve a quadratic, and then after doing that, you just have to take one derivative. Because after solving the quadratic, you immediately have the general solution. Then you take its derivative, use your initial conditions, you have a system of linear equations, which is algebra 1, and then you solve them for the two constants, c1 and c2, and you end up with your particular solution. And that's all there is to it."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "This is a scenario where we take an object that is hotter or cooler than the ambient room temperature and we want to model how fast it cools or heats up. And the way that we'll think about it is the way that Newton thought about it. And it is, Newton's described as Newton's law of cooling. Newton's law of cooling. And in a lot of ways, it's common sense. It states that the rate of change of temperature should be proportional to the difference between the temperature of the object and the ambient temperature. So let me write that in mathematical terms."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "Newton's law of cooling. And in a lot of ways, it's common sense. It states that the rate of change of temperature should be proportional to the difference between the temperature of the object and the ambient temperature. So let me write that in mathematical terms. So Newton's law of cooling tells us that the rate of change of temperature, I'll use that with a capital T, with respect to time, lowercase t, should be proportional to the difference between the temperature of the object and the ambient temperature. So that is a mathematical description of it. And once again, it's common sense."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So let me write that in mathematical terms. So Newton's law of cooling tells us that the rate of change of temperature, I'll use that with a capital T, with respect to time, lowercase t, should be proportional to the difference between the temperature of the object and the ambient temperature. So that is a mathematical description of it. And once again, it's common sense. If something is much, much, much, much hotter than the ambient temperature, the rate of change should be pretty steep. It should be declining in temperature quickly. If something is much, much, much, much cooler, it should be increasing in temperature quickly."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "And once again, it's common sense. If something is much, much, much, much hotter than the ambient temperature, the rate of change should be pretty steep. It should be declining in temperature quickly. If something is much, much, much, much cooler, it should be increasing in temperature quickly. And if something is close, well, maybe the rate, if these two things are pretty close, well, maybe this rate of change shouldn't be so big. Now, I know one thing that you're thinking. It's like, okay, if the temperature is hotter than the ambient temperature, then I should be cooling."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "If something is much, much, much, much cooler, it should be increasing in temperature quickly. And if something is close, well, maybe the rate, if these two things are pretty close, well, maybe this rate of change shouldn't be so big. Now, I know one thing that you're thinking. It's like, okay, if the temperature is hotter than the ambient temperature, then I should be cooling. My temperature should be decreasing. And a decreasing temperature would imply a negative instantaneous change. So how will this be a negative value in the case where our temperature of our object is greater than our ambient temperature?"}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "It's like, okay, if the temperature is hotter than the ambient temperature, then I should be cooling. My temperature should be decreasing. And a decreasing temperature would imply a negative instantaneous change. So how will this be a negative value in the case where our temperature of our object is greater than our ambient temperature? And the way that that would happen is you would have to have a negative K. Now, if you don't like thinking in terms of a negative K, you could just put a negative right over here, and now you would have a positive K. Now, it makes sense. If our thing is hotter, if it has a higher temperature than the ambient temperature, so this is a positive, then our rate of change will be negative. We'll be getting cooler."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So how will this be a negative value in the case where our temperature of our object is greater than our ambient temperature? And the way that that would happen is you would have to have a negative K. Now, if you don't like thinking in terms of a negative K, you could just put a negative right over here, and now you would have a positive K. Now, it makes sense. If our thing is hotter, if it has a higher temperature than the ambient temperature, so this is a positive, then our rate of change will be negative. We'll be getting cooler. If it was the other way around, if our temperature of our object is cooler than our ambient temperature, then this thing is going to be a negative. Well, then the negative of that is going to be a positive. We're assuming a positive K, and our temperature will be increasing."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "We'll be getting cooler. If it was the other way around, if our temperature of our object is cooler than our ambient temperature, then this thing is going to be a negative. Well, then the negative of that is going to be a positive. We're assuming a positive K, and our temperature will be increasing. So hopefully, this makes some intuitive sense, and our constant K could depend on the specific heat of the object, how much surface area is exposed to it, or whatever else. But now, given this, let's see if we can solve this differential equation for a general solution, and I encourage you to pause this video and do that, and I will give you a clue. This is a separable differential equation."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "We're assuming a positive K, and our temperature will be increasing. So hopefully, this makes some intuitive sense, and our constant K could depend on the specific heat of the object, how much surface area is exposed to it, or whatever else. But now, given this, let's see if we can solve this differential equation for a general solution, and I encourage you to pause this video and do that, and I will give you a clue. This is a separable differential equation. So I assume you've had a go at it, so let's now work through it together. So we just have to algebraically manipulate this, so all my Ts and DTs are on one side, or I should say, so all my capital Ts and D capital Ts are on one side, and so this is going to be a little bit more confusing, because I have a capital T and a lowercase t, capital T for temperature, lowercase t for time, but hopefully we'll be able to work through it, and then I'm going to have all my time differentials and time variables on the other side. So one thing I could do is I could divide both sides by capital T minus ambient temperature, minus T sub a."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "This is a separable differential equation. So I assume you've had a go at it, so let's now work through it together. So we just have to algebraically manipulate this, so all my Ts and DTs are on one side, or I should say, so all my capital Ts and D capital Ts are on one side, and so this is going to be a little bit more confusing, because I have a capital T and a lowercase t, capital T for temperature, lowercase t for time, but hopefully we'll be able to work through it, and then I'm going to have all my time differentials and time variables on the other side. So one thing I could do is I could divide both sides by capital T minus ambient temperature, minus T sub a. Remember, this is just going to be a constant based on what our ambient temperature is. We're going to assume our ambient temperature doesn't change as a function of time, that there's just, it's just such a big room that our cup of tea is not going to actually warm up the room, so that's just one of these assumptions that we're going to make. So if we do that, if we divide both sides by this, we are going to have, so I'm going to divide both sides, let me do this in a new color."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So one thing I could do is I could divide both sides by capital T minus ambient temperature, minus T sub a. Remember, this is just going to be a constant based on what our ambient temperature is. We're going to assume our ambient temperature doesn't change as a function of time, that there's just, it's just such a big room that our cup of tea is not going to actually warm up the room, so that's just one of these assumptions that we're going to make. So if we do that, if we divide both sides by this, we are going to have, so I'm going to divide both sides, let me do this in a new color. If I divide both sides by that, I get one over capital T minus T sub a, and let me multiply both sides times the time differential. So I'm going to have, so that DT, DT, so our temperature differential, times our temperature differential is going to be equal to negative K, negative K times our time, times our time differential. So once again, to separate the variables, all I did is divide both sides by this, and multiply both sides by that."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So if we do that, if we divide both sides by this, we are going to have, so I'm going to divide both sides, let me do this in a new color. If I divide both sides by that, I get one over capital T minus T sub a, and let me multiply both sides times the time differential. So I'm going to have, so that DT, DT, so our temperature differential, times our temperature differential is going to be equal to negative K, negative K times our time, times our time differential. So once again, to separate the variables, all I did is divide both sides by this, and multiply both sides by that. Now I can integrate both sides. We've seen this show before. So I can integrate both sides, and the integral of this is going to be the natural log of the absolute value of what we have in the denominator."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So once again, to separate the variables, all I did is divide both sides by this, and multiply both sides by that. Now I can integrate both sides. We've seen this show before. So I can integrate both sides, and the integral of this is going to be the natural log of the absolute value of what we have in the denominator. You could do U substitution if you want. If we said U is equal to capital T minus T sub a, then DU is just going to be one DT, and so this is essentially, you could say the integral of one over U DU, and this would be the natural log of the absolute value of U, and this right over here is U. So this is the natural log of the absolute value of capital T minus T sub a is equal to, and once again, I could put a constant here, but I'm going to end up with a constant on the right-hand side too, so I'm just going to merge them into the constant on the right-hand side."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So I can integrate both sides, and the integral of this is going to be the natural log of the absolute value of what we have in the denominator. You could do U substitution if you want. If we said U is equal to capital T minus T sub a, then DU is just going to be one DT, and so this is essentially, you could say the integral of one over U DU, and this would be the natural log of the absolute value of U, and this right over here is U. So this is the natural log of the absolute value of capital T minus T sub a is equal to, and once again, I could put a constant here, but I'm going to end up with a constant on the right-hand side too, so I'm just going to merge them into the constant on the right-hand side. So that is going to be equal to, now here, this is going to be negative KT, and once again, we have plus C, and now we can take, we can raise E to both of these powers, or another way of interpreting this is that E to this thing is going to be the same as that. So we can write this as the absolute value. Let me do that in that same blue color."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So this is the natural log of the absolute value of capital T minus T sub a is equal to, and once again, I could put a constant here, but I'm going to end up with a constant on the right-hand side too, so I'm just going to merge them into the constant on the right-hand side. So that is going to be equal to, now here, this is going to be negative KT, and once again, we have plus C, and now we can take, we can raise E to both of these powers, or another way of interpreting this is that E to this thing is going to be the same as that. So we can write this as the absolute value. Let me do that in that same blue color. We can write this as the absolute value of T minus T sub a is equal to E, something about E I always think of the color green, E to the negative KT plus C, plus C. This, of course, is the same thing as this is equal to E to the negative KT. We've done this multiple times before. Negative KT times E to the C, times E, let me do that in that same, so times E to the C power, and we could just call this, well, we could just call this another arbitrary constant."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "Let me do that in that same blue color. We can write this as the absolute value of T minus T sub a is equal to E, something about E I always think of the color green, E to the negative KT plus C, plus C. This, of course, is the same thing as this is equal to E to the negative KT. We've done this multiple times before. Negative KT times E to the C, times E, let me do that in that same, so times E to the C power, and we could just call this, well, we could just call this another arbitrary constant. Let's see, if we call this C1, then we could just call this whole thing C. So this, we could say, is CE, CE to the negative KT, negative KT. So at least it's starting to resemble what we did when we were modeling population, but we see it's a little bit different. Instead of just temperature on this left-hand side, we have temperature minus our ambient temperature."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "Negative KT times E to the C, times E, let me do that in that same, so times E to the C power, and we could just call this, well, we could just call this another arbitrary constant. Let's see, if we call this C1, then we could just call this whole thing C. So this, we could say, is CE, CE to the negative KT, negative KT. So at least it's starting to resemble what we did when we were modeling population, but we see it's a little bit different. Instead of just temperature on this left-hand side, we have temperature minus our ambient temperature. And so we can do a couple of things. If in a world, in a world, say, where we're dealing with a hot cup of tea, something that's hotter than the ambient temperature, so we could imagine a world where T is, let's say, greater than or equal to our ambient temperature. So that means this is hot, or it's hotter, I guess we could say."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "Instead of just temperature on this left-hand side, we have temperature minus our ambient temperature. And so we can do a couple of things. If in a world, in a world, say, where we're dealing with a hot cup of tea, something that's hotter than the ambient temperature, so we could imagine a world where T is, let's say, greater than or equal to our ambient temperature. So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value is going to be positive, or the thing inside the absolute value is going to be positive. So we don't need the absolute value, or the absolute value of it's going to be the same thing as it. And then we could just add TA to both sides, T sub A to both sides, and then we would have our temperature, and I could even write this, as a function of time, is going to be equal to this business, is going to be equal to CE, let me do that in the same color, CE to the negative KT, negative KT plus T sub A, plus T sub A."}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value is going to be positive, or the thing inside the absolute value is going to be positive. So we don't need the absolute value, or the absolute value of it's going to be the same thing as it. And then we could just add TA to both sides, T sub A to both sides, and then we would have our temperature, and I could even write this, as a function of time, is going to be equal to this business, is going to be equal to CE, let me do that in the same color, CE to the negative KT, negative KT plus T sub A, plus T sub A. All I did is I'm assuming that this, inside the absolute value, is going to be positive, so the absolute value's not going to change the value, and I added T sub A to both sides to get this. So this right over here is going to be our general solution in the case, in the case where we start with something that is hotter than the ambient room temperature, and if we want to look at the case where something is cooler than the ambient room temperature, so that's the situation, let's say T is less than our ambient room temperature, then the absolute value of T, then this thing over here is going to be negative, and so the absolute value of it's going to be the negative of that. So then this up here results in T sub A minus T, that's going to be the same thing as the absolute value, it's going to be the negative of the negative, so then that is going to be equal to, that is going to be equal to E, E to the negative K, let me do that, actually I'll do it in the same color, I think you get the point, negative KT plus, actually let me just do it, T sub A minus T is going to be equal to C, E to the negative KT, so this is equal to that, I'm just assuming that T is less than T sub A, and so then to solve for T, to solve for T, you could add T to both sides and subtract this from both sides, you would have capital T as a function of T is going to be equal to, let's see if this went on to that side and this goes over here, you would just have the, you would have T sub A minus, minus CE to the negative KT, did I do that right?"}, {"video_title": "Newton's Law of Cooling First order differential equations Khan Academy.mp3", "Sentence": "And then we could just add TA to both sides, T sub A to both sides, and then we would have our temperature, and I could even write this, as a function of time, is going to be equal to this business, is going to be equal to CE, let me do that in the same color, CE to the negative KT, negative KT plus T sub A, plus T sub A. All I did is I'm assuming that this, inside the absolute value, is going to be positive, so the absolute value's not going to change the value, and I added T sub A to both sides to get this. So this right over here is going to be our general solution in the case, in the case where we start with something that is hotter than the ambient room temperature, and if we want to look at the case where something is cooler than the ambient room temperature, so that's the situation, let's say T is less than our ambient room temperature, then the absolute value of T, then this thing over here is going to be negative, and so the absolute value of it's going to be the negative of that. So then this up here results in T sub A minus T, that's going to be the same thing as the absolute value, it's going to be the negative of the negative, so then that is going to be equal to, that is going to be equal to E, E to the negative K, let me do that, actually I'll do it in the same color, I think you get the point, negative KT plus, actually let me just do it, T sub A minus T is going to be equal to C, E to the negative KT, so this is equal to that, I'm just assuming that T is less than T sub A, and so then to solve for T, to solve for T, you could add T to both sides and subtract this from both sides, you would have capital T as a function of T is going to be equal to, let's see if this went on to that side and this goes over here, you would just have the, you would have T sub A minus, minus CE to the negative KT, did I do that right? So yep, that looks right. So this is a situation where you have something that is cooler, cooler than the ambient temperature. So this right over here, based on the logic of Newton's law of cooling, are the two, are two, or I guess these are the general solutions to that differential equation."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And as we'll see, differential equations are super useful for modeling and simulating phenomena and understanding how they operate. But we'll get into that later. For now, let's just think about, or at least look at, what a differential equation actually is. So if I were to write, so here's an example of a differential equation. If I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it, if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So if I were to write, so here's an example of a differential equation. If I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it, if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, we could also write the second derivative of y with respect to x plus two times the first derivative of y with respect to x is equal to three times y. All three of these, I guess, equations are really representing the same thing. They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself?"}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, we could also write the second derivative of y with respect to x plus two times the first derivative of y with respect to x is equal to three times y. All three of these, I guess, equations are really representing the same thing. They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself? So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions. It's not just a value or a set of values."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself? So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions. It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. And it's important to contrast this relative to a traditional equation. So let me write that down."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. And it's important to contrast this relative to a traditional equation. So let me write that down. So a traditional equation, I guess I could say, maybe I shouldn't say traditional equation. Differential equations have been around for a while. So let me write this as a, maybe an algebraic equation that you're familiar with."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So let me write that down. So a traditional equation, I guess I could say, maybe I shouldn't say traditional equation. Differential equations have been around for a while. So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'll just write a simple quadratic, say x squared, x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers or a set of numbers. We can solve this as going to be x plus two times x plus one is equal to zero."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'll just write a simple quadratic, say x squared, x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers or a set of numbers. We can solve this as going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers or a set of values or set of values that satisfy the equation. Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "We can solve this as going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers or a set of values or set of values that satisfy the equation. Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions. Now let's make that a little bit more tangible. What would a solution to something like any of these three which really represent the same thing, what would a solution actually look like? And actually let me move over to the, let me move this over a little bit."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions. Now let's make that a little bit more tangible. What would a solution to something like any of these three which really represent the same thing, what would a solution actually look like? And actually let me move over to the, let me move this over a little bit. Move this a little bit so that we can take a look at what some of these solutions could look like. Let me erase this little stuff that I have right over here. So I'm just going to give you examples of solutions here."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And actually let me move over to the, let me move this over a little bit. Move this a little bit so that we can take a look at what some of these solutions could look like. Let me erase this little stuff that I have right over here. So I'm just going to give you examples of solutions here. We'll verify that these indeed are solutions for, I guess this is really just one differential equation represented in different ways, but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution or that there's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one, so one solution, I'll call it y one, and I could even write it as y one of x to make it explicit that it is a function of x. One solution is y one of x is equal to e to the negative 3x."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So I'm just going to give you examples of solutions here. We'll verify that these indeed are solutions for, I guess this is really just one differential equation represented in different ways, but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution or that there's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one, so one solution, I'll call it y one, and I could even write it as y one of x to make it explicit that it is a function of x. One solution is y one of x is equal to e to the negative 3x. And I encourage you to pause this video right now and find the first derivative of y one or the second derivative of y one and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it, so let's work through this together. So that's y one."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "One solution is y one of x is equal to e to the negative 3x. And I encourage you to pause this video right now and find the first derivative of y one or the second derivative of y one and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it, so let's work through this together. So that's y one. So the first derivative of y one, well, this is going to be, let's see, we just have to do the chain rule here, derivative of negative 3x with respect to x is just negative three, and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x. And if we take the second derivative of y one, this is equal to, same exact idea, derivative of this is negative three times negative three, it's going to be nine e to the negative 3x. And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So that's y one. So the first derivative of y one, well, this is going to be, let's see, we just have to do the chain rule here, derivative of negative 3x with respect to x is just negative three, and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x. And if we take the second derivative of y one, this is equal to, same exact idea, derivative of this is negative three times negative three, it's going to be nine e to the negative 3x. And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function. So let's verify that. So let me, so we first have the second derivative of y, so that's that term right over there. So we have nine e to the negative 3x plus two times the first derivative."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function. So let's verify that. So let me, so we first have the second derivative of y, so that's that term right over there. So we have nine e to the negative 3x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative 3x. Notice I just took this two times the first derivative."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So we have nine e to the negative 3x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative 3x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative 3x. Three e to the negative 3x."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative 3x. Three e to the negative 3x. And let's see if that indeed is true. So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x. So y one is indeed a solution to this differential equation."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "Three e to the negative 3x. And let's see if that indeed is true. So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x. So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it, so the first derivative of this, this is pretty straightforward, is e to the x. Second derivative is one of the profound things of the exponential function. The second derivative here is also e to the x."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it, so the first derivative of this, this is pretty straightforward, is e to the x. Second derivative is one of the profound things of the exponential function. The second derivative here is also e to the x. And so if I have, so the second derivative, let me just do it in those same colors. So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to the x. This is absolutely going to be true."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "The second derivative here is also e to the x. And so if I have, so the second derivative, let me just do it in those same colors. So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to the x. This is absolutely going to be true. E to the x plus two to the x is three e to the x. So y two is also a solution to this differential equation. So that's a start."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "We're given a differential equation right over here, cosine of y plus two, this whole thing, times the derivative of y with respect to x, is equal to two x, and we're given that for a particular solution, when x is equal to one, y of one is equal to zero. And we're asked, what is x when y is equal to pi? So the first thing I like to look at when I see a differential equation is, is it separable? Can I get all the y's and dy's on one side, and can I get all the x's and dx's on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well then you get cosine of y plus two times dy is equal to two x times dx. So just like that, I've been able to, all I did is I multiplied both sides of this times dx."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "Can I get all the y's and dy's on one side, and can I get all the x's and dx's on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well then you get cosine of y plus two times dy is equal to two x times dx. So just like that, I've been able to, all I did is I multiplied both sides of this times dx. And I was able to separate the y's and the dy's from the x's and the dx's. And now I can integrate both sides. So if I integrate both sides, what am I going to get?"}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "So just like that, I've been able to, all I did is I multiplied both sides of this times dx. And I was able to separate the y's and the dy's from the x's and the dx's. And now I can integrate both sides. So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y is sine of y. And then the antiderivative of two with respect to y is two y. And that is going to be equal to, well the antiderivative of two x with respect to x is x squared."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y is sine of y. And then the antiderivative of two with respect to y is two y. And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't forget that we could say a different constant on either side, but it serves our purpose just to say plus c on one side. And so this is a general solution to the separable differential equation. And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't forget that we could say a different constant on either side, but it serves our purpose just to say plus c on one side. And so this is a general solution to the separable differential equation. And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero. So let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one. So sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero zero, two times zero zero, all of that's just going to be zero. So we get zero is equal to one plus c, or c is equal to negative one."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero. So let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one. So sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero zero, two times zero zero, all of that's just going to be zero. So we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here. Sine of y plus two y is equal to x squared, and our constant is negative one, so minus one."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "So we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here. Sine of y plus two y is equal to x squared, and our constant is negative one, so minus one. And now what is x when y is equal to pi? So sine of pi plus two times pi is equal to x squared minus one. See sine of pi is equal to zero, and so we get, see we can add one to both sides, and we get two pi plus one is equal to x squared, or we could say that x is equal to the plus or minus square root of two pi plus one."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "Before we move on past the method of undetermined coefficients, I want to make an interesting and actually a useful point. Let's say that I had the following non-homogenous differential equation. Second derivative of y minus 3 times the first derivative minus 4y is equal to, now this is where it gets interesting, 3e to the 2x plus 2 sine of x plus 4x squared. So you might say, wow, this is a tremendously complicated problem. I have the three types of functions I've been exposed to. I would have so many undetermined coefficients, it would get really unwieldy. And this is where you need to make a simplifying realization."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "So you might say, wow, this is a tremendously complicated problem. I have the three types of functions I've been exposed to. I would have so many undetermined coefficients, it would get really unwieldy. And this is where you need to make a simplifying realization. We know the three particular solutions to the following differential equations. We know the solution to second derivative minus 3 times the first derivative minus 4y. Well, this is a homogenous, right?"}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "And this is where you need to make a simplifying realization. We know the three particular solutions to the following differential equations. We know the solution to second derivative minus 3 times the first derivative minus 4y. Well, this is a homogenous, right? And we know that the solution to the homogenous equation, we did this a bunch of times, is c1 e to the 4x plus c2 e to the minus x. We know the solution to, and I'll switch colors just for variety, y prime prime minus 3y prime minus 4y is equal to just this alone, 3e to the 2x. And we saw that that solution, that particular solution there, y particular, was minus 1 half e to the 2x."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "Well, this is a homogenous, right? And we know that the solution to the homogenous equation, we did this a bunch of times, is c1 e to the 4x plus c2 e to the minus x. We know the solution to, and I'll switch colors just for variety, y prime prime minus 3y prime minus 4y is equal to just this alone, 3e to the 2x. And we saw that that solution, that particular solution there, y particular, was minus 1 half e to the 2x. And we did that using undetermined coefficients. We did that a couple of videos ago. And then, let me just write this out a couple of times."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "And we saw that that solution, that particular solution there, y particular, was minus 1 half e to the 2x. And we did that using undetermined coefficients. We did that a couple of videos ago. And then, let me just write this out a couple of times. We know the solution to this one as well. This was another particular solution we found. I think it was two videos ago."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "And then, let me just write this out a couple of times. We know the solution to this one as well. This was another particular solution we found. I think it was two videos ago. And we found that the particular solution in this case, and this was a fairly hairy problem, was minus 5 over 17x plus 3 over 17. Sorry. The particular solution was minus 5 over 17 sine of x plus 3 over 17 cosine of x."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "I think it was two videos ago. And we found that the particular solution in this case, and this was a fairly hairy problem, was minus 5 over 17x plus 3 over 17. Sorry. The particular solution was minus 5 over 17 sine of x plus 3 over 17 cosine of x. And then finally, this last polynomial, we could call it. We know the solution when that was just the right-hand side. That was this equation."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "The particular solution was minus 5 over 17 sine of x plus 3 over 17 cosine of x. And then finally, this last polynomial, we could call it. We know the solution when that was just the right-hand side. That was this equation. And there, we figured out, and this was in the last video, we figured out that the particular solution in this case was minus x squared plus 3 halves x minus 13 over 8. So we know the particular solution when 0 is on the right-hand side. We know it when just 3 e to the 2x is on the right-hand side."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "That was this equation. And there, we figured out, and this was in the last video, we figured out that the particular solution in this case was minus x squared plus 3 halves x minus 13 over 8. So we know the particular solution when 0 is on the right-hand side. We know it when just 3 e to the 2x is on the right-hand side. We know it just when 2 sine of x is on the right-hand side. And we know it just when 4x squared is on the right-hand side. So if we want, first of all, the particular solution to this non-homogeneous equation, we could just take the sum of the three particular solutions."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "We know it when just 3 e to the 2x is on the right-hand side. We know it just when 2 sine of x is on the right-hand side. And we know it just when 4x squared is on the right-hand side. So if we want, first of all, the particular solution to this non-homogeneous equation, we could just take the sum of the three particular solutions. And that makes sense, right? Because one of the particular solutions, like this one, when you put it on the left-hand side, it'll just equal this term. This particular solution, when you put it on the left-hand side, will equal this term."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "So if we want, first of all, the particular solution to this non-homogeneous equation, we could just take the sum of the three particular solutions. And that makes sense, right? Because one of the particular solutions, like this one, when you put it on the left-hand side, it'll just equal this term. This particular solution, when you put it on the left-hand side, will equal this term. And finally, this particular solution, when you put it on the left-hand side, will equal the 4x squared. And then you could add this homogenous solution to that. You put it on this side, and you'll get 0."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "This particular solution, when you put it on the left-hand side, will equal this term. And finally, this particular solution, when you put it on the left-hand side, will equal the 4x squared. And then you could add this homogenous solution to that. You put it on this side, and you'll get 0. So it won't change the right-hand side. And then you will have the most general solution, because you have these two constants that you can solve for depending on your initial condition. So the solution to this seemingly hairy differential equation is really just the sum of these four solutions."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "You put it on this side, and you'll get 0. So it won't change the right-hand side. And then you will have the most general solution, because you have these two constants that you can solve for depending on your initial condition. So the solution to this seemingly hairy differential equation is really just the sum of these four solutions. And let me clean up some space, because I want everything to be on the board at the same time. So the solution is going to be, well, I don't want that to be deleted. The solution is going to be, I'll do it in baby blue, it's going to be the solution to the homogenous c1 e to the 4x plus c2 e to the minus x minus 1 half e to the 2x."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "So the solution to this seemingly hairy differential equation is really just the sum of these four solutions. And let me clean up some space, because I want everything to be on the board at the same time. So the solution is going to be, well, I don't want that to be deleted. The solution is going to be, I'll do it in baby blue, it's going to be the solution to the homogenous c1 e to the 4x plus c2 e to the minus x minus 1 half e to the 2x. And I'll continue this line. Minus 5 over 17 sine of x plus 3 over 17 cosine of x minus x squared plus 3 halves x minus 13 over 8. And it seems daunting."}, {"video_title": "Undetermined coefficients 4 Second order differential equations Khan Academy.mp3", "Sentence": "The solution is going to be, I'll do it in baby blue, it's going to be the solution to the homogenous c1 e to the 4x plus c2 e to the minus x minus 1 half e to the 2x. And I'll continue this line. Minus 5 over 17 sine of x plus 3 over 17 cosine of x minus x squared plus 3 halves x minus 13 over 8. And it seems daunting. When you saw this, it probably looked daunting. This solution, if I told you this was a solution and you didn't know how to do undetermined coefficients, you're like, oh, I would never be able to figure out something like that. But the important realization is that you just have to find the particular solutions for each of these terms, and then sum them up, and then add them to the general solution for the homogenous equation, if this was a 0 on the right hand side, and then you get the general solution for this fairly intimidating looking second order linear non-homogenous differential equation with constant coefficients."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "For one of the first times, a mathematician's actually named something similar to what it's actually doing. You're actually convoluting the functions. And in this video, I'm not going to dive into the intuition of the convolution. Because a convolution, well, there's a lot of different ways you can look at it. It has a lot of different applications. And if you become an engineer, really of any kind, you're going to see the convolution in kind of a discrete form and a continuous form and a bunch of different ways. But in this video, I just want to make you comfortable with the idea of a convolution, especially in the context of taking Laplace transforms."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Because a convolution, well, there's a lot of different ways you can look at it. It has a lot of different applications. And if you become an engineer, really of any kind, you're going to see the convolution in kind of a discrete form and a continuous form and a bunch of different ways. But in this video, I just want to make you comfortable with the idea of a convolution, especially in the context of taking Laplace transforms. So the convolution theorem. Actually, before I even go to the convolution theorem, let me define what a convolution is. So let's say that I have some function f of t. So if I convolute f with g. So this means that I'm going to take the convolution of f and g, and this is going to be a function of t. And so far, nothing I've written should make any sense to you, because I haven't defined what this means."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But in this video, I just want to make you comfortable with the idea of a convolution, especially in the context of taking Laplace transforms. So the convolution theorem. Actually, before I even go to the convolution theorem, let me define what a convolution is. So let's say that I have some function f of t. So if I convolute f with g. So this means that I'm going to take the convolution of f and g, and this is going to be a function of t. And so far, nothing I've written should make any sense to you, because I haven't defined what this means. This is like those SAT problems where they say, like, a triangle b means a plus b over 3 while you're standing on one leg or something like that. So I need to define this in some similar way. So let me undo this silliness that I just wrote there."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's say that I have some function f of t. So if I convolute f with g. So this means that I'm going to take the convolution of f and g, and this is going to be a function of t. And so far, nothing I've written should make any sense to you, because I haven't defined what this means. This is like those SAT problems where they say, like, a triangle b means a plus b over 3 while you're standing on one leg or something like that. So I need to define this in some similar way. So let me undo this silliness that I just wrote there. The definition of the convolution, we're going to do it over a, well, there's several definitions you'll see, but the definition we're going to use in this context, there's actually one other definition you'll see in the continuous case, is the integral from 0 to t of f of t minus tau times g of tau d tau. Now this might seem like a very bizarro thing to do when you're like, Sal, how do I even compute one of these things? And to kind of give you that comfort, let's actually compute a convolution."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let me undo this silliness that I just wrote there. The definition of the convolution, we're going to do it over a, well, there's several definitions you'll see, but the definition we're going to use in this context, there's actually one other definition you'll see in the continuous case, is the integral from 0 to t of f of t minus tau times g of tau d tau. Now this might seem like a very bizarro thing to do when you're like, Sal, how do I even compute one of these things? And to kind of give you that comfort, let's actually compute a convolution. Let's say that, and it's actually hard to find some functions that are very easy to analytically compute, and you're going to find that we're going to go into a lot of trig identities to actually compute this. But if I say that f of t, if I define f of t to be equal to the sine of t, and I define cosine of t, let me do it in orange, or I define g of t to be equal to the cosine of t. Now let's convolute the two functions. So the convolution of f with g, and this is going to be a function of t, it equals this."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And to kind of give you that comfort, let's actually compute a convolution. Let's say that, and it's actually hard to find some functions that are very easy to analytically compute, and you're going to find that we're going to go into a lot of trig identities to actually compute this. But if I say that f of t, if I define f of t to be equal to the sine of t, and I define cosine of t, let me do it in orange, or I define g of t to be equal to the cosine of t. Now let's convolute the two functions. So the convolution of f with g, and this is going to be a function of t, it equals this. I'm just going to show you how to apply this integral. So it equals the integral, I'll do it in purple, the integral from 0 to t of f of t minus tau. This is my f of t, so it's going to be sine of t minus tau times g of tau."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So the convolution of f with g, and this is going to be a function of t, it equals this. I'm just going to show you how to apply this integral. So it equals the integral, I'll do it in purple, the integral from 0 to t of f of t minus tau. This is my f of t, so it's going to be sine of t minus tau times g of tau. Well, this is my g of t, so g of tau is cosine of tau. So that's the integral. And now to evaluate it, we're going to have to break out some trigonometry."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is my f of t, so it's going to be sine of t minus tau times g of tau. Well, this is my g of t, so g of tau is cosine of tau. So that's the integral. And now to evaluate it, we're going to have to break out some trigonometry. So let's do that. This almost is just a very good trigonometry and integration review. So let's evaluate this."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And now to evaluate it, we're going to have to break out some trigonometry. So let's do that. This almost is just a very good trigonometry and integration review. So let's evaluate this. But I want to evaluate this in this video, because I want to show you that this isn't just some abstract thing, that you can actually evaluate these functions. So the first thing I want to do, I mean, I don't know what the antiderivative of this is. It's tempting."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's evaluate this. But I want to evaluate this in this video, because I want to show you that this isn't just some abstract thing, that you can actually evaluate these functions. So the first thing I want to do, I mean, I don't know what the antiderivative of this is. It's tempting. You see a sine and a cosine, maybe they're the derivatives of each other, but this is a sine of t minus tau. So let me rewrite that sine of t minus tau, and we'll just use the trig identity, that the sine of t minus tau is just equal to the sine of t times the cosine of tau minus the sine of tau times the cosine of t. And actually, I just made a video where I go through all of these trig identities really just to review them for myself and actually to make a video in better quality on them as well. So if we make this substitution, this you'll find it on the inside cover of any trigonometry or calculus book, you get the convolution of f and g is equal to, I'll just write that, f star g, I'll just write it with that, is equal to the integral from 0 to t of, instead of sine of t minus tau, I'm going to write this thing right there."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It's tempting. You see a sine and a cosine, maybe they're the derivatives of each other, but this is a sine of t minus tau. So let me rewrite that sine of t minus tau, and we'll just use the trig identity, that the sine of t minus tau is just equal to the sine of t times the cosine of tau minus the sine of tau times the cosine of t. And actually, I just made a video where I go through all of these trig identities really just to review them for myself and actually to make a video in better quality on them as well. So if we make this substitution, this you'll find it on the inside cover of any trigonometry or calculus book, you get the convolution of f and g is equal to, I'll just write that, f star g, I'll just write it with that, is equal to the integral from 0 to t of, instead of sine of t minus tau, I'm going to write this thing right there. So I'm going to write the sine of t times the cosine of tau minus the sine of tau times the cosine of t. And then all of that's times the cosine of tau. I have to be careful with my taus and t's. And let's see, t and tau and tau and t, everything's working so far."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if we make this substitution, this you'll find it on the inside cover of any trigonometry or calculus book, you get the convolution of f and g is equal to, I'll just write that, f star g, I'll just write it with that, is equal to the integral from 0 to t of, instead of sine of t minus tau, I'm going to write this thing right there. So I'm going to write the sine of t times the cosine of tau minus the sine of tau times the cosine of t. And then all of that's times the cosine of tau. I have to be careful with my taus and t's. And let's see, t and tau and tau and t, everything's working so far. So let's see, so then that's dt. I have to be very careful here. Now let's distribute this cosine of tau out."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And let's see, t and tau and tau and t, everything's working so far. So let's see, so then that's dt. I have to be very careful here. Now let's distribute this cosine of tau out. And what do we get? We get this is equal to, so f convoluted with g, I guess we'll call it f star g, it's equal to the integral from 0 to t of sine of t times cosine of tau times cosine of tau. I'm just distributing this cosine of tau."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now let's distribute this cosine of tau out. And what do we get? We get this is equal to, so f convoluted with g, I guess we'll call it f star g, it's equal to the integral from 0 to t of sine of t times cosine of tau times cosine of tau. I'm just distributing this cosine of tau. So it's cosine squared of tau. And then minus, let's rewrite the cosine of t first. And I'm doing that because we're integrating with respect to tau."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'm just distributing this cosine of tau. So it's cosine squared of tau. And then minus, let's rewrite the cosine of t first. And I'm doing that because we're integrating with respect to tau. So I'm just going to write my cosine of t first. So cosine of t times sine of tau times the cosine of tau, d tau. And now since we're taking the integral of really two things subtracting from each other, let's just turn this into two separate integrals."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And I'm doing that because we're integrating with respect to tau. So I'm just going to write my cosine of t first. So cosine of t times sine of tau times the cosine of tau, d tau. And now since we're taking the integral of really two things subtracting from each other, let's just turn this into two separate integrals. So this is equal to the integral from 0 to t of sine of t times the cosine squared of tau, d tau, minus the integral from 0 to t of cosine of t times sine of tau, cosine of tau, d tau. Now what can we do? Well to simplify it more, remember we're integrating with respect to, let me be careful here, we're integrating with respect to tau."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And now since we're taking the integral of really two things subtracting from each other, let's just turn this into two separate integrals. So this is equal to the integral from 0 to t of sine of t times the cosine squared of tau, d tau, minus the integral from 0 to t of cosine of t times sine of tau, cosine of tau, d tau. Now what can we do? Well to simplify it more, remember we're integrating with respect to, let me be careful here, we're integrating with respect to tau. I wrote a t there. We're integrating with respect to tau. So all of these, this cosine of t right here, that's a constant, the sine of t is a constant."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well to simplify it more, remember we're integrating with respect to, let me be careful here, we're integrating with respect to tau. I wrote a t there. We're integrating with respect to tau. So all of these, this cosine of t right here, that's a constant, the sine of t is a constant. For all I know, t could be equal to 5. It doesn't matter that one of the boundaries of our integration is also a t. That t would be a 5. In which case, these are all just constants."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So all of these, this cosine of t right here, that's a constant, the sine of t is a constant. For all I know, t could be equal to 5. It doesn't matter that one of the boundaries of our integration is also a t. That t would be a 5. In which case, these are all just constants. We're integrating only with respect to the tau. So if cosine of 5, that's a constant. We can take it out of the integral."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "In which case, these are all just constants. We're integrating only with respect to the tau. So if cosine of 5, that's a constant. We can take it out of the integral. So this is equal to sine of t times the integral from 0 to t of cosine squared of tau, d tau, and then minus cosine of t, that's just a constant, I'm bringing it out, times the integral from 0 to t of sine of tau, cosine of tau, d tau. Now this antiderivative is pretty straightforward. You could do u substitution."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We can take it out of the integral. So this is equal to sine of t times the integral from 0 to t of cosine squared of tau, d tau, and then minus cosine of t, that's just a constant, I'm bringing it out, times the integral from 0 to t of sine of tau, cosine of tau, d tau. Now this antiderivative is pretty straightforward. You could do u substitution. And let me do it here. Instead of doing it in our head, this is a complicated problem, so we don't want to skip steps. If we set u is equal to sine of tau, then du d tau is equal to the cosine of tau, just the derivative of sine."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You could do u substitution. And let me do it here. Instead of doing it in our head, this is a complicated problem, so we don't want to skip steps. If we set u is equal to sine of tau, then du d tau is equal to the cosine of tau, just the derivative of sine. Or we could write that du is equal to cosine of tau d tau. And then of course, we'll undo the substitution before we evaluate the endpoints here. But this one's a little bit more of a conundrum."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If we set u is equal to sine of tau, then du d tau is equal to the cosine of tau, just the derivative of sine. Or we could write that du is equal to cosine of tau d tau. And then of course, we'll undo the substitution before we evaluate the endpoints here. But this one's a little bit more of a conundrum. I don't know how to take the antiderivative of cosine squared of tau. It's not obvious what that is. So to do this, we're going to break out some more trigonometric identities."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But this one's a little bit more of a conundrum. I don't know how to take the antiderivative of cosine squared of tau. It's not obvious what that is. So to do this, we're going to break out some more trigonometric identities. And in the video I just recorded, it might not be the last video in the playlist, I showed that the cosine squared of tau, I'm just using tau as an example, is equal to 1 half times 1 plus the cosine of 2 tau. And once again, this is just a trig identity that you'll find really in the inside cover of probably your calculus book. So we can make this substitution here."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So to do this, we're going to break out some more trigonometric identities. And in the video I just recorded, it might not be the last video in the playlist, I showed that the cosine squared of tau, I'm just using tau as an example, is equal to 1 half times 1 plus the cosine of 2 tau. And once again, this is just a trig identity that you'll find really in the inside cover of probably your calculus book. So we can make this substitution here. And then let's see what our integrals become. So this first one over here, let me just write it here. We get sine of t times the integral from 0 to t of this thing here."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we can make this substitution here. And then let's see what our integrals become. So this first one over here, let me just write it here. We get sine of t times the integral from 0 to t of this thing here. I could just take the 1 half out. So let me just, to keep things simple, so I'll put the 1 half out here, that's this 1 half. So 1 plus cosine of 2 tau, and all of that is d tau."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We get sine of t times the integral from 0 to t of this thing here. I could just take the 1 half out. So let me just, to keep things simple, so I'll put the 1 half out here, that's this 1 half. So 1 plus cosine of 2 tau, and all of that is d tau. That's this integral right there. And then we have this integral right here. Minus cosine of t times the integral from, let me be very clear, this is tau is equal to 0."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So 1 plus cosine of 2 tau, and all of that is d tau. That's this integral right there. And then we have this integral right here. Minus cosine of t times the integral from, let me be very clear, this is tau is equal to 0. So this is tau is equal to 0. To tau is equal to t. And then this thing right here, I did some u substitution. If u is equal to sine of t, then this becomes u."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Minus cosine of t times the integral from, let me be very clear, this is tau is equal to 0. So this is tau is equal to 0. To tau is equal to t. And then this thing right here, I did some u substitution. If u is equal to sine of t, then this becomes u. And we showed that du is equal to cosine, sorry, u is equal to sine of tau. And then we showed that du is equal to cosine tau d tau. So this thing right here is equal to du."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If u is equal to sine of t, then this becomes u. And we showed that du is equal to cosine, sorry, u is equal to sine of tau. And then we showed that du is equal to cosine tau d tau. So this thing right here is equal to du. So it's u du. And let's see if we can do anything useful now. So this integral right here, the antiderivative of this is pretty straightforward."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this thing right here is equal to du. So it's u du. And let's see if we can do anything useful now. So this integral right here, the antiderivative of this is pretty straightforward. So what are we going to get? Let me write this outside part. So we have 1 half times the sine of t. And now let me take the antiderivative of this."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this integral right here, the antiderivative of this is pretty straightforward. So what are we going to get? Let me write this outside part. So we have 1 half times the sine of t. And now let me take the antiderivative of this. This is going to be tau plus, the antiderivative of this is going to be 1 half sine of 2 tau. I mean, we could have done a u substitution. We could have said u is equal to 2 tau and all of that."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we have 1 half times the sine of t. And now let me take the antiderivative of this. This is going to be tau plus, the antiderivative of this is going to be 1 half sine of 2 tau. I mean, we could have done a u substitution. We could have said u is equal to 2 tau and all of that. But I think you could do that from recognition. And if you don't believe me, you just have to take the derivative of this. 1 half sine of 2 tau is the derivative of this."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We could have said u is equal to 2 tau and all of that. But I think you could do that from recognition. And if you don't believe me, you just have to take the derivative of this. 1 half sine of 2 tau is the derivative of this. You multiply, you take the derivative of the inside. So that's 2. So the 2 and the 1 half cancel out."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "1 half sine of 2 tau is the derivative of this. You multiply, you take the derivative of the inside. So that's 2. So the 2 and the 1 half cancel out. And then the derivative of the outside. So cosine of 2 tau. And you're going to evaluate that from 0 to t. And then we have minus cosine of t. And then we're going to have, we take the antiderivative of this."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So the 2 and the 1 half cancel out. And then the derivative of the outside. So cosine of 2 tau. And you're going to evaluate that from 0 to t. And then we have minus cosine of t. And then we're going to have, we take the antiderivative of this. Let me do this on the side. So the integral of u du, that's trivially easy. That's 1 half u squared."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And you're going to evaluate that from 0 to t. And then we have minus cosine of t. And then we're going to have, we take the antiderivative of this. Let me do this on the side. So the integral of u du, that's trivially easy. That's 1 half u squared. Now, that's 1 half u squared. But what was u to begin with? It was sine of tau."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That's 1 half u squared. Now, that's 1 half u squared. But what was u to begin with? It was sine of tau. So the antiderivative of this thing right here is 1 half u squared, but u is sine of tau. So it's going to be 1 half u, which is sine of tau squared. And we're going to evaluate that from 0 to t. And we didn't even have to do all this u substitution the way I often do it in my head."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It was sine of tau. So the antiderivative of this thing right here is 1 half u squared, but u is sine of tau. So it's going to be 1 half u, which is sine of tau squared. And we're going to evaluate that from 0 to t. And we didn't even have to do all this u substitution the way I often do it in my head. I see the sine of tau, cosine of tau. If I have a function and I have its derivative, I can treat that function just like as if I had an x there. So it would be sine squared of tau over 2, which is exactly what we have there."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we're going to evaluate that from 0 to t. And we didn't even have to do all this u substitution the way I often do it in my head. I see the sine of tau, cosine of tau. If I have a function and I have its derivative, I can treat that function just like as if I had an x there. So it would be sine squared of tau over 2, which is exactly what we have there. So it looks like we're in the home stretch. We're taking the convolution of sine of t with cosine of t. And so we get 1 half sine of t. Now, if I evaluate this thing at t, what do I get? I get t plus 1 half sine of 2t."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So it would be sine squared of tau over 2, which is exactly what we have there. So it looks like we're in the home stretch. We're taking the convolution of sine of t with cosine of t. And so we get 1 half sine of t. Now, if I evaluate this thing at t, what do I get? I get t plus 1 half sine of 2t. That's when I evaluated at t. And then from that, I need to subtract and evaluate it at 0. So minus 0 minus 1 half sine of 2 times 0, which is just sine of 0. So this part right here, this whole thing right there, what does that simplify to?"}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I get t plus 1 half sine of 2t. That's when I evaluated at t. And then from that, I need to subtract and evaluate it at 0. So minus 0 minus 1 half sine of 2 times 0, which is just sine of 0. So this part right here, this whole thing right there, what does that simplify to? Well, this is 0. Sine of 0 is 0, so this is all 0. So this first integral right there simplifies to 1 half sine of t times t plus 1 half sine of 2t."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this part right here, this whole thing right there, what does that simplify to? Well, this is 0. Sine of 0 is 0, so this is all 0. So this first integral right there simplifies to 1 half sine of t times t plus 1 half sine of 2t. Now, what does this one simplify to over here? Well, this one over here, you have minus cosine of t. And we're going to evaluate this whole thing at t. So you get 1 half sine squared of t minus 1 half sine of 0 squared, which is a 0. So that's just minus 0."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this first integral right there simplifies to 1 half sine of t times t plus 1 half sine of 2t. Now, what does this one simplify to over here? Well, this one over here, you have minus cosine of t. And we're going to evaluate this whole thing at t. So you get 1 half sine squared of t minus 1 half sine of 0 squared, which is a 0. So that's just minus 0. So so far, everything that we have written simplifies to, let me multiply it all out. So I have 1 half t sine of t, right? I'm just multiplying those out."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So that's just minus 0. So so far, everything that we have written simplifies to, let me multiply it all out. So I have 1 half t sine of t, right? I'm just multiplying those out. Plus 1 fourth sine of 2t. And then I have over here, I have minus 1 half sine squared t times cosine of t, right? I just took the minus cosine t and multiplied it through here, and I got that."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'm just multiplying those out. Plus 1 fourth sine of 2t. And then I have over here, I have minus 1 half sine squared t times cosine of t, right? I just took the minus cosine t and multiplied it through here, and I got that. Now, this is a valid answer, but I suspect that we can simplify this more, maybe using some more trigonometric identities. And this guy right there looks ripe to simplify. We know that the sine of 2t, another trig identity you'll find on the inside cover of any of your books, is equal to the sine, is 2 times the sine of t times the cosine of t. So if you substitute that there, what does our whole expression equal?"}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I just took the minus cosine t and multiplied it through here, and I got that. Now, this is a valid answer, but I suspect that we can simplify this more, maybe using some more trigonometric identities. And this guy right there looks ripe to simplify. We know that the sine of 2t, another trig identity you'll find on the inside cover of any of your books, is equal to the sine, is 2 times the sine of t times the cosine of t. So if you substitute that there, what does our whole expression equal? You get this first term. Let me scroll down a little bit. You get 1 half t times the sine of t plus 1 fourth sine of t times this thing in here."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We know that the sine of 2t, another trig identity you'll find on the inside cover of any of your books, is equal to the sine, is 2 times the sine of t times the cosine of t. So if you substitute that there, what does our whole expression equal? You get this first term. Let me scroll down a little bit. You get 1 half t times the sine of t plus 1 fourth sine of t times this thing in here. So times 2 sine of t cosine of t. Just a trig identity. Nothing more than that. And then finally I have this minus 1 half sine squared t cosine of t. No one ever said this was going to be easy, but hopefully it's instructive on some level."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You get 1 half t times the sine of t plus 1 fourth sine of t times this thing in here. So times 2 sine of t cosine of t. Just a trig identity. Nothing more than that. And then finally I have this minus 1 half sine squared t cosine of t. No one ever said this was going to be easy, but hopefully it's instructive on some level. At least it shows you that you didn't memorize your trig identities for nothing. So what is this simple? So let me rewrite the whole thing."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then finally I have this minus 1 half sine squared t cosine of t. No one ever said this was going to be easy, but hopefully it's instructive on some level. At least it shows you that you didn't memorize your trig identities for nothing. So what is this simple? So let me rewrite the whole thing. Or let me just rewrite this part. So this is equal to 1 fourth. Now I have, well let me see, 1 fourth times 2."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let me rewrite the whole thing. Or let me just rewrite this part. So this is equal to 1 fourth. Now I have, well let me see, 1 fourth times 2. So it's really 1 fourth times 2 is 1 half. And then sine squared of t. This sine times this sine, sine squared of t. Cosine of t. And then this one over here is minus 1 half sine squared of t cosine of t. And luckily for us, or lucky for us, these cancel out. And of course we had this guy out in the front."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now I have, well let me see, 1 fourth times 2. So it's really 1 fourth times 2 is 1 half. And then sine squared of t. This sine times this sine, sine squared of t. Cosine of t. And then this one over here is minus 1 half sine squared of t cosine of t. And luckily for us, or lucky for us, these cancel out. And of course we had this guy out in the front. We had this 1 half t sine t out in front. Now this guy cancels with this guy. And all we're left with through this whole hairy problem, and this is pretty satisfying, is 1 half t sine of t. So we just showed you that the convolution, if I define, let me write our result."}, {"video_title": "Introduction to the convolution Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And of course we had this guy out in the front. We had this 1 half t sine t out in front. Now this guy cancels with this guy. And all we're left with through this whole hairy problem, and this is pretty satisfying, is 1 half t sine of t. So we just showed you that the convolution, if I define, let me write our result. I feel like writing this in stone because this is so much work. But if we write that f of t is equal to sine of t and g of t is equal to cosine of t, I just showed you that the convolution of f with g, which is a function of t, which is defined as the integral from 0 to t of f of t minus tau times g of tau d tau, which was equal to, and I'll switch colors here, which was equal to the integral from 0 to t of sine of t minus tau times g of tau d tau, that all of this mess, all of this convolution, it all equals, and this is pretty satisfying, it all equals 1 half t sine of t. And the whole reason why I went through all this mess and kind of bringing out the neurons that had the trig identities memorized or having to reprove them or whatever else, is to just show you that this convolution, it is convoluted and it seems a little bit bizarre, but you really can take the convolutions of actual functions and get an actual answer. So the convolution of sine of t with cosine of t is 1 half t sine of t. So hopefully you have a little of intuition, well not intuition, but you at least have a little bit of hands on understanding of how the convolution can be calculated."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And it doesn't happen exactly like this, but it can be approximated by the unit step function. Similarly, sometimes you have nothing happening, you have nothing happening for a long period of time, nothing happens for a long period of time, and then whack, something hits you really hard and then goes away, and then nothing happens for a very long period of time, and you'll learn this in the future. You can kind of view this as an impulse, and we'll talk about unit impulse functions and all of that. So wouldn't it be neat if we had some type of function that could model this type of behavior? And our ideal function, what would happen is is that nothing happens until we get to some point, and then bam, it would get infinitely strong, but maybe it has a finite area, and then it would go back to zero and then go like that. So it'd be infinitely high right at, let's say zero right there, and then it continues. And let's say that the area under this, I mean, it becomes very, to call this a function is actually kind of pushing it, and this is beyond the math of this video, but we'll call it a function in this video."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So wouldn't it be neat if we had some type of function that could model this type of behavior? And our ideal function, what would happen is is that nothing happens until we get to some point, and then bam, it would get infinitely strong, but maybe it has a finite area, and then it would go back to zero and then go like that. So it'd be infinitely high right at, let's say zero right there, and then it continues. And let's say that the area under this, I mean, it becomes very, to call this a function is actually kind of pushing it, and this is beyond the math of this video, but we'll call it a function in this video. But what we can, you say, well, how do you even, what good is this function for? How can you even manipulate it? And I'm gonna make one more definition of this function."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And let's say that the area under this, I mean, it becomes very, to call this a function is actually kind of pushing it, and this is beyond the math of this video, but we'll call it a function in this video. But what we can, you say, well, how do you even, what good is this function for? How can you even manipulate it? And I'm gonna make one more definition of this function. So what I just do here, let's say we call this function, we represent it by the delta, and that's what we do represent this function by. It's called the Dirac delta function. And we'll just informally say, look, when it's in infinity, it pops up to infinity when x is equal to zero, and it's zero everywhere else when x is not equal to zero."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And I'm gonna make one more definition of this function. So what I just do here, let's say we call this function, we represent it by the delta, and that's what we do represent this function by. It's called the Dirac delta function. And we'll just informally say, look, when it's in infinity, it pops up to infinity when x is equal to zero, and it's zero everywhere else when x is not equal to zero. And you say, how do I deal with that? How do I take the integral of that? And to help you with that, I'm gonna make a definition."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we'll just informally say, look, when it's in infinity, it pops up to infinity when x is equal to zero, and it's zero everywhere else when x is not equal to zero. And you say, how do I deal with that? How do I take the integral of that? And to help you with that, I'm gonna make a definition. I'm gonna tell you what the integral of this is. This is part of the definition of the function. I'm gonna tell you that if I were to take the integral of this function from minus infinity to infinity, so essentially over the entire real number line, if I take the integral of this function, I'm defining it."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And to help you with that, I'm gonna make a definition. I'm gonna tell you what the integral of this is. This is part of the definition of the function. I'm gonna tell you that if I were to take the integral of this function from minus infinity to infinity, so essentially over the entire real number line, if I take the integral of this function, I'm defining it. I'm defining it to be equal to one. I'm defining this. Now, you might say, Sal, you didn't prove it to me."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'm gonna tell you that if I were to take the integral of this function from minus infinity to infinity, so essentially over the entire real number line, if I take the integral of this function, I'm defining it. I'm defining it to be equal to one. I'm defining this. Now, you might say, Sal, you didn't prove it to me. No, I'm defining it. I'm telling you that this delta of x is a function such that its integral is one. So it has this infinitely narrow base that goes infinitely high, and the area under this, I'm telling you, is of area one."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, you might say, Sal, you didn't prove it to me. No, I'm defining it. I'm telling you that this delta of x is a function such that its integral is one. So it has this infinitely narrow base that goes infinitely high, and the area under this, I'm telling you, is of area one. And you're like, hey, Sal, that's a crazy function. I want a little bit better understanding of how someone can construct a function like this. So let's see if we can satisfy that a little bit more."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So it has this infinitely narrow base that goes infinitely high, and the area under this, I'm telling you, is of area one. And you're like, hey, Sal, that's a crazy function. I want a little bit better understanding of how someone can construct a function like this. So let's see if we can satisfy that a little bit more. But then once that's satisfied, then we're gonna start taking the Laplace transform of this, and then we'll start manipulating it and whatnot. Let's see if we can complete this delta right here. Let's say that I constructed another function."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's see if we can satisfy that a little bit more. But then once that's satisfied, then we're gonna start taking the Laplace transform of this, and then we'll start manipulating it and whatnot. Let's see if we can complete this delta right here. Let's say that I constructed another function. Let's call it d sub tau. And this is all just to satisfy this craving for maybe a better intuition for how this Dirac delta function can be constructed. And let's say my d sub tau of, well, let me put it in, it's a function of t, because I wanna, everything we're doing in Laplace transform world, everything's been a function of t. So let's say that it equals, let's say that it equals one over two tau, and you'll see why I'm picking these numbers the way I am."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's say that I constructed another function. Let's call it d sub tau. And this is all just to satisfy this craving for maybe a better intuition for how this Dirac delta function can be constructed. And let's say my d sub tau of, well, let me put it in, it's a function of t, because I wanna, everything we're doing in Laplace transform world, everything's been a function of t. So let's say that it equals, let's say that it equals one over two tau, and you'll see why I'm picking these numbers the way I am. One over two tau when t is less than tau and greater than minus tau. And let's say it's zero everywhere else. Everywhere else."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And let's say my d sub tau of, well, let me put it in, it's a function of t, because I wanna, everything we're doing in Laplace transform world, everything's been a function of t. So let's say that it equals, let's say that it equals one over two tau, and you'll see why I'm picking these numbers the way I am. One over two tau when t is less than tau and greater than minus tau. And let's say it's zero everywhere else. Everywhere else. So this type of, this equation, this is more reasonable. This will actually look like a combination of unit step functions, and we can actually define it as a combination of unit step functions. So if I draw, that's my x-axis."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Everywhere else. So this type of, this equation, this is more reasonable. This will actually look like a combination of unit step functions, and we can actually define it as a combination of unit step functions. So if I draw, that's my x-axis. That's my x-axis. And then if I put my y-axis right here, that's my y-axis, y, and this is, sorry, this is the t-axis. To get out of that habit, this is a t-axis, and I mean, we could call it, well, we could call it the y-axis or the f of t-axis or whatever we wanna call it."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if I draw, that's my x-axis. That's my x-axis. And then if I put my y-axis right here, that's my y-axis, y, and this is, sorry, this is the t-axis. To get out of that habit, this is a t-axis, and I mean, we could call it, well, we could call it the y-axis or the f of t-axis or whatever we wanna call it. That's the dependent variable. So what's going to happen here? It's gonna be zero everywhere until we get to minus t, and then at minus t, we're gonna jump up to some level."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "To get out of that habit, this is a t-axis, and I mean, we could call it, well, we could call it the y-axis or the f of t-axis or whatever we wanna call it. That's the dependent variable. So what's going to happen here? It's gonna be zero everywhere until we get to minus t, and then at minus t, we're gonna jump up to some level. So let me put that point here. So this is minus tau, and this is plus tau. Right?"}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It's gonna be zero everywhere until we get to minus t, and then at minus t, we're gonna jump up to some level. So let me put that point here. So this is minus tau, and this is plus tau. Right? Minus tau and plus tau. So it's gonna be zero everywhere, and then at minus tau, we jump to this level, and then we stay constant at that level until we get to plus tau. And that level, I'm saying, is one over two tau."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Right? Minus tau and plus tau. So it's gonna be zero everywhere, and then at minus tau, we jump to this level, and then we stay constant at that level until we get to plus tau. And that level, I'm saying, is one over two tau. So this point right here on the dependent axis, this is one over two tau. So why did I construct this function this way? Well, let's think about it."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And that level, I'm saying, is one over two tau. So this point right here on the dependent axis, this is one over two tau. So why did I construct this function this way? Well, let's think about it. What happens if I take the integral, let me write a nicer integral sign. If I took the integral from minus infinity to infinity of d sub tau of t, dt. What is this going to be equal to?"}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, let's think about it. What happens if I take the integral, let me write a nicer integral sign. If I took the integral from minus infinity to infinity of d sub tau of t, dt. What is this going to be equal to? Well, if you just, I mean, if the integral's just the area under this curve, this is a pretty straightforward thing to calculate. Right? You just look at this, and you say, well, this is, first of all, it's zero everywhere else."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What is this going to be equal to? Well, if you just, I mean, if the integral's just the area under this curve, this is a pretty straightforward thing to calculate. Right? You just look at this, and you say, well, this is, first of all, it's zero everywhere else. It's zero everywhere else, and it's only the area right here. I mean, I could write this, I could rewrite this integral as the integral from minus tau to tau. We don't care, infinity, minus infinity or positive infinity, because there's no area under any of those points of one over two tau d tau."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You just look at this, and you say, well, this is, first of all, it's zero everywhere else. It's zero everywhere else, and it's only the area right here. I mean, I could write this, I could rewrite this integral as the integral from minus tau to tau. We don't care, infinity, minus infinity or positive infinity, because there's no area under any of those points of one over two tau d tau. We could write it, dt, sorry, one over two tau dt. So we could write it this way too, right? Because we can just take the boundaries from here to here, because we get nothing, the whole other, whether t goes to positive infinity or minus infinity, and then over that boundary, the function is a constant, one over two tau."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We don't care, infinity, minus infinity or positive infinity, because there's no area under any of those points of one over two tau d tau. We could write it, dt, sorry, one over two tau dt. So we could write it this way too, right? Because we can just take the boundaries from here to here, because we get nothing, the whole other, whether t goes to positive infinity or minus infinity, and then over that boundary, the function is a constant, one over two tau. So we could just take this integral, and either way we evaluate it, we don't even have to know calculus to know what this integral's gonna evaluate to. This is just the area under this, the area under this, which is just the base. What's the base?"}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Because we can just take the boundaries from here to here, because we get nothing, the whole other, whether t goes to positive infinity or minus infinity, and then over that boundary, the function is a constant, one over two tau. So we could just take this integral, and either way we evaluate it, we don't even have to know calculus to know what this integral's gonna evaluate to. This is just the area under this, the area under this, which is just the base. What's the base? The base is two tau, right? The base is two tau. You have one tau here and then another tau there."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What's the base? The base is two tau, right? The base is two tau. You have one tau here and then another tau there. So it's equal to two tau times your height. And your height, I just said, is one over two tau. One over two tau."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You have one tau here and then another tau there. So it's equal to two tau times your height. And your height, I just said, is one over two tau. One over two tau. So your area for this function, or for this integral, is going to be one. You could evaluate this, you could get this is going to be equal to, you take the antiderivative of one over two tau, you get, I'll do this just to satiate your curiosity, t over two tau, and you have to evaluate this from minus tau to tau. You put tau in there, you get tau over two tau, and then minus minus tau over two tau."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "One over two tau. So your area for this function, or for this integral, is going to be one. You could evaluate this, you could get this is going to be equal to, you take the antiderivative of one over two tau, you get, I'll do this just to satiate your curiosity, t over two tau, and you have to evaluate this from minus tau to tau. You put tau in there, you get tau over two tau, and then minus minus tau over two tau. And then you get tau plus tau over two tau, that's two tau over two tau, which is equal to one. Maybe I'm beating a dead horse. I think you're satisfied that the area under this is going to be one, regardless of what tau was."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You put tau in there, you get tau over two tau, and then minus minus tau over two tau. And then you get tau plus tau over two tau, that's two tau over two tau, which is equal to one. Maybe I'm beating a dead horse. I think you're satisfied that the area under this is going to be one, regardless of what tau was. I kept this abstract. Now, if I take smaller and smaller values of tau, what's going to happen? If my new tau is going to be, let's say here, let's say my new tau is going to be there."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I think you're satisfied that the area under this is going to be one, regardless of what tau was. I kept this abstract. Now, if I take smaller and smaller values of tau, what's going to happen? If my new tau is going to be, let's say here, let's say my new tau is going to be there. I'm just going to pick my new tau there. Then my one over two tau, tau is now a smaller number, so when it's in the denominator, my one over two tau is going to be something like this. It's going to be something like this."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If my new tau is going to be, let's say here, let's say my new tau is going to be there. I'm just going to pick my new tau there. Then my one over two tau, tau is now a smaller number, so when it's in the denominator, my one over two tau is going to be something like this. It's going to be something like this. I'm just saying, if I pick smaller and smaller tau, so if I pick an even smaller tau than that, then my height is going to have to be higher. My one over two tau is going to have to even be higher than that. I think you see where I'm going with this."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It's going to be something like this. I'm just saying, if I pick smaller and smaller tau, so if I pick an even smaller tau than that, then my height is going to have to be higher. My one over two tau is going to have to even be higher than that. I think you see where I'm going with this. What happens as the limit as tau approaches zero? What is the limit as tau approaches zero of my little d sub tau function? What's the limit of this?"}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I think you see where I'm going with this. What happens as the limit as tau approaches zero? What is the limit as tau approaches zero of my little d sub tau function? What's the limit of this? Well, these things are going to go infinitely close to zero, but this is the limit. They're never going to be quite at zero. Your height here is going to go infinitely high, but the whole time, I said no matter what my tau is, because it was defined in very arbitrary terms, my area is always going to be one."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What's the limit of this? Well, these things are going to go infinitely close to zero, but this is the limit. They're never going to be quite at zero. Your height here is going to go infinitely high, but the whole time, I said no matter what my tau is, because it was defined in very arbitrary terms, my area is always going to be one. So you're going to end up with your Dirac delta function. Let me write it, I was going to write it in x again. Your Dirac delta function is a function of t. Because of this, if you ask, what's the limit as tau approaches zero of the integral from minus infinity to infinity of d sub tau of t, dt?"}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Your height here is going to go infinitely high, but the whole time, I said no matter what my tau is, because it was defined in very arbitrary terms, my area is always going to be one. So you're going to end up with your Dirac delta function. Let me write it, I was going to write it in x again. Your Dirac delta function is a function of t. Because of this, if you ask, what's the limit as tau approaches zero of the integral from minus infinity to infinity of d sub tau of t, dt? Well, this should still be one, right? Because this thing right here, this evaluates to one. So as you take the limit as tau approaches zero, and I'm being very generous with my definitions of limits and whatnot."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Your Dirac delta function is a function of t. Because of this, if you ask, what's the limit as tau approaches zero of the integral from minus infinity to infinity of d sub tau of t, dt? Well, this should still be one, right? Because this thing right here, this evaluates to one. So as you take the limit as tau approaches zero, and I'm being very generous with my definitions of limits and whatnot. I'm not being very rigorous, but I think you can kind of understand the intuition where I'm going. This is going to be equal to one. And so by the same, I guess, intuitive argument, you could say that the limit as from minus infinity to infinity of our Dirac delta function of t, dt, is also going to be one."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So as you take the limit as tau approaches zero, and I'm being very generous with my definitions of limits and whatnot. I'm not being very rigorous, but I think you can kind of understand the intuition where I'm going. This is going to be equal to one. And so by the same, I guess, intuitive argument, you could say that the limit as from minus infinity to infinity of our Dirac delta function of t, dt, is also going to be one. And likewise, Dirac delta function, I mean, this thing pops up to infinity at t is equal to zero, right? This thing, if I were to draw my x-axis, my x-axis like that, and then right at t equals zero, my Dirac delta function pops up like that. And you normally draw it like that, and you normally draw it so it goes up to one to kind of depict its area, but you actually put an arrow there, and so this is your Dirac delta function."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And so by the same, I guess, intuitive argument, you could say that the limit as from minus infinity to infinity of our Dirac delta function of t, dt, is also going to be one. And likewise, Dirac delta function, I mean, this thing pops up to infinity at t is equal to zero, right? This thing, if I were to draw my x-axis, my x-axis like that, and then right at t equals zero, my Dirac delta function pops up like that. And you normally draw it like that, and you normally draw it so it goes up to one to kind of depict its area, but you actually put an arrow there, and so this is your Dirac delta function. But what happens if you want to shift it? What happens if you want to shift it? What would, how would I represent, how would I represent my, let's say I want to do t minus three."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And you normally draw it like that, and you normally draw it so it goes up to one to kind of depict its area, but you actually put an arrow there, and so this is your Dirac delta function. But what happens if you want to shift it? What happens if you want to shift it? What would, how would I represent, how would I represent my, let's say I want to do t minus three. What would the graph of this be? Well, this would just be shifting it to the right by three. For example, when t equals three, this will become the Dirac delta of zero."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What would, how would I represent, how would I represent my, let's say I want to do t minus three. What would the graph of this be? Well, this would just be shifting it to the right by three. For example, when t equals three, this will become the Dirac delta of zero. So this graph, this graph will just look like this. This will be my x-axis. Let's say that this is my y-axis."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "For example, when t equals three, this will become the Dirac delta of zero. So this graph, this graph will just look like this. This will be my x-axis. Let's say that this is my y-axis. Let me just make that one. And let me just draw some points here. So it's one, two, three."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's say that this is my y-axis. Let me just make that one. And let me just draw some points here. So it's one, two, three. That's t is equal to three. Did I say that was x-axis? That's my t-axis."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So it's one, two, three. That's t is equal to three. Did I say that was x-axis? That's my t-axis. So this is t equal to three. And what I'm gonna do here, it's the Dirac delta function, it's gonna be zero everywhere. Everywhere."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That's my t-axis. So this is t equal to three. And what I'm gonna do here, it's the Dirac delta function, it's gonna be zero everywhere. Everywhere. Zero everywhere, but then right at three, it goes infinitely high. And obviously we don't have enough paper to draw an infinitely high spike right there. So what we do is we draw an arrow."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Everywhere. Zero everywhere, but then right at three, it goes infinitely high. And obviously we don't have enough paper to draw an infinitely high spike right there. So what we do is we draw an arrow. We draw an arrow there. And the arrow, we usually draw the magnitude of the area under that spike. So we do it like this."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So what we do is we draw an arrow. We draw an arrow there. And the arrow, we usually draw the magnitude of the area under that spike. So we do it like this. And let me be clear, this is not telling me that the function just goes to one and then spikes back down. This tells me that the area under the function is equal to one. This spike would have to be infinitely high to have any area, considering it has no, an infinitely small base."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we do it like this. And let me be clear, this is not telling me that the function just goes to one and then spikes back down. This tells me that the area under the function is equal to one. This spike would have to be infinitely high to have any area, considering it has no, an infinitely small base. So the area under this impulse function or under this Dirac delta function. Now this one right here is t minus three. But your area under this is still going to be one."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This spike would have to be infinitely high to have any area, considering it has no, an infinitely small base. So the area under this impulse function or under this Dirac delta function. Now this one right here is t minus three. But your area under this is still going to be one. And that's why I made the arrow go to one. If I were to, let's say I wanted to graph, let me do it in another color. Let's say I wanted to graph two times the Dirac delta of t minus two."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But your area under this is still going to be one. And that's why I made the arrow go to one. If I were to, let's say I wanted to graph, let me do it in another color. Let's say I wanted to graph two times the Dirac delta of t minus two. How would I graph this? Well, I would go to t minus two. When t is equal to two, you get the Dirac delta of zero."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's say I wanted to graph two times the Dirac delta of t minus two. How would I graph this? Well, I would go to t minus two. When t is equal to two, you get the Dirac delta of zero. So that's where you'd have your spike. And we're multiplying it by two. So you would do a spike twice as high like this."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "When t is equal to two, you get the Dirac delta of zero. So that's where you'd have your spike. And we're multiplying it by two. So you would do a spike twice as high like this. Now both of these go to infinity, but this goes twice as high to infinity. I know this is all being a little ridiculous now, but the idea here is that the area under this curve should be twice the area under this curve. And that's why we make the arrow go to two, to say that the area under this arrow is two."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So you would do a spike twice as high like this. Now both of these go to infinity, but this goes twice as high to infinity. I know this is all being a little ridiculous now, but the idea here is that the area under this curve should be twice the area under this curve. And that's why we make the arrow go to two, to say that the area under this arrow is two. The spike would have to go infinitely high. So this is all a little abstract, but this is a useful way to model things that are kind of very jarring. That all of a sudden, obviously nothing actually behaves like this, but there are a lot of phenomena in physics or the real world that kind of have this spiky behavior."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And that's why we make the arrow go to two, to say that the area under this arrow is two. The spike would have to go infinitely high. So this is all a little abstract, but this is a useful way to model things that are kind of very jarring. That all of a sudden, obviously nothing actually behaves like this, but there are a lot of phenomena in physics or the real world that kind of have this spiky behavior. Instead of trying to say, oh, what does that spike exactly look like? We say, hey, that's a Dirac delta function. And we'll dictate its impulse by something like this."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That all of a sudden, obviously nothing actually behaves like this, but there are a lot of phenomena in physics or the real world that kind of have this spiky behavior. Instead of trying to say, oh, what does that spike exactly look like? We say, hey, that's a Dirac delta function. And we'll dictate its impulse by something like this. And just to give you a little bit of motivation behind this, and I was gonna go here in the last video, but then I kind of decided not to. But I'm just gonna show it, because I've been doing a lot of differential equations and I've been giving you no motivation for how this applies in the real world. But you can imagine, if I have just a, say, a wall, and then I have a spring attached to some mass, some mass right there."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we'll dictate its impulse by something like this. And just to give you a little bit of motivation behind this, and I was gonna go here in the last video, but then I kind of decided not to. But I'm just gonna show it, because I've been doing a lot of differential equations and I've been giving you no motivation for how this applies in the real world. But you can imagine, if I have just a, say, a wall, and then I have a spring attached to some mass, some mass right there. And let's say that this is the natural state of the spring. So the spring would wanna be here. So it's been stretched a distance y from its kind of natural, where it wants to go."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But you can imagine, if I have just a, say, a wall, and then I have a spring attached to some mass, some mass right there. And let's say that this is the natural state of the spring. So the spring would wanna be here. So it's been stretched a distance y from its kind of natural, where it wants to go. And let's say I have some external force, external force right here. Let's say I have some external force right here on the spring. And of course, let's say it's ice on ice, there's no friction in all of this."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So it's been stretched a distance y from its kind of natural, where it wants to go. And let's say I have some external force, external force right here. Let's say I have some external force right here on the spring. And of course, let's say it's ice on ice, there's no friction in all of this. I just wanna show you that I can represent this behavior of this system with a differential equation. And actually things like the unit step function and the Dirac delta function actually start to become useful in this type of an environment. So we know that, we know that F is equal to mass times acceleration."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And of course, let's say it's ice on ice, there's no friction in all of this. I just wanna show you that I can represent this behavior of this system with a differential equation. And actually things like the unit step function and the Dirac delta function actually start to become useful in this type of an environment. So we know that, we know that F is equal to mass times acceleration. That's the basic physics right there. Now, what are all of the forces on this mass right here? Well, you have this force right here, and we'll say this in the positive rightward direction."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we know that, we know that F is equal to mass times acceleration. That's the basic physics right there. Now, what are all of the forces on this mass right here? Well, you have this force right here, and we'll say this in the positive rightward direction. So it's that force. Then you have a minus force from the spring, right? The force from the spring is Hooke's law."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, you have this force right here, and we'll say this in the positive rightward direction. So it's that force. Then you have a minus force from the spring, right? The force from the spring is Hooke's law. It's proportional to how far it's been stretched from its kind of natural point. So its force in that direction is going to be ky, or you could call it minus ky, because it's going in the opposite direction of what we've already said is a positive direction. So the net forces on this is F minus ky, and that's equal to the mass of our, it's equal to the mass of our object times its acceleration."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "The force from the spring is Hooke's law. It's proportional to how far it's been stretched from its kind of natural point. So its force in that direction is going to be ky, or you could call it minus ky, because it's going in the opposite direction of what we've already said is a positive direction. So the net forces on this is F minus ky, and that's equal to the mass of our, it's equal to the mass of our object times its acceleration. Now, what's its acceleration? If its position is y, so if y is equal to position, if we take the derivative of y with respect to t, y prime, which we could also say dy dt, this is going to be its velocity. And then if we take the derivative of that, y prime prime, which is equal to d squared y with respect to dt squared, this is equal to acceleration."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So the net forces on this is F minus ky, and that's equal to the mass of our, it's equal to the mass of our object times its acceleration. Now, what's its acceleration? If its position is y, so if y is equal to position, if we take the derivative of y with respect to t, y prime, which we could also say dy dt, this is going to be its velocity. And then if we take the derivative of that, y prime prime, which is equal to d squared y with respect to dt squared, this is equal to acceleration. Acceleration. So instead of writing a, we could write y prime prime. Y prime prime."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then if we take the derivative of that, y prime prime, which is equal to d squared y with respect to dt squared, this is equal to acceleration. Acceleration. So instead of writing a, we could write y prime prime. Y prime prime. And so if we just put this on the other side of the equation, what do we get? We get the force, this force, not just this force, this is just F equals ma, but this force is equal to the mass of our object times the acceleration of the object plus whatever the spring constant is for this spring, plus k times our position, times y. So if you had no outside force, if this was zero, you'd have a homogeneous differential equation, and in that case, the spring would just start moving on its own."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Y prime prime. And so if we just put this on the other side of the equation, what do we get? We get the force, this force, not just this force, this is just F equals ma, but this force is equal to the mass of our object times the acceleration of the object plus whatever the spring constant is for this spring, plus k times our position, times y. So if you had no outside force, if this was zero, you'd have a homogeneous differential equation, and in that case, the spring would just start moving on its own. But now this F, all of a sudden, kind of the non-homogeneous term, it's what the outside force you're applying to this mass. So if this outside force was some type of Dirac delta function, so let's say it's t minus two is equal to our mass times y prime prime plus our spring constant times y. This is saying that at time is equal to two seconds, we're just gonna jar this thing to the right, and it's going to have an, and I'll talk more about it, it's gonna have an impulse of two."}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if you had no outside force, if this was zero, you'd have a homogeneous differential equation, and in that case, the spring would just start moving on its own. But now this F, all of a sudden, kind of the non-homogeneous term, it's what the outside force you're applying to this mass. So if this outside force was some type of Dirac delta function, so let's say it's t minus two is equal to our mass times y prime prime plus our spring constant times y. This is saying that at time is equal to two seconds, we're just gonna jar this thing to the right, and it's going to have an, and I'll talk more about it, it's gonna have an impulse of two. Its force times time is going to be, or its impulse is gonna have a one. So it's, and I don't wanna get too much into the physics here, but its impulse or its change in momentum is going to be of magnitude one, depending on what our units are. But anyway, I just wanted to take that slight diversion because you might be saying, Sal is introducing me to these weird exotic functions, what are they ever going to be good for?"}, {"video_title": "Dirac delta function Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is saying that at time is equal to two seconds, we're just gonna jar this thing to the right, and it's going to have an, and I'll talk more about it, it's gonna have an impulse of two. Its force times time is going to be, or its impulse is gonna have a one. So it's, and I don't wanna get too much into the physics here, but its impulse or its change in momentum is going to be of magnitude one, depending on what our units are. But anyway, I just wanted to take that slight diversion because you might be saying, Sal is introducing me to these weird exotic functions, what are they ever going to be good for? But this is good for the idea of at some time, you just jar this thing by some magnitude and then let go. And you do it kind of infinitely fast, but you do it with some, enough to change the momentum of this in a well-defined way. Anyway, in the next video, we'll continue with the Dirac delta function, we'll figure out its Laplace transform and see what it does to the Laplace transforms of other functions."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And when you actually try to solve this differential equation you try to find an n of t that satisfies this, we found that an exponential would work. An exponential satisfies this differential equation. And it would look like this visually. It would look like this visually, where you're starting at a population of n naught. This is the time axis, this is the population axis. And as time increases, population increases exponentially. Now we said there's an issue there."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "It would look like this visually, where you're starting at a population of n naught. This is the time axis, this is the population axis. And as time increases, population increases exponentially. Now we said there's an issue there. What if Thomas Malthus is right? That the environment can't support, let's say that the environment can't support, let me do this in a new color, let's say that the environment really can't support more than k, more than a population, more than a population of k. Then clearly the population can't just go and go right through the ceiling. They're not going to be able to have food or water or resources or whatever it might be."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Now we said there's an issue there. What if Thomas Malthus is right? That the environment can't support, let's say that the environment can't support, let me do this in a new color, let's say that the environment really can't support more than k, more than a population, more than a population of k. Then clearly the population can't just go and go right through the ceiling. They're not going to be able to have food or water or resources or whatever it might be. They might generate too much pollution. Who knows what it might be. And so this first stab at modeling population doesn't quite do the trick, especially if you are kind of in Malthus' camp."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "They're not going to be able to have food or water or resources or whatever it might be. They might generate too much pollution. Who knows what it might be. And so this first stab at modeling population doesn't quite do the trick, especially if you are kind of in Malthus' camp. And that's where P.F., and once again, I'm sure I'm mispronouncing the name, Verhulst, who is going to come into the picture, because he read Malthus' work and said, well, yeah, I think I can do a pretty good job of modeling the type of behavior that Malthus is talking about. And he says, you know, what we really want, what we really want is something, let me write it, so the rate, let's try to model, let's set up another differential equation. Let's set up another differential equation."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And so this first stab at modeling population doesn't quite do the trick, especially if you are kind of in Malthus' camp. And that's where P.F., and once again, I'm sure I'm mispronouncing the name, Verhulst, who is going to come into the picture, because he read Malthus' work and said, well, yeah, I think I can do a pretty good job of modeling the type of behavior that Malthus is talking about. And he says, you know, what we really want, what we really want is something, let me write it, so the rate, let's try to model, let's set up another differential equation. Let's set up another differential equation. And now let's say, okay, instead of, you know, if N is substantially smaller than what the environment can support, then yeah, that makes sense to have exponential growth. That makes sense to have exponential growth. But maybe we can dampen this, or maybe we can bring this growth to zero as N approaches, as N approaches K. And so how can we actually modify this?"}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Let's set up another differential equation. And now let's say, okay, instead of, you know, if N is substantially smaller than what the environment can support, then yeah, that makes sense to have exponential growth. That makes sense to have exponential growth. But maybe we can dampen this, or maybe we can bring this growth to zero as N approaches, as N approaches K. And so how can we actually modify this? Maybe we can multiply it by something that for when N is small, when N is much smaller than K, this term right over here is going to be close to one. And when N is close to K, this term is close to zero. So let me write that."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "But maybe we can dampen this, or maybe we can bring this growth to zero as N approaches, as N approaches K. And so how can we actually modify this? Maybe we can multiply it by something that for when N is small, when N is much smaller than K, this term right over here is going to be close to one. And when N is close to K, this term is close to zero. So let me write that. When N, so these are our goals for this term right over here. When N is much smaller, so much smaller, much smaller than K, so now the population is not constrained at all. People can have babies, and those babies can be fed, and then they can have babies, et cetera, et cetera."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So let me write that. When N, so these are our goals for this term right over here. When N is much smaller, so much smaller, much smaller than K, so now the population is not constrained at all. People can have babies, and those babies can be fed, and then they can have babies, et cetera, et cetera. Then this thing should be close to one. And so then you have essentially our old model. But then as N approaches K, when N, as N approaches K, then this thing should approach, then this term, this term or this expression, should approach zero."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "People can have babies, and those babies can be fed, and then they can have babies, et cetera, et cetera. Then this thing should be close to one. And so then you have essentially our old model. But then as N approaches K, when N, as N approaches K, then this thing should approach, then this term, this term or this expression, should approach zero. And what that does is as N approaches the natural limit, the ceiling to population, then no matter what this is doing, if this thing is approaching zero, that's going to make the actual rate of growth approach zero. So food is going to be more scarce. It's going to be harder to find things."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "But then as N approaches K, when N, as N approaches K, then this thing should approach, then this term, this term or this expression, should approach zero. And what that does is as N approaches the natural limit, the ceiling to population, then no matter what this is doing, if this thing is approaching zero, that's going to make the actual rate of growth approach zero. So food is going to be more scarce. It's going to be harder to find things. And so what can I construct here dealing with N and K that will have these properties? And for fun, you might actually want to pause the video and see if you can construct a fairly simple algebraic statement using N and K, and maybe the number one if you find the need, for an expression that has these properties. Well, let's see."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "It's going to be harder to find things. And so what can I construct here dealing with N and K that will have these properties? And for fun, you might actually want to pause the video and see if you can construct a fairly simple algebraic statement using N and K, and maybe the number one if you find the need, for an expression that has these properties. Well, let's see. What if we start with a one, we start with a one, and we subtract N over K? We subtract N over, my K is in pink, over K. Does this have those properties? Well, yeah, sure it does."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Well, let's see. What if we start with a one, we start with a one, and we subtract N over K? We subtract N over, my K is in pink, over K. Does this have those properties? Well, yeah, sure it does. When N is really small, or I should say when it's a small fraction of K, then one mind, this is going to be a small fraction, then this whole thing is going to be pretty close to one. It's going to be a little bit less than one. And when N approaches K, as N gets closer and closer and closer to K, then this thing right over here is going to approach one, which means this whole expression is going to approach zero, which is exactly what we wanted."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Well, yeah, sure it does. When N is really small, or I should say when it's a small fraction of K, then one mind, this is going to be a small fraction, then this whole thing is going to be pretty close to one. It's going to be a little bit less than one. And when N approaches K, as N gets closer and closer and closer to K, then this thing right over here is going to approach one, which means this whole expression is going to approach zero, which is exactly what we wanted. And this thing right over here is actually, and this is used in tons of applications, not just in population modeling, but that's kind of one of its first reasons or motivations. This differential equation right over here is actually quite famous. It's called the logistic differential equation."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And when N approaches K, as N gets closer and closer and closer to K, then this thing right over here is going to approach one, which means this whole expression is going to approach zero, which is exactly what we wanted. And this thing right over here is actually, and this is used in tons of applications, not just in population modeling, but that's kind of one of its first reasons or motivations. This differential equation right over here is actually quite famous. It's called the logistic differential equation. Logistic differential equation. Logistic differential equation. And in the next video, we're actually going to solve this."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "It's called the logistic differential equation. Logistic differential equation. Logistic differential equation. And in the next video, we're actually going to solve this. And this is a separable differential equation. You can actually solve it just using standard techniques of integration. It's a little bit hairier than this one, so we're going to work through it together."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And in the next video, we're actually going to solve this. And this is a separable differential equation. You can actually solve it just using standard techniques of integration. It's a little bit hairier than this one, so we're going to work through it together. And we're going to look at the solution. The solution to the logistic differential equation is a logistic function, which, once again, is really, essentially models population in this way. But before we actually solve for it, let's just try to interpret this differential equation and think about what the shape of this function might look like."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "It's a little bit hairier than this one, so we're going to work through it together. And we're going to look at the solution. The solution to the logistic differential equation is a logistic function, which, once again, is really, essentially models population in this way. But before we actually solve for it, let's just try to interpret this differential equation and think about what the shape of this function might look like. And to do that, actually, let me, it's nice to see the faces. So let me draw some axes here. Let me draw some axes here."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "But before we actually solve for it, let's just try to interpret this differential equation and think about what the shape of this function might look like. And to do that, actually, let me, it's nice to see the faces. So let me draw some axes here. Let me draw some axes here. So that's my time axis. That is my population axis. Let me scroll up a little bit, because sometimes the subtitles show up around here, and then people can't see what's going on."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Let me draw some axes here. So that's my time axis. That is my population axis. Let me scroll up a little bit, because sometimes the subtitles show up around here, and then people can't see what's going on. So let's think about a couple of permutations, a couple of situations. So if our initial, if our initial, if our n at time equals zero, remember, n is a function of t. If at time equals zero, n is equal to zero. So if n is equal to zero, then this term is going to be zero, and then your rate of change is going to be zero, and so you're not going to add any population."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Let me scroll up a little bit, because sometimes the subtitles show up around here, and then people can't see what's going on. So let's think about a couple of permutations, a couple of situations. So if our initial, if our initial, if our n at time equals zero, remember, n is a function of t. If at time equals zero, n is equal to zero. So if n is equal to zero, then this term is going to be zero, and then your rate of change is going to be zero, and so you're not going to add any population. And that's good, because if your population is zero, how are you going to actually be able to add population? There's no one there to have children. So there's actually one constant solution to this differential equation, which is just n of t, that is n of t, is equal to zero."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So if n is equal to zero, then this term is going to be zero, and then your rate of change is going to be zero, and so you're not going to add any population. And that's good, because if your population is zero, how are you going to actually be able to add population? There's no one there to have children. So there's actually one constant solution to this differential equation, which is just n of t, that is n of t, is equal to zero. And that's neat that this satisfies the logistic differential equation. Hey, if your population starts at zero, if n sub naught is zero, then you're just going to be at zero forever. Well, that's actually what would happen in real life."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So there's actually one constant solution to this differential equation, which is just n of t, that is n of t, is equal to zero. And that's neat that this satisfies the logistic differential equation. Hey, if your population starts at zero, if n sub naught is zero, then you're just going to be at zero forever. Well, that's actually what would happen in real life. There's no one there to have kids. Now let's think about another situation. What if our population, what if n naught is equal to k?"}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Well, that's actually what would happen in real life. There's no one there to have kids. Now let's think about another situation. What if our population, what if n naught is equal to k? What happens if n naught is equal to, what happens if n naught is equal, so that's k right over there. What happens if at time equals zero, this is our population? Well, if n is equal to k, then this is one minus one, then this thing is zero, and so our rate of population change is going to be zero."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "What if our population, what if n naught is equal to k? What happens if n naught is equal to, what happens if n naught is equal, so that's k right over there. What happens if at time equals zero, this is our population? Well, if n is equal to k, then this is one minus one, then this thing is zero, and so our rate of population change is going to be zero. So essentially, if my population is zero, then after a little bit of time, my population is still the same k. If my rate of change of population is zero, that means my population is staying constant. And so my population is just going to stay there at k. And that's actually believable. Malthus would actually probably say that you're going to have, maybe it grows a little bit beyond the capacity of the environment, and then you have some flood, or some hurricane, or some famine, and it goes around."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Well, if n is equal to k, then this is one minus one, then this thing is zero, and so our rate of population change is going to be zero. So essentially, if my population is zero, then after a little bit of time, my population is still the same k. If my rate of change of population is zero, that means my population is staying constant. And so my population is just going to stay there at k. And that's actually believable. Malthus would actually probably say that you're going to have, maybe it grows a little bit beyond the capacity of the environment, and then you have some flood, or some hurricane, or some famine, and it goes around. But for our purposes, you can never model anything perfectly, for our purposes, that's pretty good. You're at the limit of what the environment can handle, so you just kind of stay there. So that's actually another constant solution, that n of t, if it starts, and now you can kind of appreciate why initial conditions are important."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Malthus would actually probably say that you're going to have, maybe it grows a little bit beyond the capacity of the environment, and then you have some flood, or some hurricane, or some famine, and it goes around. But for our purposes, you can never model anything perfectly, for our purposes, that's pretty good. You're at the limit of what the environment can handle, so you just kind of stay there. So that's actually another constant solution, that n of t, if it starts, and now you can kind of appreciate why initial conditions are important. If you start at zero, you're going to stay at zero. If you start at k, you're going to stay at k. So that is n of t just stays at k. But now let's think of a more interesting scenario. Let's assume an initial population that's someplace between zero and k. So this is going to be, I'm going to assume initial population that is someplace, it's greater than zero, so there are people to actually have children, and it is less than k, so we aren't fully maxing out the environment, or the land, or the food, or the water, or whatever it might be."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So that's actually another constant solution, that n of t, if it starts, and now you can kind of appreciate why initial conditions are important. If you start at zero, you're going to stay at zero. If you start at k, you're going to stay at k. So that is n of t just stays at k. But now let's think of a more interesting scenario. Let's assume an initial population that's someplace between zero and k. So this is going to be, I'm going to assume initial population that is someplace, it's greater than zero, so there are people to actually have children, and it is less than k, so we aren't fully maxing out the environment, or the land, or the food, or the water, or whatever it might be. So what's going to happen? So, and once again, I'm just going to kind of sketch it, and then we're going to actually solve it in the next video. So when n is a lot less than k, when it's a small fraction of k, you're going to, it's going to, you know, this term is going to be the main one that's influencing it, because this is a small fraction of k. I mean, even the way I drew it, it looks like it's about a, I don't know, it looks like it's about a sixth, or a seventh, or an eighth of k. So it's one minus one eighth."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Let's assume an initial population that's someplace between zero and k. So this is going to be, I'm going to assume initial population that is someplace, it's greater than zero, so there are people to actually have children, and it is less than k, so we aren't fully maxing out the environment, or the land, or the food, or the water, or whatever it might be. So what's going to happen? So, and once again, I'm just going to kind of sketch it, and then we're going to actually solve it in the next video. So when n is a lot less than k, when it's a small fraction of k, you're going to, it's going to, you know, this term is going to be the main one that's influencing it, because this is a small fraction of k. I mean, even the way I drew it, it looks like it's about a, I don't know, it looks like it's about a sixth, or a seventh, or an eighth of k. So it's one minus one eighth. So it's seven eighths. It's going to be seven eighths times this. So really, this is what's going to dictate, this is what's going to dictate what our rate of growth is."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So when n is a lot less than k, when it's a small fraction of k, you're going to, it's going to, you know, this term is going to be the main one that's influencing it, because this is a small fraction of k. I mean, even the way I drew it, it looks like it's about a, I don't know, it looks like it's about a sixth, or a seventh, or an eighth of k. So it's one minus one eighth. So it's seven eighths. It's going to be seven eighths times this. So really, this is what's going to dictate, this is what's going to dictate what our rate of growth is. And if this is kind of dictating it, we're kind of looking more of a, well, let's just think of it this way. As the population grows, the rate of change is going to grow. So it's going to look, it's going to look something, it's going to look something like this."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So really, this is what's going to dictate, this is what's going to dictate what our rate of growth is. And if this is kind of dictating it, we're kind of looking more of a, well, let's just think of it this way. As the population grows, the rate of change is going to grow. So it's going to look, it's going to look something, it's going to look something like this. As our population gets larger, our slope is getting higher, and it's getting steeper and steeper. But then as n approaches k, then this thing is going to become, this is going to be close to one minus, close to one, and so this is going to become a very small number. It's going to make this whole thing approach zero."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So it's going to look, it's going to look something, it's going to look something like this. As our population gets larger, our slope is getting higher, and it's getting steeper and steeper. But then as n approaches k, then this thing is going to become, this is going to be close to one minus, close to one, and so this is going to become a very small number. It's going to make this whole thing approach zero. So as n approaches k, the whole thing, the rate of change, is going to flatten out, and we're going to asymptote towards k. And so the solution, the solution to the logistic differential equation should look something like this, depending on what your initial conditions are. If your initial condition is here, maybe it does something like this. If your initial condition's here, maybe it does something like, something like this."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "It's going to make this whole thing approach zero. So as n approaches k, the whole thing, the rate of change, is going to flatten out, and we're going to asymptote towards k. And so the solution, the solution to the logistic differential equation should look something like this, depending on what your initial conditions are. If your initial condition is here, maybe it does something like this. If your initial condition's here, maybe it does something like, something like this. And once again, and this is what's fun about differential equations, before even doing the fancy math, you can kind of get an intuition, just by thinking through the differential equation, of what it is likely, what it is likely to be. Down here, or when n is much smaller than k, its rate of increase is increasing as n increases, and over here, as n gets close to k, its rate of increase is decreasing. Now let's actually, in the next video, actually solve for what n, a solution to this, and see if that confirms what our intuition is."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "At the point negative one comma one, I would draw a short segment of slope blank. And like always, pause this video and see if you can fill out these three blanks. Well, when you're, the short segments that you're trying to draw to construct this slope field, you figure out their slope based on the differential equation. So you're saying when x is equal to negative one and y is equal to one, what is the derivative of y with respect to x? And that's what this differential equation tells us. So for this first case, the derivative of y with respect to x is going to be equal to y, which is one, minus two times x. X is negative one. So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So you're saying when x is equal to negative one and y is equal to one, what is the derivative of y with respect to x? And that's what this differential equation tells us. So for this first case, the derivative of y with respect to x is going to be equal to y, which is one, minus two times x. X is negative one. So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two. So the derivative of y with respect to x at this point is going to be three. So I would draw a short line segment or a short segment of slope three. And we keep going at the point zero comma two."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two. So the derivative of y with respect to x at this point is going to be three. So I would draw a short line segment or a short segment of slope three. And we keep going at the point zero comma two. Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero. Well, that's just going to be two. And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "And we keep going at the point zero comma two. Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero. Well, that's just going to be two. And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two. Two times two, three minus four is equal to four. Three minus four is equal to negative one. And that's all that problem asks us to do."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two. Two times two, three minus four is equal to four. Three minus four is equal to negative one. And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like, I'll try to draw it real fast. So let's see, let me make sure I have space for all of these points here. So that's my coordinate axes."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like, I'll try to draw it real fast. So let's see, let me make sure I have space for all of these points here. So that's my coordinate axes. And I want to get the point zero comma two. So that's zero comma two. Actually, I want to go all the way to two comma three, so let me get some space here."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So that's my coordinate axes. And I want to get the point zero comma two. So that's zero comma two. Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so for this first one, and this exercise isn't asking us to do it, but I'm just making it very clear how we would construct the slope field. So the point negative one comma one, negative one comma one, a short segment of slope three."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so for this first one, and this exercise isn't asking us to do it, but I'm just making it very clear how we would construct the slope field. So the point negative one comma one, negative one comma one, a short segment of slope three. So slope three would look something like that. Then at the point zero comma two, a slope of two. Zero comma two, the slope is going to be two, which looks something like that."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So the point negative one comma one, negative one comma one, a short segment of slope three. So slope three would look something like that. Then at the point zero comma two, a slope of two. Zero comma two, the slope is going to be two, which looks something like that. And then at the point two comma three, at two comma three, a short segment of slope negative one. So two comma three, a segment of slope negative one. It would look something like that."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "It's been over a year since I last did a video in the differential equations playlist, and I thought I would start taking up making a couple of videos, and I think where I left off, I said that I would do a non-homogeneous linear equation using the Laplace transform. So let's do one as a bit of a warmup now that we've had a, or at least I've had a one-year hiatus. Maybe you're watching these continuously, so you're probably more warmed up than I am. So if we have the equation, the second derivative of y plus y is equal to sine of two t, and we're given some initial conditions here. The initial conditions are y of zero is equal to two, and y prime of zero is equal to one. And where we left off, and you probably remember this, you probably recently watched the last video. To solve these, we take the Laplace transforms of all the sides."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So if we have the equation, the second derivative of y plus y is equal to sine of two t, and we're given some initial conditions here. The initial conditions are y of zero is equal to two, and y prime of zero is equal to one. And where we left off, and you probably remember this, you probably recently watched the last video. To solve these, we take the Laplace transforms of all the sides. We solve for the Laplace transform of the function, and then we take the inverse Laplace transform. If that doesn't make sense, then let's just do it in this video, and hopefully the example will clarify all confusion. So in the last video, it was either the last one or the previous one, I showed you that the Laplace transform, the Laplace transform of the second derivative of y is equal to s squared times the Laplace transform of y, and we keep lowering the degree on s, so minus s times y of zero."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "To solve these, we take the Laplace transforms of all the sides. We solve for the Laplace transform of the function, and then we take the inverse Laplace transform. If that doesn't make sense, then let's just do it in this video, and hopefully the example will clarify all confusion. So in the last video, it was either the last one or the previous one, I showed you that the Laplace transform, the Laplace transform of the second derivative of y is equal to s squared times the Laplace transform of y, and we keep lowering the degree on s, so minus s times y of zero. You can kind of think of it as taking the derivative. This is an integral. It's not exactly the antiderivative of this, but the Laplace transform, it is an integral."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So in the last video, it was either the last one or the previous one, I showed you that the Laplace transform, the Laplace transform of the second derivative of y is equal to s squared times the Laplace transform of y, and we keep lowering the degree on s, so minus s times y of zero. You can kind of think of it as taking the derivative. This is an integral. It's not exactly the antiderivative of this, but the Laplace transform, it is an integral. The transform is an integral. So y of zero is kind of a one derivative away from that, and then minus y prime of zero, and then we could also rewrite this, and this is just a purely notational issue. I could write this instead of writing the Laplace transform of y all the time."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "It's not exactly the antiderivative of this, but the Laplace transform, it is an integral. The transform is an integral. So y of zero is kind of a one derivative away from that, and then minus y prime of zero, and then we could also rewrite this, and this is just a purely notational issue. I could write this instead of writing the Laplace transform of y all the time. I could write this as s squared times capital Y of s, because this is going to be a function of s, not a function of y, minus s times y of zero minus y prime y of zero. These are going to be numbers, right? These aren't functions."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "I could write this instead of writing the Laplace transform of y all the time. I could write this as s squared times capital Y of s, because this is going to be a function of s, not a function of y, minus s times y of zero minus y prime y of zero. These are going to be numbers, right? These aren't functions. These are the function evaluated at zero or the function evaluated, or the derivative of the function evaluated at zero, and we know what these values are. Y of zero right here is two, and y prime of zero is one. It was given to us."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "These aren't functions. These are the function evaluated at zero or the function evaluated, or the derivative of the function evaluated at zero, and we know what these values are. Y of zero right here is two, and y prime of zero is one. It was given to us. So if we take the Laplace transforms of both sides of this equation, first we're going to want to take the Laplace transform of this term right there, which we've really just done. The Laplace transform of the second derivative is s squared times Laplace transform of the function, which we write as capital Y of s, minus this, minus 2s. They gave us that initial condition."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "It was given to us. So if we take the Laplace transforms of both sides of this equation, first we're going to want to take the Laplace transform of this term right there, which we've really just done. The Laplace transform of the second derivative is s squared times Laplace transform of the function, which we write as capital Y of s, minus this, minus 2s. They gave us that initial condition. Minus 2s and then minus one, right? This term right here is just one, so minus one. So that's the term right there."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "They gave us that initial condition. Minus 2s and then minus one, right? This term right here is just one, so minus one. So that's the term right there. Then we want to take the Laplace transform of y by itself. So this is just plus y of s, right? The Laplace transform of y, so I'll just rewrite Laplace transform of y. I'm just rewriting it in this notation."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So that's the term right there. Then we want to take the Laplace transform of y by itself. So this is just plus y of s, right? The Laplace transform of y, so I'll just rewrite Laplace transform of y. I'm just rewriting it in this notation. Y of s. It's good to get used to either one. And then we want to take, this is going to be equal to the Laplace transform of sine of 2t. And I showed you in a video last year of what we showed what the Laplace transform of sine of at is, but I'll write it down here just so you remember it."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "The Laplace transform of y, so I'll just rewrite Laplace transform of y. I'm just rewriting it in this notation. Y of s. It's good to get used to either one. And then we want to take, this is going to be equal to the Laplace transform of sine of 2t. And I showed you in a video last year of what we showed what the Laplace transform of sine of at is, but I'll write it down here just so you remember it. Laplace transform of the sine of at is equal to a over s squared plus a squared. Right? a over s squared plus a squared."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "And I showed you in a video last year of what we showed what the Laplace transform of sine of at is, but I'll write it down here just so you remember it. Laplace transform of the sine of at is equal to a over s squared plus a squared. Right? a over s squared plus a squared. So the Laplace transform of sine of 2t, here a is two. This is going to be two over s squared plus four. So if we take the Laplace transform of both sides of this, the right-hand side is going to be two over s squared plus four."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "a over s squared plus a squared. So the Laplace transform of sine of 2t, here a is two. This is going to be two over s squared plus four. So if we take the Laplace transform of both sides of this, the right-hand side is going to be two over s squared plus four. Now what we can do is we can separate out all the y of s terms. And so we can factor, well, I guess we could say factor out their coefficients. So that's a y of s term."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So if we take the Laplace transform of both sides of this, the right-hand side is going to be two over s squared plus four. Now what we can do is we can separate out all the y of s terms. And so we can factor, well, I guess we could say factor out their coefficients. So that's a y of s term. That's a y of s term. And so we could write the left-hand side here as s squared, that's that term, plus one, the coefficient on that term, s squared plus one times y of s. Let me do it in green. So this is a y of s and this is a y of s times y of s. And then we have the non-y of s terms, these two right here."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So that's a y of s term. That's a y of s term. And so we could write the left-hand side here as s squared, that's that term, plus one, the coefficient on that term, s squared plus one times y of s. Let me do it in green. So this is a y of s and this is a y of s times y of s. And then we have the non-y of s terms, these two right here. So minus 2s minus one is equal to two over s squared plus four. We can add 2s plus one to both sides to essentially move this to the right-hand side. And we're left with, we are left with, s squared plus one times y of s, times y of s, is equal to two over s squared plus four plus 2s plus one."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So this is a y of s and this is a y of s times y of s. And then we have the non-y of s terms, these two right here. So minus 2s minus one is equal to two over s squared plus four. We can add 2s plus one to both sides to essentially move this to the right-hand side. And we're left with, we are left with, s squared plus one times y of s, times y of s, is equal to two over s squared plus four plus 2s plus one. Now we can divide both sides of this equation by s squared plus one and we get the Laplace transform of y, y of s, is equal to two, let me switch colors, it's equal to two over s squared plus four, times this thing right here, I'm dividing the whole, both sides of this equation by this term right there. So times s squared plus one, it's in the denominator, so I'm dividing by it, plus 2s plus one, I have to divide both of those terms by the s squared plus one. Divided by s squared plus one, divided by s squared plus one."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "And we're left with, we are left with, s squared plus one times y of s, times y of s, is equal to two over s squared plus four plus 2s plus one. Now we can divide both sides of this equation by s squared plus one and we get the Laplace transform of y, y of s, is equal to two, let me switch colors, it's equal to two over s squared plus four, times this thing right here, I'm dividing the whole, both sides of this equation by this term right there. So times s squared plus one, it's in the denominator, so I'm dividing by it, plus 2s plus one, I have to divide both of those terms by the s squared plus one. Divided by s squared plus one, divided by s squared plus one. Now, in order to be able to take the inverse Laplace transform of this, I need to get it in some type of simple fraction form, these are actually easier to do, but this one's a little bit difficult. I want to do some partial fraction decomposition to break this up into maybe simpler fractions. And since both of these, so what I want to do, I'm going to do a little bit of an aside here."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Divided by s squared plus one, divided by s squared plus one. Now, in order to be able to take the inverse Laplace transform of this, I need to get it in some type of simple fraction form, these are actually easier to do, but this one's a little bit difficult. I want to do some partial fraction decomposition to break this up into maybe simpler fractions. And since both of these, so what I want to do, I'm going to do a little bit of an aside here. And this really is the hardest part of these problems, is the algebra of breaking this thing up. So, since we're going to break this up, I'm going to break this up, so let me write this this way, two over s squared plus four, times s squared plus one. I'm going to break this up into two fractions, this is the partial fraction decomposition, one fraction is s squared plus four, and then the other fraction is s squared plus one."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "And since both of these, so what I want to do, I'm going to do a little bit of an aside here. And this really is the hardest part of these problems, is the algebra of breaking this thing up. So, since we're going to break this up, I'm going to break this up, so let me write this this way, two over s squared plus four, times s squared plus one. I'm going to break this up into two fractions, this is the partial fraction decomposition, one fraction is s squared plus four, and then the other fraction is s squared plus one. And since both of these, both of these denominators are of degree two, the numerators are going to have, they're going to be of degree one, so they're going to be some, let me write it this way, a, this one will be a s plus b, and then this one will be c s plus d. This is just pure algebra here, this is just partial fraction decomposition, I've made a couple of videos on it, and I'm saying that I'm assuming that this expression right here can be broken up into two expressions of this form, and now I need to solve for a, b, c, and d. So let's see how we can do that. So if I were to start with these two and add them up, what do I get? I would have to multiply these times, so my denominator, my common denominator would be this thing again, it would be s squared plus four, times s squared plus one, and now I'm going to have to multiply the a s plus b, a s plus b, times this s squared plus one, times s squared plus one, this has its right now, if these two terms would cancel out, you'll just get this term, but I need to add it to this right here."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "I'm going to break this up into two fractions, this is the partial fraction decomposition, one fraction is s squared plus four, and then the other fraction is s squared plus one. And since both of these, both of these denominators are of degree two, the numerators are going to have, they're going to be of degree one, so they're going to be some, let me write it this way, a, this one will be a s plus b, and then this one will be c s plus d. This is just pure algebra here, this is just partial fraction decomposition, I've made a couple of videos on it, and I'm saying that I'm assuming that this expression right here can be broken up into two expressions of this form, and now I need to solve for a, b, c, and d. So let's see how we can do that. So if I were to start with these two and add them up, what do I get? I would have to multiply these times, so my denominator, my common denominator would be this thing again, it would be s squared plus four, times s squared plus one, and now I'm going to have to multiply the a s plus b, a s plus b, times this s squared plus one, times s squared plus one, this has its right now, if these two terms would cancel out, you'll just get this term, but I need to add it to this right here. So you get plus c s plus d, times this term right here, times s squared plus four, and now let's see what we could do to match up the terms here with this number two right here. So let's multiply all of this out. So a s times s squared is a s to the third, a s times one is plus a s, a s times s squared, so plus b s squared, and then you have b times one is plus b, and then you have c s times s squared, that's c s to the third, and then c s times four, so it's plus four c s, and then let's see, these problems are tiring, and I also have a cold, so this is especially tiring, so I'm going to soldier forward."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "I would have to multiply these times, so my denominator, my common denominator would be this thing again, it would be s squared plus four, times s squared plus one, and now I'm going to have to multiply the a s plus b, a s plus b, times this s squared plus one, times s squared plus one, this has its right now, if these two terms would cancel out, you'll just get this term, but I need to add it to this right here. So you get plus c s plus d, times this term right here, times s squared plus four, and now let's see what we could do to match up the terms here with this number two right here. So let's multiply all of this out. So a s times s squared is a s to the third, a s times one is plus a s, a s times s squared, so plus b s squared, and then you have b times one is plus b, and then you have c s times s squared, that's c s to the third, and then c s times four, so it's plus four c s, and then let's see, these problems are tiring, and I also have a cold, so this is especially tiring, so I'm going to soldier forward. Where was I? So I've multiplied the c's times each of these, I have to multiply the d's, so plus d s squared, that's d times that one, plus d times four, so plus four d, so that's all of them, and I just wrote it this way so I have the common degree terms under each other. So if I were to add the entire numerator, I get, and I'll just switch colors somewhat arbitrarily, I get a plus c times s to the third, plus, let me write the s squared term next, plus b plus d times s squared, and I'll write this s term, plus a plus four c times s, plus b plus four d. This is just the numerator, this is when I just added these two things up, this whole thing up here simplifies to this."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So a s times s squared is a s to the third, a s times one is plus a s, a s times s squared, so plus b s squared, and then you have b times one is plus b, and then you have c s times s squared, that's c s to the third, and then c s times four, so it's plus four c s, and then let's see, these problems are tiring, and I also have a cold, so this is especially tiring, so I'm going to soldier forward. Where was I? So I've multiplied the c's times each of these, I have to multiply the d's, so plus d s squared, that's d times that one, plus d times four, so plus four d, so that's all of them, and I just wrote it this way so I have the common degree terms under each other. So if I were to add the entire numerator, I get, and I'll just switch colors somewhat arbitrarily, I get a plus c times s to the third, plus, let me write the s squared term next, plus b plus d times s squared, and I'll write this s term, plus a plus four c times s, plus b plus four d. This is just the numerator, this is when I just added these two things up, this whole thing up here simplifies to this. I don't know if the word simplifies is appropriate, but it becomes this expression right here, and that's just the numerator. The denominator is still what we had written before, the denominator is still the s squared plus four times the s squared plus one. Of course I have to show that this is a fraction, and this is going to be equal to this thing over here, two over s squared plus four times s squared plus one."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So if I were to add the entire numerator, I get, and I'll just switch colors somewhat arbitrarily, I get a plus c times s to the third, plus, let me write the s squared term next, plus b plus d times s squared, and I'll write this s term, plus a plus four c times s, plus b plus four d. This is just the numerator, this is when I just added these two things up, this whole thing up here simplifies to this. I don't know if the word simplifies is appropriate, but it becomes this expression right here, and that's just the numerator. The denominator is still what we had written before, the denominator is still the s squared plus four times the s squared plus one. Of course I have to show that this is a fraction, and this is going to be equal to this thing over here, two over s squared plus four times s squared plus one. Now, why did I go through this whole mess right here? Well, the reason why I went through it is because we should be able to solve for a, b, c, and d. So let's see, a plus c, this is the coefficient on the s cubed term. Do we see any s cubed terms here?"}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Of course I have to show that this is a fraction, and this is going to be equal to this thing over here, two over s squared plus four times s squared plus one. Now, why did I go through this whole mess right here? Well, the reason why I went through it is because we should be able to solve for a, b, c, and d. So let's see, a plus c, this is the coefficient on the s cubed term. Do we see any s cubed terms here? No, we see no s cubed terms here, so a plus c, let me write this down, a plus c must be equal to zero, because we see nothing here that has an s to the third. b plus d is the coefficient on the s squared term. Do we see any s squared terms here?"}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Do we see any s cubed terms here? No, we see no s cubed terms here, so a plus c, let me write this down, a plus c must be equal to zero, because we see nothing here that has an s to the third. b plus d is the coefficient on the s squared term. Do we see any s squared terms here? No, so b plus d must be equal to zero. a plus four c are the coefficient on the s term. I see no s term over here, so a plus four c must also be equal to zero."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Do we see any s squared terms here? No, so b plus d must be equal to zero. a plus four c are the coefficient on the s term. I see no s term over here, so a plus four c must also be equal to zero. And then finally, we look at just the constant terms, and we do have a constant term on the left-hand side of this equation. We have two, so b plus four d, didn't want to make it that thick, b plus four d must be equal to two. So let's see if we can do any, this seems like these linear equations are pretty easy to solve for."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "I see no s term over here, so a plus four c must also be equal to zero. And then finally, we look at just the constant terms, and we do have a constant term on the left-hand side of this equation. We have two, so b plus four d, didn't want to make it that thick, b plus four d must be equal to two. So let's see if we can do any, this seems like these linear equations are pretty easy to solve for. Let's subtract this from this. So a, or let me subtract the bottom one from the top one. So a minus a, that's zero a, and then c minus four c minus three c is equal to zero, and so you get c is equal to zero."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So let's see if we can do any, this seems like these linear equations are pretty easy to solve for. Let's subtract this from this. So a, or let me subtract the bottom one from the top one. So a minus a, that's zero a, and then c minus four c minus three c is equal to zero, and so you get c is equal to zero. If c is equal to zero, a plus c is equal to zero, a must be equal to zero. Let's do the same thing here. Let's subtract this from that."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So a minus a, that's zero a, and then c minus four c minus three c is equal to zero, and so you get c is equal to zero. If c is equal to zero, a plus c is equal to zero, a must be equal to zero. Let's do the same thing here. Let's subtract this from that. So you get b minus b is zero, and then minus three d, that's just d minus four d, and then zero minus two is equal to minus two, and then you get d is equal to two thirds. Minus two divided by minus three is two thirds, and then this isn't a minus here, I wrote that there later, we said b plus d is equal to zero, so b must be the opposite of d. We could write b is equal to minus d, or b is equal to minus two thirds. Let's remember all of this and go back to our original problem, because we've kind of, so actually let me just be clear."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Let's subtract this from that. So you get b minus b is zero, and then minus three d, that's just d minus four d, and then zero minus two is equal to minus two, and then you get d is equal to two thirds. Minus two divided by minus three is two thirds, and then this isn't a minus here, I wrote that there later, we said b plus d is equal to zero, so b must be the opposite of d. We could write b is equal to minus d, or b is equal to minus two thirds. Let's remember all of this and go back to our original problem, because we've kind of, so actually let me just be clear. So we can rewrite two over s squared plus four times s squared plus one. We can rewrite this as, well a is zero, b is minus two thirds, so this is equal to minus two thirds over s squared plus four, and then c is zero, we figured that out, and then d is two thirds, so plus two thirds over s squared plus one. So all of that work that I just did, that was just to break up this piece right here."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Let's remember all of this and go back to our original problem, because we've kind of, so actually let me just be clear. So we can rewrite two over s squared plus four times s squared plus one. We can rewrite this as, well a is zero, b is minus two thirds, so this is equal to minus two thirds over s squared plus four, and then c is zero, we figured that out, and then d is two thirds, so plus two thirds over s squared plus one. So all of that work that I just did, that was just to break up this piece right here. That was just to break up that piece right there, and of course we have these other two pieces here that we forgot about. So after all of this work, what do we have? I'm going to make sure I don't make a careless mistake here."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So all of that work that I just did, that was just to break up this piece right here. That was just to break up that piece right there, and of course we have these other two pieces here that we forgot about. So after all of this work, what do we have? I'm going to make sure I don't make a careless mistake here. We get the Laplace transform of y. As you can see, the algebra is the hardest part here. Is equal to this first term, I'm just going back, this first term which I've now decomposed into this."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "I'm going to make sure I don't make a careless mistake here. We get the Laplace transform of y. As you can see, the algebra is the hardest part here. Is equal to this first term, I'm just going back, this first term which I've now decomposed into this. So it's minus, let me write it this way, minus one third, and I think you're going to see in a second why I'm writing this way, minus one third times two over s squared plus four, and then plus two thirds times one over s squared plus one, and you're probably saying, Sal, why are you writing it this way? Well you can already immediately see that this is a Laplace transform of sine of two t. This is a Laplace transform of sine of t. So I wanted to write this two here, because this is two, this is two squared, this is one, this is one squared. So I wanted to write it in this form."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Is equal to this first term, I'm just going back, this first term which I've now decomposed into this. So it's minus, let me write it this way, minus one third, and I think you're going to see in a second why I'm writing this way, minus one third times two over s squared plus four, and then plus two thirds times one over s squared plus one, and you're probably saying, Sal, why are you writing it this way? Well you can already immediately see that this is a Laplace transform of sine of two t. This is a Laplace transform of sine of t. So I wanted to write this two here, because this is two, this is two squared, this is one, this is one squared. So I wanted to write it in this form. This was just the first term. We had two more terms to worry about. We don't want to make a careless mistake."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "So I wanted to write it in this form. This was just the first term. We had two more terms to worry about. We don't want to make a careless mistake. I have two s over s squared plus one, so let me write that down. So plus two times s over s squared plus one, plus last one, plus one over s squared plus one, plus one over s squared plus one. Now we just take the inverse Laplace transform of the whole thing, and then we'll know what y is."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "We don't want to make a careless mistake. I have two s over s squared plus one, so let me write that down. So plus two times s over s squared plus one, plus last one, plus one over s squared plus one, plus one over s squared plus one. Now we just take the inverse Laplace transform of the whole thing, and then we'll know what y is. Let me just write, just to remember, the Laplace transform, so this is going to be a little inverse, this is going to be sine of two t. Let me just write, just so we have it here, so you know I'm not doing some type of voodoo. The Laplace transform of sine of a t is equal to a over s squared plus a squared, and the Laplace transform of cosine of a t is equal to s over s squared plus a squared. Let's just remember those two things when we take the inverse Laplace transform of both sides of this equation."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Now we just take the inverse Laplace transform of the whole thing, and then we'll know what y is. Let me just write, just to remember, the Laplace transform, so this is going to be a little inverse, this is going to be sine of two t. Let me just write, just so we have it here, so you know I'm not doing some type of voodoo. The Laplace transform of sine of a t is equal to a over s squared plus a squared, and the Laplace transform of cosine of a t is equal to s over s squared plus a squared. Let's just remember those two things when we take the inverse Laplace transform of both sides of this equation. The inverse Laplace transform of the Laplace transform of y, well, that's just y. This y, maybe we'll write it as a function of t, is equal to, well, this is the Laplace transform of sine of two t. You can just do some pattern matching right here. If a is equal to two, then this would be the Laplace transform of sine of two t, so it's minus 1 3rd times sine of two t plus 2 3rds times, this is the Laplace transform of sine of t. If you just make a is equal to one, sine of t's Laplace transform is one over s squared plus one, so plus 2 3rds times the sine of t. Let me do the next one in blue, just because it was already written in blue, plus 2 times, this is the Laplace transform of cosine of t. If you make a is equal to one, then the cosine t Laplace transform is s over s squared plus one, so 2 times cosine of t. Then one last term, plus, this is just like this one over here, this is just the Laplace transform of sine of t. Plus sine of t. We're almost done."}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "Let's just remember those two things when we take the inverse Laplace transform of both sides of this equation. The inverse Laplace transform of the Laplace transform of y, well, that's just y. This y, maybe we'll write it as a function of t, is equal to, well, this is the Laplace transform of sine of two t. You can just do some pattern matching right here. If a is equal to two, then this would be the Laplace transform of sine of two t, so it's minus 1 3rd times sine of two t plus 2 3rds times, this is the Laplace transform of sine of t. If you just make a is equal to one, sine of t's Laplace transform is one over s squared plus one, so plus 2 3rds times the sine of t. Let me do the next one in blue, just because it was already written in blue, plus 2 times, this is the Laplace transform of cosine of t. If you make a is equal to one, then the cosine t Laplace transform is s over s squared plus one, so 2 times cosine of t. Then one last term, plus, this is just like this one over here, this is just the Laplace transform of sine of t. Plus sine of t. We're almost done. We're essentially done, but there's a little bit more simplification we can do. I have 2 3rds times the sine of t here, and then I have another one sine of t here, so I can add the 2 3rds to the one. What's 2 3rds plus one, or 3 3rds?"}, {"video_title": "Using the Laplace transform to solve a nonhomogeneous eq Laplace transform Khan Academy.mp3", "Sentence": "If a is equal to two, then this would be the Laplace transform of sine of two t, so it's minus 1 3rd times sine of two t plus 2 3rds times, this is the Laplace transform of sine of t. If you just make a is equal to one, sine of t's Laplace transform is one over s squared plus one, so plus 2 3rds times the sine of t. Let me do the next one in blue, just because it was already written in blue, plus 2 times, this is the Laplace transform of cosine of t. If you make a is equal to one, then the cosine t Laplace transform is s over s squared plus one, so 2 times cosine of t. Then one last term, plus, this is just like this one over here, this is just the Laplace transform of sine of t. Plus sine of t. We're almost done. We're essentially done, but there's a little bit more simplification we can do. I have 2 3rds times the sine of t here, and then I have another one sine of t here, so I can add the 2 3rds to the one. What's 2 3rds plus one, or 3 3rds? It's 5 3rds, so I can write y of t is equal to minus 1 3rd sine of two t plus these two terms I'm just going to add up, plus 5 3rds sine of t, and then I have this last term here, plus 2 cosine of t. This was a hairy problem, a lot of work, and we saw that the hardest part really was just the partial fraction decomposition that we did up here, and not making any careless mistakes, but at the end we got a pretty neat answer that's not too complicated, that satisfies this non-homogeneous differential equation. We were able to incorporate the boundary conditions as we did it. Anyway, hopefully you found that vaguely satisfying."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So now let's try to solve some. And this first class of differential equations I'll introduce you to, they're called separable equations. And I think what you'll find is that we're not learning really anything new. Using just your first year calculus, derivative and integrating skills, you can solve a separable equation. And the reason why they're called separable is because you can actually separate the x and y terms and integrate them separately to get the solution of the differential equation. So let's separable equations. So let's do a couple, and I think you'll get the point."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "Using just your first year calculus, derivative and integrating skills, you can solve a separable equation. And the reason why they're called separable is because you can actually separate the x and y terms and integrate them separately to get the solution of the differential equation. So let's separable equations. So let's do a couple, and I think you'll get the point. These often are really more of exercises in algebra than anything else. So the first separable differential equation is dy over dx is equal to x squared over 1 minus y squared. And actually this is a good time to just review our terminology."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So let's do a couple, and I think you'll get the point. These often are really more of exercises in algebra than anything else. So the first separable differential equation is dy over dx is equal to x squared over 1 minus y squared. And actually this is a good time to just review our terminology. So first of all, what is the order of this differential equation? Well, the highest derivative in it is just the first derivative, so the order is equal to 1. So it's first order."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And actually this is a good time to just review our terminology. So first of all, what is the order of this differential equation? Well, the highest derivative in it is just the first derivative, so the order is equal to 1. So it's first order. It's ordinary because we only have a regular derivative, no partial derivatives here. And then is this linear or nonlinear? Well, at first you say, oh, well, you know, this looks linear."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So it's first order. It's ordinary because we only have a regular derivative, no partial derivatives here. And then is this linear or nonlinear? Well, at first you say, oh, well, you know, this looks linear. I'm not multiplying the derivative times anything else. But if you look carefully, something interesting is going on. First of all, you have a y squared, and y is a dependent variable."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "Well, at first you say, oh, well, you know, this looks linear. I'm not multiplying the derivative times anything else. But if you look carefully, something interesting is going on. First of all, you have a y squared, and y is a dependent variable. y is a function of x. So to have the y squared, that makes it nonlinear. And even if this was a y, if you were to actually multiply both sides of this equation times 1 minus y and get it in the form that I showed you in the previous equation, you would have 1 minus y squared."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "First of all, you have a y squared, and y is a dependent variable. y is a function of x. So to have the y squared, that makes it nonlinear. And even if this was a y, if you were to actually multiply both sides of this equation times 1 minus y and get it in the form that I showed you in the previous equation, you would have 1 minus y squared. Actually, this is actually the first step of what we have to do anyway, so I'll write it down. So if I'm just multiplying both sides of this equation times 1 minus y squared, you get 1 minus y squared times dy dx is equal to x squared. And then you immediately see that you're actually, even if this wasn't a squared here, you'd be multiplying the y times dy dx, and that also makes it nonlinear because you're multiplying the dependent variable times the derivative of itself."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And even if this was a y, if you were to actually multiply both sides of this equation times 1 minus y and get it in the form that I showed you in the previous equation, you would have 1 minus y squared. Actually, this is actually the first step of what we have to do anyway, so I'll write it down. So if I'm just multiplying both sides of this equation times 1 minus y squared, you get 1 minus y squared times dy dx is equal to x squared. And then you immediately see that you're actually, even if this wasn't a squared here, you'd be multiplying the y times dy dx, and that also makes it nonlinear because you're multiplying the dependent variable times the derivative of itself. So that also makes this a nonlinear equation. But anyway, let's get back to solving this. So this was the first step."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And then you immediately see that you're actually, even if this wasn't a squared here, you'd be multiplying the y times dy dx, and that also makes it nonlinear because you're multiplying the dependent variable times the derivative of itself. So that also makes this a nonlinear equation. But anyway, let's get back to solving this. So this was the first step. Let's multiply both sides by 1 minus y squared. And the real end goal is just to separate the y's and the x's and then integrate both sides. So I'm almost there."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So this was the first step. Let's multiply both sides by 1 minus y squared. And the real end goal is just to separate the y's and the x's and then integrate both sides. So I'm almost there. So now what I want to do is I want to multiply both sides of this equation times dx. So I have a dx here and get rid of this dx there. I'm going to go here."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So I'm almost there. So now what I want to do is I want to multiply both sides of this equation times dx. So I have a dx here and get rid of this dx there. I'm going to go here. I don't want to waste too much space. So you get 1 minus y squared dy is equal to x squared dx. I have separated the x and y variables and the differentials."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I'm going to go here. I don't want to waste too much space. So you get 1 minus y squared dy is equal to x squared dx. I have separated the x and y variables and the differentials. All I did is I multiplied both sides of this equation times dx to get here. Now I can just integrate both sides. So let's do that."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I have separated the x and y variables and the differentials. All I did is I multiplied both sides of this equation times dx to get here. Now I can just integrate both sides. So let's do that. So whatever you do to one side of the equation, you have to do to the other. That's true with regular equations or differential equations. So we're going to integrate both sides."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So let's do that. So whatever you do to one side of the equation, you have to do to the other. That's true with regular equations or differential equations. So we're going to integrate both sides. So what's the integral of this expression with respect to y? The integral of 1 is y. The integral of y squared, well that's minus y to the third over 3."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So we're going to integrate both sides. So what's the integral of this expression with respect to y? The integral of 1 is y. The integral of y squared, well that's minus y to the third over 3. And I'll write the plus c here just to kind of show you something. But you really don't have to write a plus c on both sides. I'll call that plus the constant due to y, the y integration."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "The integral of y squared, well that's minus y to the third over 3. And I'll write the plus c here just to kind of show you something. But you really don't have to write a plus c on both sides. I'll call that plus the constant due to y, the y integration. You'll never see this in a calculus class, but I just want to make a point here. Is equal to, I just want to show you this, our plus c's never disappeared from when we were taking our traditional antiderivatives. And what's the derivative of this?"}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I'll call that plus the constant due to y, the y integration. You'll never see this in a calculus class, but I just want to make a point here. Is equal to, I just want to show you this, our plus c's never disappeared from when we were taking our traditional antiderivatives. And what's the derivative of this? Well that's x to the third over 3. And this is also going to have a plus c due to the x variable. Now the reason why I did this magenta one in magenta, I labeled it like that, is because you really just have to write a plus c on one side of the equation."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And what's the derivative of this? Well that's x to the third over 3. And this is also going to have a plus c due to the x variable. Now the reason why I did this magenta one in magenta, I labeled it like that, is because you really just have to write a plus c on one side of the equation. And if that doesn't make a lot of sense, let's subtract this c from both sides and we get y minus, let me scroll down a little bit, y, my y looks like a g, y minus y to the third over 3 is equal to x to the third over 3 plus the constant when we took the antiderivative of the x, minus the constant of the antiderivative when we took the y. But these two constants, they're just, I mean we don't know what they are, they're arbitrary constants, so we could just write a general c here. So you could have just, you have to have a constant, but it doesn't have to be on both sides of this equation because they're arbitrary."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "Now the reason why I did this magenta one in magenta, I labeled it like that, is because you really just have to write a plus c on one side of the equation. And if that doesn't make a lot of sense, let's subtract this c from both sides and we get y minus, let me scroll down a little bit, y, my y looks like a g, y minus y to the third over 3 is equal to x to the third over 3 plus the constant when we took the antiderivative of the x, minus the constant of the antiderivative when we took the y. But these two constants, they're just, I mean we don't know what they are, they're arbitrary constants, so we could just write a general c here. So you could have just, you have to have a constant, but it doesn't have to be on both sides of this equation because they're arbitrary. Cx minus Cy, well that's still just another constant. And then if we wanted to simplify this equation more, we can multiply both sides of this by 3, just to make it look nicer. And you get 3y minus y to the third is equal to x to the third plus, well I could write 3c here, but once again, c is an arbitrary constant, right?"}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So you could have just, you have to have a constant, but it doesn't have to be on both sides of this equation because they're arbitrary. Cx minus Cy, well that's still just another constant. And then if we wanted to simplify this equation more, we can multiply both sides of this by 3, just to make it look nicer. And you get 3y minus y to the third is equal to x to the third plus, well I could write 3c here, but once again, c is an arbitrary constant, right? So 3 times an arbitrary constant, that's just another arbitrary constant. So I'll write the c there. And there you have it."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And you get 3y minus y to the third is equal to x to the third plus, well I could write 3c here, but once again, c is an arbitrary constant, right? So 3 times an arbitrary constant, that's just another arbitrary constant. So I'll write the c there. And there you have it. We have solved this differential equation, although it's in implicit form right now, and it's fairly hard to get it out of implicit form. We could put the c on one side, so the solution could be 3y minus y to the third minus x to the third is equal to c. Some people might like that a little bit better, but that's the solution. And notice, the solution, just like when you take an antiderivative, the solution is a class of implicit functions in this case."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And there you have it. We have solved this differential equation, although it's in implicit form right now, and it's fairly hard to get it out of implicit form. We could put the c on one side, so the solution could be 3y minus y to the third minus x to the third is equal to c. Some people might like that a little bit better, but that's the solution. And notice, the solution, just like when you take an antiderivative, the solution is a class of implicit functions in this case. And why is it a class? Because we have that constant there. Depending on what number you pick there, it will be another solution."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And notice, the solution, just like when you take an antiderivative, the solution is a class of implicit functions in this case. And why is it a class? Because we have that constant there. Depending on what number you pick there, it will be another solution. But any constant there will satisfy the original differential equation. Which was up here. This was the original differential equation."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "Depending on what number you pick there, it will be another solution. But any constant there will satisfy the original differential equation. Which was up here. This was the original differential equation. And if you want to solve for that constant, someone has to give you an initial condition. Someone has to say, well, when x is 2, y is 3. And then you could solve for c. Anyway, let's do another one that gives us an initial condition."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "This was the original differential equation. And if you want to solve for that constant, someone has to give you an initial condition. Someone has to say, well, when x is 2, y is 3. And then you could solve for c. Anyway, let's do another one that gives us an initial condition. So this one's a little bit... I don't want to... I'll start over."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And then you could solve for c. Anyway, let's do another one that gives us an initial condition. So this one's a little bit... I don't want to... I'll start over. Clear image, invert colors. So I have optimal space. So this one is the first derivative of y with respect to x is equal to 3x squared plus 4x plus 2 over 2 times y minus 1."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I'll start over. Clear image, invert colors. So I have optimal space. So this one is the first derivative of y with respect to x is equal to 3x squared plus 4x plus 2 over 2 times y minus 1. This is a parentheses, not an absolute value. And they give us initial conditions. They say that y of 0 is equal to negative 1."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So this one is the first derivative of y with respect to x is equal to 3x squared plus 4x plus 2 over 2 times y minus 1. This is a parentheses, not an absolute value. And they give us initial conditions. They say that y of 0 is equal to negative 1. So once we solve this differential equation, and this is a separable differential equation, then we can use this initial condition, when x is 0, y is 1, to figure out the constant. So let's first separate this equation. So let's multiply both sides by 2 times y minus 1."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "They say that y of 0 is equal to negative 1. So once we solve this differential equation, and this is a separable differential equation, then we can use this initial condition, when x is 0, y is 1, to figure out the constant. So let's first separate this equation. So let's multiply both sides by 2 times y minus 1. And you get 2 times y minus 1 times dy dx is equal to 3x squared plus 4x plus 2. Multiply both sides times dx. This is really just an exercise in algebra."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So let's multiply both sides by 2 times y minus 1. And you get 2 times y minus 1 times dy dx is equal to 3x squared plus 4x plus 2. Multiply both sides times dx. This is really just an exercise in algebra. You get... and I can multiply this one out too. You get 2y minus 2, that's just this, dy. I multiply both sides times dx, so that equals 3x squared plus 4x plus 2 dx."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "This is really just an exercise in algebra. You get... and I can multiply this one out too. You get 2y minus 2, that's just this, dy. I multiply both sides times dx, so that equals 3x squared plus 4x plus 2 dx. I have separated the equations. I've separated the independent from the dependent variable and their relative differentials. And so now I can integrate."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I multiply both sides times dx, so that equals 3x squared plus 4x plus 2 dx. I have separated the equations. I've separated the independent from the dependent variable and their relative differentials. And so now I can integrate. And I can integrate in magenta. I can integrate. What's the antiderivative of this expression with respect to y?"}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And so now I can integrate. And I can integrate in magenta. I can integrate. What's the antiderivative of this expression with respect to y? Well, let's just see. It's y squared minus 2y. I won't write the plus C. I'll just do it on the right-hand side."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "What's the antiderivative of this expression with respect to y? Well, let's just see. It's y squared minus 2y. I won't write the plus C. I'll just do it on the right-hand side. That is equal to 3x squared. Well, the antiderivative is x to the third plus antiderivative is 2x squared plus 2x plus C. And that C kind of takes care of the constant for both sides of the equation. And hopefully you understand y from the last example."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I won't write the plus C. I'll just do it on the right-hand side. That is equal to 3x squared. Well, the antiderivative is x to the third plus antiderivative is 2x squared plus 2x plus C. And that C kind of takes care of the constant for both sides of the equation. And hopefully you understand y from the last example. But we can solve for C using the initial condition y of 0 is equal to negative 1. So let's see. When x is 0, y is negative 1."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "And hopefully you understand y from the last example. But we can solve for C using the initial condition y of 0 is equal to negative 1. So let's see. When x is 0, y is negative 1. So let's put y as negative 1. So we get negative 1 squared minus 2 times negative 1. That's the value of y."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "When x is 0, y is negative 1. So let's put y as negative 1. So we get negative 1 squared minus 2 times negative 1. That's the value of y. It's equal to when x is equal to 0. So when x is equal to 0, that's 0 to the third plus 2 times 0 squared plus 2 times 0 plus C. So this is fairly straightforward. All of these, this is all 0."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "That's the value of y. It's equal to when x is equal to 0. So when x is equal to 0, that's 0 to the third plus 2 times 0 squared plus 2 times 0 plus C. So this is fairly straightforward. All of these, this is all 0. This is, let's see, negative 1 squared. That's 1 minus 2 times minus 1. That's plus 2."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "All of these, this is all 0. This is, let's see, negative 1 squared. That's 1 minus 2 times minus 1. That's plus 2. It's equal to C. And we get C is equal to 3. So the implicit exact solution, the solution of our differential equation, remember now it's not a class because they gave us an initial condition, is y squared minus 2y is equal to x to the third plus 2x squared plus 2x plus 3. We figured out that's what C was."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "That's plus 2. It's equal to C. And we get C is equal to 3. So the implicit exact solution, the solution of our differential equation, remember now it's not a class because they gave us an initial condition, is y squared minus 2y is equal to x to the third plus 2x squared plus 2x plus 3. We figured out that's what C was. And actually if you want, you could write this in an explicit form by completing the square. This is just algebra at this point. You're done."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "We figured out that's what C was. And actually if you want, you could write this in an explicit form by completing the square. This is just algebra at this point. You're done. This is an implicit form. If you wanted to make it explicit, you could add 1 to both sides. I'm just completing the square here."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "You're done. This is an implicit form. If you wanted to make it explicit, you could add 1 to both sides. I'm just completing the square here. So y squared minus 2y plus 1. If I add 1 to that side, I'd have to add 1 to this side. So it becomes x to the third plus 2x squared plus 2x plus 4."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I'm just completing the square here. So y squared minus 2y plus 1. If I add 1 to that side, I'd have to add 1 to this side. So it becomes x to the third plus 2x squared plus 2x plus 4. I just added 1 to both sides of this equation. Why did I do that? Because I wanted this side to be a perfect square in terms of y."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "So it becomes x to the third plus 2x squared plus 2x plus 4. I just added 1 to both sides of this equation. Why did I do that? Because I wanted this side to be a perfect square in terms of y. Then I could rewrite this side as y minus 1 squared is equal to x to the third plus 2x squared plus 2x plus 4. Then I could say y minus 1 is equal to the plus or minus square root of x to the third plus 2x squared plus 2x plus 4. I could add 1 to both sides and then I could get y is equal to 1 plus or minus the square root of x to the third plus 2x squared plus 2x plus 4."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "Because I wanted this side to be a perfect square in terms of y. Then I could rewrite this side as y minus 1 squared is equal to x to the third plus 2x squared plus 2x plus 4. Then I could say y minus 1 is equal to the plus or minus square root of x to the third plus 2x squared plus 2x plus 4. I could add 1 to both sides and then I could get y is equal to 1 plus or minus the square root of x to the third plus 2x squared plus 2x plus 4. It has plus or minus here. If we had to pick one of the two, we'd go back to the initial condition. Our initial condition told us that y of 0 is equal to negative 1."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "I could add 1 to both sides and then I could get y is equal to 1 plus or minus the square root of x to the third plus 2x squared plus 2x plus 4. It has plus or minus here. If we had to pick one of the two, we'd go back to the initial condition. Our initial condition told us that y of 0 is equal to negative 1. If we put 0 here for x, we get y is equal to 1 plus or minus 0 plus 4. So 1 plus or minus 4. If y is going to be equal to negative 1, so we get y is equal to 1 plus or minus 2."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "Our initial condition told us that y of 0 is equal to negative 1. If we put 0 here for x, we get y is equal to 1 plus or minus 0 plus 4. So 1 plus or minus 4. If y is going to be equal to negative 1, so we get y is equal to 1 plus or minus 2. If this is going to be equal to negative 1, then this has to be 1 minus 2. So the explicit form that satisfies our initial condition, and we're getting a little geeky here, you can get rid of the plus. It's 1 minus this whole thing."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "If y is going to be equal to negative 1, so we get y is equal to 1 plus or minus 2. If this is going to be equal to negative 1, then this has to be 1 minus 2. So the explicit form that satisfies our initial condition, and we're getting a little geeky here, you can get rid of the plus. It's 1 minus this whole thing. That's what satisfies our initial condition. You could figure out where it's satisfied because in order for this, over what domain is it satisfied? That's satisfied when this term is positive."}, {"video_title": "Old separable differential equations introduction Khan Academy.mp3", "Sentence": "It's 1 minus this whole thing. That's what satisfies our initial condition. You could figure out where it's satisfied because in order for this, over what domain is it satisfied? That's satisfied when this term is positive. This becomes negative and you get it undefined in reals and all of that. Anyway, I've run out of time. See you in the next video."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And if I wanted to take the derivative of this with respect to x, this is equal to the partial of xi with respect to x plus the partial of xi with respect to y times dy dx. And in the last video, I didn't prove it to you, but I hopefully gave you a little bit of intuition that you can believe me. But maybe one day I'll prove it a little bit more rigorously. But you can find proofs on the web if you're interested for the chain rule with partial derivatives. So let's put that aside and let's explore another property of partial derivatives. And then we're ready to get the intuition behind exact equations. Because what you're going to find, it's fairly straightforward to solve exact equations."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "But you can find proofs on the web if you're interested for the chain rule with partial derivatives. So let's put that aside and let's explore another property of partial derivatives. And then we're ready to get the intuition behind exact equations. Because what you're going to find, it's fairly straightforward to solve exact equations. But the intuition is a little bit more, well, I don't want to say it's difficult. Because if you have the intuition, you have it. So what if I had, say, the same function xi, and I were to take the partial derivative of xi with respect to x first."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Because what you're going to find, it's fairly straightforward to solve exact equations. But the intuition is a little bit more, well, I don't want to say it's difficult. Because if you have the intuition, you have it. So what if I had, say, the same function xi, and I were to take the partial derivative of xi with respect to x first. I'll just write xi. I don't have to write x and y every time. And then I were to take the partial derivative with respect to y."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So what if I had, say, the same function xi, and I were to take the partial derivative of xi with respect to x first. I'll just write xi. I don't have to write x and y every time. And then I were to take the partial derivative with respect to y. So just as a notation, this you could write as, you could kind of view it as you're multiplying the operators. So it could be written like this. The partial del squared times xi, or del squared xi, over del y, del, or curly dx."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And then I were to take the partial derivative with respect to y. So just as a notation, this you could write as, you could kind of view it as you're multiplying the operators. So it could be written like this. The partial del squared times xi, or del squared xi, over del y, del, or curly dx. And that can also be written as, and this is my preferred notation, because it doesn't have all this extra junk everywhere. You could just say, well, we took the partial with respect to x first. So this just means the partial of xi with respect to x."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "The partial del squared times xi, or del squared xi, over del y, del, or curly dx. And that can also be written as, and this is my preferred notation, because it doesn't have all this extra junk everywhere. You could just say, well, we took the partial with respect to x first. So this just means the partial of xi with respect to x. And then we took the partial with respect to y. So that's one situation to consider. What happens when we take the partial with respect to x and then y?"}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So this just means the partial of xi with respect to x. And then we took the partial with respect to y. So that's one situation to consider. What happens when we take the partial with respect to x and then y? So with respect to x, you hold y constant to get just the partial with respect to x, ignore the y there. And then you hold the x constant, and you take the partial with respect to y. So what's the difference between that and if we were to switch the order?"}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "What happens when we take the partial with respect to x and then y? So with respect to x, you hold y constant to get just the partial with respect to x, ignore the y there. And then you hold the x constant, and you take the partial with respect to y. So what's the difference between that and if we were to switch the order? So what happens if we were to, I'll do it in a different color. If we had xi, and we were to take the partial with respect to y first, and then we were to take the partial with respect to x. So just the notation, just you're comfortable with it."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So what's the difference between that and if we were to switch the order? So what happens if we were to, I'll do it in a different color. If we had xi, and we were to take the partial with respect to y first, and then we were to take the partial with respect to x. So just the notation, just you're comfortable with it. That would be, so partial x, partial y. And this is the operator. And it might be a little confusing that here between these two notations, even though they're the same thing, the order is mixed."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So just the notation, just you're comfortable with it. That would be, so partial x, partial y. And this is the operator. And it might be a little confusing that here between these two notations, even though they're the same thing, the order is mixed. That's just because it's just a different way of thinking about it. This says, OK, partial first with respect to x, then y. This views it more as the operator."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And it might be a little confusing that here between these two notations, even though they're the same thing, the order is mixed. That's just because it's just a different way of thinking about it. This says, OK, partial first with respect to x, then y. This views it more as the operator. So we took the partial of x first, and then we took y. Like you're multiplying the operators. But anyway, so this can also be written as the partial of y, and then we took the partial of that with respect to x."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "This views it more as the operator. So we took the partial of x first, and then we took y. Like you're multiplying the operators. But anyway, so this can also be written as the partial of y, and then we took the partial of that with respect to x. Now I'm going to tell you right now that if each of the first partials are continuous, and most of the functions we've dealt with in a normal domain, as long as there aren't any discontinuities or holes or something strange in the function definition, they usually are continuous. And especially in a first year calculus or differential course, we're probably going to be dealing with continuous functions in our domain. If both of these functions are continuous, if both of the first partials are continuous, then these two are going to be equal to each other."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "But anyway, so this can also be written as the partial of y, and then we took the partial of that with respect to x. Now I'm going to tell you right now that if each of the first partials are continuous, and most of the functions we've dealt with in a normal domain, as long as there aren't any discontinuities or holes or something strange in the function definition, they usually are continuous. And especially in a first year calculus or differential course, we're probably going to be dealing with continuous functions in our domain. If both of these functions are continuous, if both of the first partials are continuous, then these two are going to be equal to each other. So xi of xy is going to be equal to xi of yx. Now, we can use this knowledge, which is the chain rule using partial derivatives, and this knowledge to now solve a certain class of differential equations, first order differential equations, called exact equations. And what does an exact equation look like?"}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "If both of these functions are continuous, if both of the first partials are continuous, then these two are going to be equal to each other. So xi of xy is going to be equal to xi of yx. Now, we can use this knowledge, which is the chain rule using partial derivatives, and this knowledge to now solve a certain class of differential equations, first order differential equations, called exact equations. And what does an exact equation look like? An exact equation looks like this. It's always the color picking is the hard part. So let's say I have, this is my differential equation."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And what does an exact equation look like? An exact equation looks like this. It's always the color picking is the hard part. So let's say I have, this is my differential equation. I have some function of x and y. So I don't know, it could be x squared times cosine of y or something, I don't know. It could be any function of x and y."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So let's say I have, this is my differential equation. I have some function of x and y. So I don't know, it could be x squared times cosine of y or something, I don't know. It could be any function of x and y. Plus some function of x and y, we'll call that n, times dy dx is equal to 0. This is, well, I don't know it's an exact equation yet, but if you saw something of this form, your first impulse should be, oh, well, actually, your very first impulse is, is this separable? And you should try to play around with the algebra a little bit and see if it's separable, because that's always the most straightforward way."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "It could be any function of x and y. Plus some function of x and y, we'll call that n, times dy dx is equal to 0. This is, well, I don't know it's an exact equation yet, but if you saw something of this form, your first impulse should be, oh, well, actually, your very first impulse is, is this separable? And you should try to play around with the algebra a little bit and see if it's separable, because that's always the most straightforward way. If it's not separable, but you can still put it in this form, you say, hey, is it an exact equation? And what's an exact equation? Well, look immediately."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And you should try to play around with the algebra a little bit and see if it's separable, because that's always the most straightforward way. If it's not separable, but you can still put it in this form, you say, hey, is it an exact equation? And what's an exact equation? Well, look immediately. This pattern right here looks an awful lot like this pattern. What if m was the partial of xi with respect to x? What if xi with respect to x is equal to m?"}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Well, look immediately. This pattern right here looks an awful lot like this pattern. What if m was the partial of xi with respect to x? What if xi with respect to x is equal to m? What if this was xi with respect to x, and what if this was xi with respect to y? So xi with respect to y is equal to n. What if? I'm just saying, we don't know for sure, right?"}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "What if xi with respect to x is equal to m? What if this was xi with respect to x, and what if this was xi with respect to y? So xi with respect to y is equal to n. What if? I'm just saying, we don't know for sure, right? If you just see this someplace randomly, you won't know for sure that this is the partial with respect to x of some function, and this is the partial with respect to y of some function. But we're just saying, what if? If this were true, then we could rewrite this as the partial of xi with respect to x plus the partial of xi with respect to y times dy dx equal to 0."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "I'm just saying, we don't know for sure, right? If you just see this someplace randomly, you won't know for sure that this is the partial with respect to x of some function, and this is the partial with respect to y of some function. But we're just saying, what if? If this were true, then we could rewrite this as the partial of xi with respect to x plus the partial of xi with respect to y times dy dx equal to 0. And this right here, the left side right there, that's the same thing as this. This is just the derivative of xi with respect to x using the partial derivative chain rule. So you could rewrite it."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "If this were true, then we could rewrite this as the partial of xi with respect to x plus the partial of xi with respect to y times dy dx equal to 0. And this right here, the left side right there, that's the same thing as this. This is just the derivative of xi with respect to x using the partial derivative chain rule. So you could rewrite it. This is just the derivative of xi with respect to x. And xi is a function of xy is equal to 0. So if you see a differential equation, it has this form, and you're saying, boy, I can't separate it, but maybe it's an exact equation."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So you could rewrite it. This is just the derivative of xi with respect to x. And xi is a function of xy is equal to 0. So if you see a differential equation, it has this form, and you're saying, boy, I can't separate it, but maybe it's an exact equation. And frankly, if that was what was recently covered before the current exam, it probably is an exact equation. But if you see this form, you say, boy, maybe it's an exact equation. If it is an exact equation, and I'll show you how to test it in a second using this information, then this can be written as the derivative of some function xi, where this is the partial of xi with respect to x."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So if you see a differential equation, it has this form, and you're saying, boy, I can't separate it, but maybe it's an exact equation. And frankly, if that was what was recently covered before the current exam, it probably is an exact equation. But if you see this form, you say, boy, maybe it's an exact equation. If it is an exact equation, and I'll show you how to test it in a second using this information, then this can be written as the derivative of some function xi, where this is the partial of xi with respect to x. This is the partial of xi with respect to y. And then if you could write it like this, and you take the derivative of both sides, and you would get xi of xy is equal to c as the solution. So there are two things that we should be caring about that you might be saying, OK, Sal, you've walked through xi's and partials and all this."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "If it is an exact equation, and I'll show you how to test it in a second using this information, then this can be written as the derivative of some function xi, where this is the partial of xi with respect to x. This is the partial of xi with respect to y. And then if you could write it like this, and you take the derivative of both sides, and you would get xi of xy is equal to c as the solution. So there are two things that we should be caring about that you might be saying, OK, Sal, you've walked through xi's and partials and all this. One, how do I know that it's an exact equation? And then if it is an exact equation, which tells us that there is some xi, then how do I solve for the xi? So the way to figure out is it an exact equation is to use this information right here."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So there are two things that we should be caring about that you might be saying, OK, Sal, you've walked through xi's and partials and all this. One, how do I know that it's an exact equation? And then if it is an exact equation, which tells us that there is some xi, then how do I solve for the xi? So the way to figure out is it an exact equation is to use this information right here. We know that if xi and its derivatives are continuous over some domain, that when you take the partial with respect to x and then y, that's the same thing as doing it in the other order. So we said, this is the partial with respect to x, right? This is the partial with respect to x, and this is the partial with respect to y."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So the way to figure out is it an exact equation is to use this information right here. We know that if xi and its derivatives are continuous over some domain, that when you take the partial with respect to x and then y, that's the same thing as doing it in the other order. So we said, this is the partial with respect to x, right? This is the partial with respect to x, and this is the partial with respect to y. So if this is an exact equation, if we were to take the partial of this with respect to y, if we were to say, if we were to take the partial of m with respect to y, so the partial of xi with respect to x is equal to m. If we were to take the partial of those with respect to y, so we could just rewrite that as that, then that should be equal to the partial of n with respect to x, right? The partial of xi with respect to y is equal to n. So if we take the partial with respect to x of both of these, we know from this that these should be equal if xi and its partials are continuous over that domain. So then this will also be equal."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "This is the partial with respect to x, and this is the partial with respect to y. So if this is an exact equation, if we were to take the partial of this with respect to y, if we were to say, if we were to take the partial of m with respect to y, so the partial of xi with respect to x is equal to m. If we were to take the partial of those with respect to y, so we could just rewrite that as that, then that should be equal to the partial of n with respect to x, right? The partial of xi with respect to y is equal to n. So if we take the partial with respect to x of both of these, we know from this that these should be equal if xi and its partials are continuous over that domain. So then this will also be equal. So that is actually the test to test if this is an exact equation. So let me rewrite all of that again and summarize it a little bit. So if you see something of the form m of xy plus n of xy times dy dx is equal to 0."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So then this will also be equal. So that is actually the test to test if this is an exact equation. So let me rewrite all of that again and summarize it a little bit. So if you see something of the form m of xy plus n of xy times dy dx is equal to 0. And then you take the partial derivative of m with respect to y, and then you take the partial derivative of n with respect to x, and they are equal to each other, then, and it's actually if and only if, so it goes both ways, this is an exact equation, an exact differential equation. This is an exact equation. And if it's an exact equation, that tells us that there exists a xi such that the derivative of xi of xy is equal to 0, or xi of xy is equal to c is a solution of this equation."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So if you see something of the form m of xy plus n of xy times dy dx is equal to 0. And then you take the partial derivative of m with respect to y, and then you take the partial derivative of n with respect to x, and they are equal to each other, then, and it's actually if and only if, so it goes both ways, this is an exact equation, an exact differential equation. This is an exact equation. And if it's an exact equation, that tells us that there exists a xi such that the derivative of xi of xy is equal to 0, or xi of xy is equal to c is a solution of this equation. And the partial derivative of xi with respect to x is equal to m, and the partial derivative of xi with respect to y is equal to n. And I'll show you in the next video how to actually use this information to solve for xi. So here are some things I want to point out. This is going to be the partial derivative of our xi with respect to x, but when we take the kind of exact test, we take it with respect to y, because we want to get that mixed derivative."}, {"video_title": "Exact equations intuition 2 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And if it's an exact equation, that tells us that there exists a xi such that the derivative of xi of xy is equal to 0, or xi of xy is equal to c is a solution of this equation. And the partial derivative of xi with respect to x is equal to m, and the partial derivative of xi with respect to y is equal to n. And I'll show you in the next video how to actually use this information to solve for xi. So here are some things I want to point out. This is going to be the partial derivative of our xi with respect to x, but when we take the kind of exact test, we take it with respect to y, because we want to get that mixed derivative. Similarly, this is going to be the partial derivative of xi with respect to y, but when we do the test, we take the partial of it with respect to x, so we get that mixed derivative. This is with respect to y and then with respect to x, so you get this. Anyway, I know that might be a little bit involved, but if you understood everything I did, I think you'll have the intuition behind why the methodology of exact equations works."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "OK, I filled your brain with a bunch of partial derivatives and xi's with respect to x's and y's. I think now it's time to actually do it with a real differential equation and make things a little bit more concrete. So let's say I have the differential equation y cosine of x plus 2xe to the y plus sine of x plus x squared e to the y minus 1 times y prime is equal to 0. Well, you could probably already, your brain is already, hopefully, in exact differential equations mode. But if you were to just see this pattern in general, where you see a function of x and y here, this is just some function of x and y. And then you have another function of x and y times y prime or times dy d of x. Your brain should immediately say if this isn't separable, and I'm not going to try to make it separable, just because that'll take a lot of time."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Well, you could probably already, your brain is already, hopefully, in exact differential equations mode. But if you were to just see this pattern in general, where you see a function of x and y here, this is just some function of x and y. And then you have another function of x and y times y prime or times dy d of x. Your brain should immediately say if this isn't separable, and I'm not going to try to make it separable, just because that'll take a lot of time. But if it's not separable, your brain said, oh, maybe this is an exact equation. And you say, let me test whether this is an exact equation. So if this is an exact equation, this is our function m, which is a function of x and y."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Your brain should immediately say if this isn't separable, and I'm not going to try to make it separable, just because that'll take a lot of time. But if it's not separable, your brain said, oh, maybe this is an exact equation. And you say, let me test whether this is an exact equation. So if this is an exact equation, this is our function m, which is a function of x and y. And this is our function n, which is a function of x and y. Now the test is to see if the partial of this with respect to y is equal to the partial of this with respect to x. So let's see."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So if this is an exact equation, this is our function m, which is a function of x and y. And this is our function n, which is a function of x and y. Now the test is to see if the partial of this with respect to y is equal to the partial of this with respect to x. So let's see. The partial of m with respect to y is equal to, let's see, y is, so this cosine of x is just a constant, so it's just cosine of x plus. Now, what's the derivative? Well, 2x is just a constant."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So let's see. The partial of m with respect to y is equal to, let's see, y is, so this cosine of x is just a constant, so it's just cosine of x plus. Now, what's the derivative? Well, 2x is just a constant. What's the derivative of e to the y with respect to y? Well, it's just e to the y, right? So we have the constant on the outside, 2x times the derivative with respect to y, so it's 2x e to the y."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Well, 2x is just a constant. What's the derivative of e to the y with respect to y? Well, it's just e to the y, right? So we have the constant on the outside, 2x times the derivative with respect to y, so it's 2x e to the y. Fair enough. Now what is the partial derivative of this with respect to x? So n sub x, or the partial of n with respect to x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So we have the constant on the outside, 2x times the derivative with respect to y, so it's 2x e to the y. Fair enough. Now what is the partial derivative of this with respect to x? So n sub x, or the partial of n with respect to x. So what's the derivative of sine of x with respect to x? Well, that's easy. That's cosine of x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So n sub x, or the partial of n with respect to x. So what's the derivative of sine of x with respect to x? Well, that's easy. That's cosine of x. Plus 2x times e to the y, right? e to the y is just a constant because y is a constant when we're taking the partial with respect to x. So plus 2x e to the y."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "That's cosine of x. Plus 2x times e to the y, right? e to the y is just a constant because y is a constant when we're taking the partial with respect to x. So plus 2x e to the y. And then minus 1, the derivative of a constant with respect to anything is going to be 0. So the derivative of n, the partial of n with respect to x is cosine of x plus 2x e to the y, which lo and behold is the same thing as the derivative, the partial of m with respect to y. So there we have it."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So plus 2x e to the y. And then minus 1, the derivative of a constant with respect to anything is going to be 0. So the derivative of n, the partial of n with respect to x is cosine of x plus 2x e to the y, which lo and behold is the same thing as the derivative, the partial of m with respect to y. So there we have it. We've shown that m of y is equal to, or the partial of m with respect to y is equal to the partial of n with respect to x, which tells us that this is an exact equation. Now given that this is an exact equation, given that this is an exact equation. Oh, yeah, my wife snuck up behind me."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So there we have it. We've shown that m of y is equal to, or the partial of m with respect to y is equal to the partial of n with respect to x, which tells us that this is an exact equation. Now given that this is an exact equation, given that this is an exact equation. Oh, yeah, my wife snuck up behind me. I was wondering whether I thought there was some critter in my house or something. Anyway, so we know that this is an exact equation. So what does that tell us?"}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Oh, yeah, my wife snuck up behind me. I was wondering whether I thought there was some critter in my house or something. Anyway, so we know that this is an exact equation. So what does that tell us? Well, that tells us that there's some xi where the partial derivative of xi with respect to x is equal to m, and the partial derivative of xi with respect to y is equal to n. And if we know that xi, then we can rewrite our differential equation as the derivative of xi with respect to x is equal to 0. So let's solve for xi. So we know that the partial of xi with respect to x is equal to m. So we could write that."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So what does that tell us? Well, that tells us that there's some xi where the partial derivative of xi with respect to x is equal to m, and the partial derivative of xi with respect to y is equal to n. And if we know that xi, then we can rewrite our differential equation as the derivative of xi with respect to x is equal to 0. So let's solve for xi. So we know that the partial of xi with respect to x is equal to m. So we could write that. We could write the partial of xi with respect to x is equal to m, which is y cosine of x plus 2xe to the y. That's just here. That's my m of x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So we know that the partial of xi with respect to x is equal to m. So we could write that. We could write the partial of xi with respect to x is equal to m, which is y cosine of x plus 2xe to the y. That's just here. That's my m of x. We could have done it the other way. We could have said the partial of xi with respect to y is this thing over here. But let's just do it with x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "That's my m of x. We could have done it the other way. We could have said the partial of xi with respect to y is this thing over here. But let's just do it with x. Now, to at least get kind of a first approximation of what xi is, not an approximation, but to start to get a sense of it, let's take the derivative of both sides with respect to sorry, take the anti-derivative, take the integral of both sides with respect to x. So if you take the derivative of this with respect to x, if you integrate, sorry, if you were to take the anti-derivative of this with respect to x. So let me just write that down."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "But let's just do it with x. Now, to at least get kind of a first approximation of what xi is, not an approximation, but to start to get a sense of it, let's take the derivative of both sides with respect to sorry, take the anti-derivative, take the integral of both sides with respect to x. So if you take the derivative of this with respect to x, if you integrate, sorry, if you were to take the anti-derivative of this with respect to x. So let me just write that down. The partial with respect to x, we're going to take the integrate with respect to x. That is going to be equal to the integral of this whole thing with respect to x. Cosine of x plus 2xe to the y. We're integrating with respect to x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So let me just write that down. The partial with respect to x, we're going to take the integrate with respect to x. That is going to be equal to the integral of this whole thing with respect to x. Cosine of x plus 2xe to the y. We're integrating with respect to x. And normally when you integrate with respect to x, you'd say, OK, plus c. But it actually could be a plus. Since this was a partial with respect to x, we could have had some function of y here in general. Because y, we treat it as a constant, right?"}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "We're integrating with respect to x. And normally when you integrate with respect to x, you'd say, OK, plus c. But it actually could be a plus. Since this was a partial with respect to x, we could have had some function of y here in general. Because y, we treat it as a constant, right? And that makes sense. Because if you were to take the partial of both sides of this with respect to x, if you were to take the partial of a function that is only a function of y with respect to x, you would have gotten a 0 here. So when you take the anti-derivative, we're like, oh well, there might have been some function of y here that we lost when we took the partial with respect to x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Because y, we treat it as a constant, right? And that makes sense. Because if you were to take the partial of both sides of this with respect to x, if you were to take the partial of a function that is only a function of y with respect to x, you would have gotten a 0 here. So when you take the anti-derivative, we're like, oh well, there might have been some function of y here that we lost when we took the partial with respect to x. So anyway, this will simplify to xi. Xi is going to be equal to the integral with respect to x, or the anti-derivative with respect to x here, plus some function of y that we might have lost when we took the partial with respect to x. So let's do that."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So when you take the anti-derivative, we're like, oh well, there might have been some function of y here that we lost when we took the partial with respect to x. So anyway, this will simplify to xi. Xi is going to be equal to the integral with respect to x, or the anti-derivative with respect to x here, plus some function of y that we might have lost when we took the partial with respect to x. So let's do that. Let's figure out this integral. I'll do it in blue. So y is just a constant."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So let's do that. Let's figure out this integral. I'll do it in blue. So y is just a constant. So the anti-derivative y cosine of x is just y sine of x plus e to the y is constant. So 2x. The anti-derivative of 2x with respect to x is x squared."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So y is just a constant. So the anti-derivative y cosine of x is just y sine of x plus e to the y is constant. So 2x. The anti-derivative of 2x with respect to x is x squared. So it's x squared e to the y. And then plus some function of y. And if you want to verify this is true, take the partial of this with respect to x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "The anti-derivative of 2x with respect to x is x squared. So it's x squared e to the y. And then plus some function of y. And if you want to verify this is true, take the partial of this with respect to x. If you take the partial of this with respect to x, you're going to get this in here, which is our function m up here, and then when you take the partial of this with respect to x, you'll get 0 and it'll get lost. OK. So we're almost there."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "And if you want to verify this is true, take the partial of this with respect to x. If you take the partial of this with respect to x, you're going to get this in here, which is our function m up here, and then when you take the partial of this with respect to x, you'll get 0 and it'll get lost. OK. So we're almost there. We've almost figured out our xi, but we still need to figure out this function of y. Well, we know that if we take the partial of this with respect to y, since this is an exact equation, we should get this. We should get our n function."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So we're almost there. We've almost figured out our xi, but we still need to figure out this function of y. Well, we know that if we take the partial of this with respect to y, since this is an exact equation, we should get this. We should get our n function. So let's do that. So the partial, I'll switch notation just to expose you to it. The partial of xi with respect to y is going to be equal to."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "We should get our n function. So let's do that. So the partial, I'll switch notation just to expose you to it. The partial of xi with respect to y is going to be equal to. So here, y sine of x, sine of x is just a constant, y is just y, so the derivative with respect to y is just sine of x. Plus, derivative of e to the y is e to the y, x squared is just a constant, so it's just x squared e to the y. Plus, what's the partial of f of y with respect to y?"}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "The partial of xi with respect to y is going to be equal to. So here, y sine of x, sine of x is just a constant, y is just y, so the derivative with respect to y is just sine of x. Plus, derivative of e to the y is e to the y, x squared is just a constant, so it's just x squared e to the y. Plus, what's the partial of f of y with respect to y? It's going to be f prime of y. And we also, so what did we do? We took m, we integrated with respect to x, and we said, well, we might have lost some function of y, so we added that to it."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Plus, what's the partial of f of y with respect to y? It's going to be f prime of y. And we also, so what did we do? We took m, we integrated with respect to x, and we said, well, we might have lost some function of y, so we added that to it. And then we took the partial of that xi that we've almost constructed, and we took the partial of that with respect to y. Now, we know, since this is exact, that that is going to equal our n. So our n is up there. Cosine of x plus, so that's going to be equal to, I want to make sure I can read it up there, to our n, right?"}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "We took m, we integrated with respect to x, and we said, well, we might have lost some function of y, so we added that to it. And then we took the partial of that xi that we've almost constructed, and we took the partial of that with respect to y. Now, we know, since this is exact, that that is going to equal our n. So our n is up there. Cosine of x plus, so that's going to be equal to, I want to make sure I can read it up there, to our n, right? Oh no, sorry, n is up here. Sine of x, let me write that, sine of x plus x squared e to the y minus 1. So sine of x plus x squared e to the y minus 1."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Cosine of x plus, so that's going to be equal to, I want to make sure I can read it up there, to our n, right? Oh no, sorry, n is up here. Sine of x, let me write that, sine of x plus x squared e to the y minus 1. So sine of x plus x squared e to the y minus 1. Plus x squared e to the y minus 1. That was just our n from our original differential equation. And now we can solve for f prime of y."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So sine of x plus x squared e to the y minus 1. Plus x squared e to the y minus 1. That was just our n from our original differential equation. And now we can solve for f prime of y. So let's see. We get sine of x plus x squared e to the y plus f prime of y is equal to sine of x plus x squared e to the y minus 1. So let's see, we can delete sine of x from both sides."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "And now we can solve for f prime of y. So let's see. We get sine of x plus x squared e to the y plus f prime of y is equal to sine of x plus x squared e to the y minus 1. So let's see, we can delete sine of x from both sides. We can delete x squared e to the y from both sides. And then what are we left with? We're left with f prime of y is equal to 1."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So let's see, we can delete sine of x from both sides. We can delete x squared e to the y from both sides. And then what are we left with? We're left with f prime of y is equal to 1. And then we're left with f of y is equal to, well it equals y plus some constant c. So what is our xi now? We wrote our xi up here and we had this f of y here, so we can rewrite it now. So xi is a function of x and y."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "We're left with f prime of y is equal to 1. And then we're left with f of y is equal to, well it equals y plus some constant c. So what is our xi now? We wrote our xi up here and we had this f of y here, so we can rewrite it now. So xi is a function of x and y. We've actually pretty much almost done solving it. Xi is a function of x and y is equal to y sine of x plus x squared e to the y plus y. Oh sorry, this is f prime of y minus 1. So this is a minus 1."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So xi is a function of x and y. We've actually pretty much almost done solving it. Xi is a function of x and y is equal to y sine of x plus x squared e to the y plus y. Oh sorry, this is f prime of y minus 1. So this is a minus 1. So this is a minus y plus c. So this is going to be a minus y plus c. So we solved for xi. And so what does that tell us? Well we said that original differential equation up here, using the partial derivative chain rule, that original differential equation, can be rewritten now as the derivative dx of xi is equal to, xi is a function of x and y, is equal to 0, or if you were to integrate both sides of this, you would get that xi of xy is equal to c is a solution of that differential equation."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "So this is a minus 1. So this is a minus y plus c. So this is going to be a minus y plus c. So we solved for xi. And so what does that tell us? Well we said that original differential equation up here, using the partial derivative chain rule, that original differential equation, can be rewritten now as the derivative dx of xi is equal to, xi is a function of x and y, is equal to 0, or if you were to integrate both sides of this, you would get that xi of xy is equal to c is a solution of that differential equation. So if we were to set this as equal to c, that's the differential equation. So we could say y sine of x plus x squared e to the y minus y. Now we could say plus this c, plus this c, you call that c1 is equal to c2."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Well we said that original differential equation up here, using the partial derivative chain rule, that original differential equation, can be rewritten now as the derivative dx of xi is equal to, xi is a function of x and y, is equal to 0, or if you were to integrate both sides of this, you would get that xi of xy is equal to c is a solution of that differential equation. So if we were to set this as equal to c, that's the differential equation. So we could say y sine of x plus x squared e to the y minus y. Now we could say plus this c, plus this c, you call that c1 is equal to c2. Well you could subtract the c's from both sides and just be left with a c at the end. But anyway, we have solved this exact equation. One, first by recognizing it was exact, by taking the partial of this with respect to y and seeing if that was equal to the partial of n with respect to x."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Now we could say plus this c, plus this c, you call that c1 is equal to c2. Well you could subtract the c's from both sides and just be left with a c at the end. But anyway, we have solved this exact equation. One, first by recognizing it was exact, by taking the partial of this with respect to y and seeing if that was equal to the partial of n with respect to x. Once we saw that they were equal, we're like, OK, this is going to be exact. So let's figure out xi. Since this is exact, m is going to be the partial of xi with respect to x. n is the partial of xi with respect to y."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "One, first by recognizing it was exact, by taking the partial of this with respect to y and seeing if that was equal to the partial of n with respect to x. Once we saw that they were equal, we're like, OK, this is going to be exact. So let's figure out xi. Since this is exact, m is going to be the partial of xi with respect to x. n is the partial of xi with respect to y. Then to figure out y, we integrated m with respect to x, and we got this. But since we said, oh, well, instead of a plus c, it could have been a function of y there, because we took the partial with respect to x, and this might have been lost. To figure out the function of y, we then took our xi that we figured out, took the partial of that with respect to y, got this, and we said this was an exact equation."}, {"video_title": "Exact equations example 1 First order differential equations Khan Academy.mp3", "Sentence": "Since this is exact, m is going to be the partial of xi with respect to x. n is the partial of xi with respect to y. Then to figure out y, we integrated m with respect to x, and we got this. But since we said, oh, well, instead of a plus c, it could have been a function of y there, because we took the partial with respect to x, and this might have been lost. To figure out the function of y, we then took our xi that we figured out, took the partial of that with respect to y, got this, and we said this was an exact equation. So this is going to equal our n of xy. We set those equal to each other, and then we solved for f of y. And then we had our final xi."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "In the last video, we showed that the Laplace transform of f prime of t is equal to s times the Laplace transform of our function f minus f of 0. Now what we're going to do here is actually use this property that we showed is true and use it to fill in some more of the entries in our Laplace transform table that you'll probably have to memorize sooner or later if you use Laplace transforms a lot. But we already learned that the Laplace transform of sine of at is equal to, and we did a very hairy integration by parts problems to show that that is equal to a over s squared plus a squared. So let's use these two things we know to figure out what the Laplace transform of cosine of at is. So the Laplace transform of cosine of at is equal to what? Well, if we assume that the Laplace transform of cosine of at is the derivative of some function, what is it the derivative of? If I were to, let me do it on the side."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So let's use these two things we know to figure out what the Laplace transform of cosine of at is. So the Laplace transform of cosine of at is equal to what? Well, if we assume that the Laplace transform of cosine of at is the derivative of some function, what is it the derivative of? If I were to, let me do it on the side. If f prime of t is equal to cosine of at, what is a potential f of t? What is a potential f of t? Well, it's the antiderivative, and we can just forget about the constant because we just have to know n f of t for which this is true."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "If I were to, let me do it on the side. If f prime of t is equal to cosine of at, what is a potential f of t? What is a potential f of t? Well, it's the antiderivative, and we can just forget about the constant because we just have to know n f of t for which this is true. So what's the antiderivative of cosine of at? It's 1 over a sine of at. So if this is f prime of t, then that is equal to s times the Laplace transform of its antiderivative, or 1 over a sine of at minus the antiderivative evaluated at 0."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "Well, it's the antiderivative, and we can just forget about the constant because we just have to know n f of t for which this is true. So what's the antiderivative of cosine of at? It's 1 over a sine of at. So if this is f prime of t, then that is equal to s times the Laplace transform of its antiderivative, or 1 over a sine of at minus the antiderivative evaluated at 0. Minus 1 over a sine of, well, a times 0 is 0. Well, sine of 0 is 0, so this whole term goes away. So this is equal to, well, this is a constant, right?"}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So if this is f prime of t, then that is equal to s times the Laplace transform of its antiderivative, or 1 over a sine of at minus the antiderivative evaluated at 0. Minus 1 over a sine of, well, a times 0 is 0. Well, sine of 0 is 0, so this whole term goes away. So this is equal to, well, this is a constant, right? This 1 over a, and we showed that the Laplace transform is a linear operator, so we can take it out. So this is equal to s over a times the Laplace transform of sine of at, and that is equal to s over a times a over f squared plus a squared, and the a's cancel out, and that was much simpler than the integration by parts we had to do to figure this out. So then we get that the Laplace transform of cosine of at is equal to s over s squared plus a squared."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So this is equal to, well, this is a constant, right? This 1 over a, and we showed that the Laplace transform is a linear operator, so we can take it out. So this is equal to s over a times the Laplace transform of sine of at, and that is equal to s over a times a over f squared plus a squared, and the a's cancel out, and that was much simpler than the integration by parts we had to do to figure this out. So then we get that the Laplace transform of cosine of at is equal to s over s squared plus a squared. And in three minutes, we filled in another table in our Laplace transform table. And now we have the two most important trig functions. Let's keep going."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So then we get that the Laplace transform of cosine of at is equal to s over s squared plus a squared. And in three minutes, we filled in another table in our Laplace transform table. And now we have the two most important trig functions. Let's keep going. We haven't really done much with polynomials. We know a couple of things. We know that the Laplace transform of 1 is equal to 1 over s, so let's see if we could use this and the fact that the Laplace transform of f prime is equal to s times the Laplace transform of f minus f of 0."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "Let's keep going. We haven't really done much with polynomials. We know a couple of things. We know that the Laplace transform of 1 is equal to 1 over s, so let's see if we could use this and the fact that the Laplace transform of f prime is equal to s times the Laplace transform of f minus f of 0. Or another way, let's rearrange this. Like if we know f, how can we figure out its Laplace transforms in terms of f prime and f of 0? So let's just rearrange this equation so we get the Laplace transform of f prime."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "We know that the Laplace transform of 1 is equal to 1 over s, so let's see if we could use this and the fact that the Laplace transform of f prime is equal to s times the Laplace transform of f minus f of 0. Or another way, let's rearrange this. Like if we know f, how can we figure out its Laplace transforms in terms of f prime and f of 0? So let's just rearrange this equation so we get the Laplace transform of f prime. I could write of t, but that gets monotonous. Plus f of 0 is equal to s times the Laplace transform of f, divide both sides by s. Let me put the Laplace transform of f. I'm also going to switch the sides so I get the Laplace transform of f is equal to 1 over s. I'm just dividing both sides by s. So 1 over s times this. Times the Laplace transform of my derivative plus my function evaluated at 0."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So let's just rearrange this equation so we get the Laplace transform of f prime. I could write of t, but that gets monotonous. Plus f of 0 is equal to s times the Laplace transform of f, divide both sides by s. Let me put the Laplace transform of f. I'm also going to switch the sides so I get the Laplace transform of f is equal to 1 over s. I'm just dividing both sides by s. So 1 over s times this. Times the Laplace transform of my derivative plus my function evaluated at 0. And let's see if we can use this and this to figure out some more useful Laplace transforms. Well, what is the Laplace transform of f of t is equal to t? Well, let's just use this property."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "Times the Laplace transform of my derivative plus my function evaluated at 0. And let's see if we can use this and this to figure out some more useful Laplace transforms. Well, what is the Laplace transform of f of t is equal to t? Well, let's just use this property. This is going to be equal to 1 over s times the Laplace transform of the derivative. Well, what's the derivative of t? The derivative of t is 1."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "Well, let's just use this property. This is going to be equal to 1 over s times the Laplace transform of the derivative. Well, what's the derivative of t? The derivative of t is 1. So it's the Laplace transform of 1 minus f of 0. When t equals 0, this becomes 0. Minus 0."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "The derivative of t is 1. So it's the Laplace transform of 1 minus f of 0. When t equals 0, this becomes 0. Minus 0. So the Laplace transform of t is equal to 1 over s times the Laplace transform of 1. Well, that's just 1 over s. So it's 1 over s squared minus 0. Interesting."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "Minus 0. So the Laplace transform of t is equal to 1 over s times the Laplace transform of 1. Well, that's just 1 over s. So it's 1 over s squared minus 0. Interesting. Laplace transform of 1 is 1 over s. Laplace transform of t is 1 over s squared. Let's figure out what the Laplace transform of t squared is. And I'll do this one in green."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "Interesting. Laplace transform of 1 is 1 over s. Laplace transform of t is 1 over s squared. Let's figure out what the Laplace transform of t squared is. And I'll do this one in green. Maybe we'll see a pattern emerge. The Laplace transform of t squared, well, it equals 1 over s times the Laplace transform of its derivative. So what's its derivative?"}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "And I'll do this one in green. Maybe we'll see a pattern emerge. The Laplace transform of t squared, well, it equals 1 over s times the Laplace transform of its derivative. So what's its derivative? Times the Laplace transform of 2t plus this evaluated 0. Well, that's just 0. So this is equal to, well, we could just take this constant out."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So what's its derivative? Times the Laplace transform of 2t plus this evaluated 0. Well, that's just 0. So this is equal to, well, we could just take this constant out. This is equal to 2 over s times the Laplace transform of t. Well, what does that equal? That is equal to, we just solved it, 1 over s squared. So it's 2 over s times 1 over s squared."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So this is equal to, well, we could just take this constant out. This is equal to 2 over s times the Laplace transform of t. Well, what does that equal? That is equal to, we just solved it, 1 over s squared. So it's 2 over s times 1 over s squared. So it's equal to 2 over s to the third. Fascinating. Well, let me ask you a, well, let me just do t to the third."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So it's 2 over s times 1 over s squared. So it's equal to 2 over s to the third. Fascinating. Well, let me ask you a, well, let me just do t to the third. And I think then you'll see the pattern. The pattern will emerge. The Laplace transform."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "Well, let me ask you a, well, let me just do t to the third. And I think then you'll see the pattern. The pattern will emerge. The Laplace transform. And this is actually kind of fun. I recommend you do it. It's somehow satisfying."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "The Laplace transform. And this is actually kind of fun. I recommend you do it. It's somehow satisfying. It's much more satisfying than integration by parts. So the Laplace transform of 2 to the third is 1 over s times the Laplace transform of its derivative, which is 3t squared, which is, take the constant out, because it's a linear operator, 3 over s times the Laplace transform of t squared. So it equals what?"}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "It's somehow satisfying. It's much more satisfying than integration by parts. So the Laplace transform of 2 to the third is 1 over s times the Laplace transform of its derivative, which is 3t squared, which is, take the constant out, because it's a linear operator, 3 over s times the Laplace transform of t squared. So it equals what? What's the Laplace transform of t squared? It's 2 over s to the third. So this equals 3 times 2 over what?"}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So it equals what? What's the Laplace transform of t squared? It's 2 over s to the third. So this equals 3 times 2 over what? s to the fourth. And you could put a t over n here and use an inductive argument to figure out a general formula. And that general formula is, and I think you see the pattern here, whatever my exponent is, the Laplace transform has an s in the denominator with one larger exponent."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So this equals 3 times 2 over what? s to the fourth. And you could put a t over n here and use an inductive argument to figure out a general formula. And that general formula is, and I think you see the pattern here, whatever my exponent is, the Laplace transform has an s in the denominator with one larger exponent. And then the numerator is the factorial of my exponent. So in general, and this is one more entry in our Laplace transform table, the Laplace transform of t to the nth power is equal to n factorial over s to the n plus 1. That's a parentheses."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "And that general formula is, and I think you see the pattern here, whatever my exponent is, the Laplace transform has an s in the denominator with one larger exponent. And then the numerator is the factorial of my exponent. So in general, and this is one more entry in our Laplace transform table, the Laplace transform of t to the nth power is equal to n factorial over s to the n plus 1. That's a parentheses. I guess I didn't have to write those parentheses. It just confuses it. But anyway, this sometimes looks like a fairly, you know when you see this in a Laplace transform table, it seems intimidating."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "That's a parentheses. I guess I didn't have to write those parentheses. It just confuses it. But anyway, this sometimes looks like a fairly, you know when you see this in a Laplace transform table, it seems intimidating. Oh boy, I have n's and I have n factorials and all that. But it's just saying with this pattern we showed, whatever, you know, t to the third, increase it by 1. So s to the fourth, put it in the denominator and take 3 factorial on the numerator, which is 6."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "But anyway, this sometimes looks like a fairly, you know when you see this in a Laplace transform table, it seems intimidating. Oh boy, I have n's and I have n factorials and all that. But it's just saying with this pattern we showed, whatever, you know, t to the third, increase it by 1. So s to the fourth, put it in the denominator and take 3 factorial on the numerator, which is 6. And that's all it is. So we have, in using the derivative property of a Laplace transform, we figured out the Laplace transform of cosine of at and the Laplace transform of any, well really any polynomial, right? Because it's a linear operator."}, {"video_title": "Laplace transform of cos t and polynomials Laplace transform Khan Academy.mp3", "Sentence": "So s to the fourth, put it in the denominator and take 3 factorial on the numerator, which is 6. And that's all it is. So we have, in using the derivative property of a Laplace transform, we figured out the Laplace transform of cosine of at and the Laplace transform of any, well really any polynomial, right? Because it's a linear operator. So now we know t to the nth power, t to the whatever power and we can multiply it by constants. So we know the basic trig functions, we know the exponential function, and we know how to take the Laplace transform of polynomials. See you in the next video."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "So I encourage you to pause this video and see if you can figure this out on your own. All right, now let's work through it together. So some of you might have immediately said, hey, this is the form of a differential equation where the solution is going to be an exponential and you just got right to it. But I'm not gonna go straight to that. I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way. So when I say it's separable, that means we can separate all the y's, d y's on one side and all the x's, d x's on the other side. And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "But I'm not gonna go straight to that. I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way. So when I say it's separable, that means we can separate all the y's, d y's on one side and all the x's, d x's on the other side. And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x. One, two, three, d x. Now on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable differential equations."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x. One, two, three, d x. Now on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable differential equations. Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here. Now I could add a constant, but I'm gonna add a constant on the right hand side so there's no reason to add two arbitrary constants on both sides. I could just add one on one side."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "That's what people talk about when they say separable differential equations. Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here. Now I could add a constant, but I'm gonna add a constant on the right hand side so there's no reason to add two arbitrary constants on both sides. I could just add one on one side. So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there. And now let's think about it a little bit. Well, we can rewrite this in exponential form."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "I could just add one on one side. So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there. And now let's think about it a little bit. Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're gonna be different values, but we're just trying to just get a sense of what the structure of this thing looks like. So we could say this is going to be some constant times e to the three x. So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're gonna be different values, but we're just trying to just get a sense of what the structure of this thing looks like. So we could say this is going to be some constant times e to the three x. So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet. We're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the three x or y is equal to negative c e to the three x. Well, we've kept it in general terms."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet. We're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the three x or y is equal to negative c e to the three x. Well, we've kept it in general terms. I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here. And so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration, whether we're going positive or negative. So let's do that."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we've kept it in general terms. I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here. And so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration, whether we're going positive or negative. So let's do that. So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So I could write it like that. And we get two is equal to c times e to the third power, three times one."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So I could write it like that. And we get two is equal to c times e to the third power, three times one. And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in. And our particular solution is going to be y is equal to c. C is two e to the negative third power times e to the three x. Now I have, I'm taking the product of two things with the same base."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "And we get two is equal to c times e to the third power, three times one. And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in. And our particular solution is going to be y is equal to c. C is two e to the negative third power times e to the three x. Now I have, I'm taking the product of two things with the same base. I can add the exponents. So I could say y is equal to two times e to the three x, e to the three x, and then I'll add the exponents to three x minus three. And there you go."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "But anyway, the problem we have here, it's the derivative of y with respect to x is equal to x squared plus 3y squared, I'm writing it a little bit small so that I'll run a space, divided by 2xy. So with all of these homogenous equations, and obviously we don't know if it's homogenous yet, so we have to try to write it as a function of y divided by x. It looks like we could do that if we divide the top and the bottom by x squared. So if we just multiply, we do it in a different color. 1 over x squared, or x to the negative 2, over 1 over x squared, right? We're essentially just multiplying by 1. And then we get that is equal to what?"}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So if we just multiply, we do it in a different color. 1 over x squared, or x to the negative 2, over 1 over x squared, right? We're essentially just multiplying by 1. And then we get that is equal to what? 1 plus 3y squared over x squared, divided by 2. If you divide x divided by x squared, you just get a 1 over x, so 2 times y over x. Or we could just rewrite the whole thing, and we get this is just equal to 1 plus 3y over x squared, divided by 2 times y over x."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "And then we get that is equal to what? 1 plus 3y squared over x squared, divided by 2. If you divide x divided by x squared, you just get a 1 over x, so 2 times y over x. Or we could just rewrite the whole thing, and we get this is just equal to 1 plus 3y over x squared, divided by 2 times y over x. So yes, this is a homogenous equation, because we were able to write it as a function of y divided by x. So now we can do the substitution with v, and hopefully this is starting to become a little bit of second nature to you. So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "Or we could just rewrite the whole thing, and we get this is just equal to 1 plus 3y over x squared, divided by 2 times y over x. So yes, this is a homogenous equation, because we were able to write it as a function of y divided by x. So now we can do the substitution with v, and hopefully this is starting to become a little bit of second nature to you. So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v. This is just the product rule. Plus x times the derivative of v with respect to x. And now we can substitute."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v. This is just the product rule. Plus x times the derivative of v with respect to x. And now we can substitute. The derivative of y with respect to x is just this. And then the right-hand side of the equation is this, but we can substitute v for y over x. So let's do that."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "And now we can substitute. The derivative of y with respect to x is just this. And then the right-hand side of the equation is this, but we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv, dvx, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So let's do that. And so we get v plus x. Instead of dv, dvx, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared. We're making the substitution v is equal to y over x. 3v squared, all of that over 2v. Now let's see what we can do."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "v prime is equal to 1 plus 3v squared. We're making the substitution v is equal to y over x. 3v squared, all of that over 2v. Now let's see what we can do. This is where we just put our algebra hat on and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. Let's see, we'll get 2v squared plus 2xvv prime."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "Now let's see what we can do. This is where we just put our algebra hat on and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. Let's see, we'll get 2v squared plus 2xvv prime. Write 2v times that. Yep, that's 2xvv prime. Is equal to 1 plus 3v squared."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "Let's see, we'll get 2v squared plus 2xvv prime. Write 2v times that. Yep, that's 2xvv prime. Is equal to 1 plus 3v squared. Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xvv prime is equal to 1 plus, let's see, we're subtracting 2v squared from both sides, so we just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "Is equal to 1 plus 3v squared. Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xvv prime is equal to 1 plus, let's see, we're subtracting 2v squared from both sides, so we just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left-hand side. So we get 2xvv prime divided by 1 plus v squared is equal to 1, and let's divide both sides by x. So we get the x's on the other side."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left-hand side. So we get 2xvv prime divided by 1 plus v squared is equal to 1, and let's divide both sides by x. So we get the x's on the other side. So then we get 2v, and now I'll switch back to the other notation, instead of v prime I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to, I'm dividing both sides by x. Notice I didn't write the x on this side, so that is equal to 1 over x."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So we get the x's on the other side. So then we get 2v, and now I'll switch back to the other notation, instead of v prime I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to, I'm dividing both sides by x. Notice I didn't write the x on this side, so that is equal to 1 over x. And then if we just multiply both sides of this times dx, we've separated the two variables and we can integrate both sides, so let's do that. Let's go up here, I'll switch to a different color so that you know I'm working on a different column now. So multiply both sides by dx, I get 2v over 1 plus v squared dv is equal to 1 over x dx."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "Notice I didn't write the x on this side, so that is equal to 1 over x. And then if we just multiply both sides of this times dx, we've separated the two variables and we can integrate both sides, so let's do that. Let's go up here, I'll switch to a different color so that you know I'm working on a different column now. So multiply both sides by dx, I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this?"}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So multiply both sides by dx, I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first you might think, oh boy, this is complicated, this is difficult, maybe some type of trig function, but you'll see that it's kind of just the reverse chain rule. We have a function here, 1 plus v squared, an expression here, we have its derivative sitting right there. So the antiderivative of this, and you could actually make a substitution if you like, you could say u is equal to 1 plus v squared, then du is equal to 2v dv, and then you'd end up saying that the antiderivative is just the natural log of u, or in this case, the antiderivative of this is just the natural log of 1 plus v squared."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "And what's the integral of this? At first you might think, oh boy, this is complicated, this is difficult, maybe some type of trig function, but you'll see that it's kind of just the reverse chain rule. We have a function here, 1 plus v squared, an expression here, we have its derivative sitting right there. So the antiderivative of this, and you could actually make a substitution if you like, you could say u is equal to 1 plus v squared, then du is equal to 2v dv, and then you'd end up saying that the antiderivative is just the natural log of u, or in this case, the antiderivative of this is just the natural log of 1 plus v squared. We don't have to even write an absolute value there, because that's always going to be a positive value. So the natural log of 1 plus v squared. And I hope I didn't confuse you."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So the antiderivative of this, and you could actually make a substitution if you like, you could say u is equal to 1 plus v squared, then du is equal to 2v dv, and then you'd end up saying that the antiderivative is just the natural log of u, or in this case, the antiderivative of this is just the natural log of 1 plus v squared. We don't have to even write an absolute value there, because that's always going to be a positive value. So the natural log of 1 plus v squared. And I hope I didn't confuse you. That's how I think about it. I say if I have an expression and I have its derivative multiplied there, then I can just take the antiderivative of the whole expression, and I don't have to worry about what's inside of it. So if this was a 1 over an x or 1 over u, it's just the natural log of it."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "And I hope I didn't confuse you. That's how I think about it. I say if I have an expression and I have its derivative multiplied there, then I can just take the antiderivative of the whole expression, and I don't have to worry about what's inside of it. So if this was a 1 over an x or 1 over u, it's just the natural log of it. So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully it'll make a little bit more sense."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So if this was a 1 over an x or 1 over u, it's just the natural log of it. So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully it'll make a little bit more sense. But anyway, that's the left-hand side, and then that equals, well, this one's easy. That's the natural log, say the absolute value of x, plus, we could say plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that as the natural log of the absolute value of some constant c, right? I mean, this is still some arbitrary constant c. So we could rewrite this whole equation as the natural log of 1 plus v squared is equal to, when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "And hopefully it'll make a little bit more sense. But anyway, that's the left-hand side, and then that equals, well, this one's easy. That's the natural log, say the absolute value of x, plus, we could say plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that as the natural log of the absolute value of some constant c, right? I mean, this is still some arbitrary constant c. So we could rewrite this whole equation as the natural log of 1 plus v squared is equal to, when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of. The natural log of, we could say, the absolute value of cx. And so the natural log of this is equal to the natural log of this, right? So we could say that 1 plus v squared is equal to cx."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "I mean, this is still some arbitrary constant c. So we could rewrite this whole equation as the natural log of 1 plus v squared is equal to, when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of. The natural log of, we could say, the absolute value of cx. And so the natural log of this is equal to the natural log of this, right? So we could say that 1 plus v squared is equal to cx. And now we can unsubstitute. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So we could say that 1 plus v squared is equal to cx. And now we can unsubstitute. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's see. Let's multiply both sides of this equation times x squared."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's see. Let's multiply both sides of this equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "Let's multiply both sides of this equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done. If we want to put all of the variable terms on the left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function or curve or whatever you want to call it is the solution to our original homogenous first order differential equation. So there you go."}, {"video_title": "First order homogeneous equations 2 First order differential equations Khan Academy.mp3", "Sentence": "And we're essentially done. If we want to put all of the variable terms on the left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function or curve or whatever you want to call it is the solution to our original homogenous first order differential equation. So there you go. I will see you in the next video. And now we're actually going to do something. We're going to start embarking on higher order differential equations, and actually these are more useful and in some ways easier to do than the homogenous and the exact equations that we've been doing so far."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And we already found some constant solutions. We can think through that a little bit, just as a little bit of a review from the last few videos. So this is the t-axis and this is the n-axis. We already saw that if n of zero, if at time equals zero, our population is zero, there's no one to reproduce, and this differential equation is consistent with that because if n is zero, this thing is going to be zero, and so our rate of change is going to be zero with respect to time, and so our population just won't change. It'll just stay at zero, which is nice because that's what would actually happen in a real population. So that gives us the constant solution that n of t is equal to zero. That's one solution to this differential equation."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "We already saw that if n of zero, if at time equals zero, our population is zero, there's no one to reproduce, and this differential equation is consistent with that because if n is zero, this thing is going to be zero, and so our rate of change is going to be zero with respect to time, and so our population just won't change. It'll just stay at zero, which is nice because that's what would actually happen in a real population. So that gives us the constant solution that n of t is equal to zero. That's one solution to this differential equation. Not that interesting. A zero population will never grow or change. The other constant solution is, well, what if our population started at the maximum of what the environment could sustain?"}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "That's one solution to this differential equation. Not that interesting. A zero population will never grow or change. The other constant solution is, well, what if our population started at the maximum of what the environment could sustain? And in that situation, this term is going to be k over k, which is one. One minus one is zero, and so in there, the population would also not change. It would just stay at k, and so the rate of change would just stay at zero."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "The other constant solution is, well, what if our population started at the maximum of what the environment could sustain? And in that situation, this term is going to be k over k, which is one. One minus one is zero, and so in there, the population would also not change. It would just stay at k, and so the rate of change would just stay at zero. So that's another constant solution that we start at the maximum population and then this differential equation tells us a scenario that never changes. But we said, hey, look, well, there could be something interesting that happens if our initial condition, which we called n sub naught, that's our n of zero, time equals zero if we start someplace in between, maybe closer, much below our eventual ceiling, I guess, because if we're a lot below our eventual ceiling, this thing is going to be a small fraction. This thing is going to be close to one, and so the rate of change is going to be pretty close to being proportional to n, which is going to, and when we're thinking that, we can kind of, it might look something like that."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "It would just stay at k, and so the rate of change would just stay at zero. So that's another constant solution that we start at the maximum population and then this differential equation tells us a scenario that never changes. But we said, hey, look, well, there could be something interesting that happens if our initial condition, which we called n sub naught, that's our n of zero, time equals zero if we start someplace in between, maybe closer, much below our eventual ceiling, I guess, because if we're a lot below our eventual ceiling, this thing is going to be a small fraction. This thing is going to be close to one, and so the rate of change is going to be pretty close to being proportional to n, which is going to, and when we're thinking that, we can kind of, it might look something like that. As n increases, our rate of change increases, but then as n approaches k, this thing is going to approach one, this thing is going to approach zero, and so it's going to overwhelm this and our rate of change is going to approach zero. And so we could imagine a scenario where we asymptote towards k. And let's see if we can solve this to find this, this n of t, this n of t, to find the actual analytic expression for this n of t right over here, because this is interesting. This is what could be used to model populations that would be more consistent with a Malthusian mindset."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "This thing is going to be close to one, and so the rate of change is going to be pretty close to being proportional to n, which is going to, and when we're thinking that, we can kind of, it might look something like that. As n increases, our rate of change increases, but then as n approaches k, this thing is going to approach one, this thing is going to approach zero, and so it's going to overwhelm this and our rate of change is going to approach zero. And so we could imagine a scenario where we asymptote towards k. And let's see if we can solve this to find this, this n of t, this n of t, to find the actual analytic expression for this n of t right over here, because this is interesting. This is what could be used to model populations that would be more consistent with a Malthusian mindset. So let's see if we can do that. And to do that, we just have to realize this is a completely different, this is a separable differential equation, and we're assuming as a function of t, we're going to solve for an n of t that satisfies this, and so we just have to separate it from the explicit t's, but there are no explicit t's here, so it's quite easy to do. So what I'm going to do is I'm going to take this part right over here, I'm going to divide both sides by that."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "This is what could be used to model populations that would be more consistent with a Malthusian mindset. So let's see if we can do that. And to do that, we just have to realize this is a completely different, this is a separable differential equation, and we're assuming as a function of t, we're going to solve for an n of t that satisfies this, and so we just have to separate it from the explicit t's, but there are no explicit t's here, so it's quite easy to do. So what I'm going to do is I'm going to take this part right over here, I'm going to divide both sides by that. I'm going to leave the r on the right-hand side. That'll make things, I think, a little bit easier as we try to solve for n of t, so let me just do that. So this is going to be equal to one over n times one minus n over k. One minus n over k times dn dt, times dn dt is equal to r, is equal to r, and another way we could think about it, oh, actually, let me just continue to tackle it this way."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So what I'm going to do is I'm going to take this part right over here, I'm going to divide both sides by that. I'm going to leave the r on the right-hand side. That'll make things, I think, a little bit easier as we try to solve for n of t, so let me just do that. So this is going to be equal to one over n times one minus n over k. One minus n over k times dn dt, times dn dt is equal to r, is equal to r, and another way we could think about it, oh, actually, let me just continue to tackle it this way. So we get that, and now what I want to do is take the antiderivative of both sides with respect to t. Well, this is pretty straightforward. That's just going to be rt times some constant, but what's this going to be? This is kind of this messy thing, and maybe if we could break this out, if we can expand this into two fractions, do a little bit of partial fraction expansion, maybe we can come up with an expression that's a little bit easier to find the antiderivative of, and so I'm hoping that I can find an a and a b where a over n plus b over one minus n over k is equal to this business, is equal to one over n times one minus n over k. Let's see if we can find an a and a b."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So this is going to be equal to one over n times one minus n over k. One minus n over k times dn dt, times dn dt is equal to r, is equal to r, and another way we could think about it, oh, actually, let me just continue to tackle it this way. So we get that, and now what I want to do is take the antiderivative of both sides with respect to t. Well, this is pretty straightforward. That's just going to be rt times some constant, but what's this going to be? This is kind of this messy thing, and maybe if we could break this out, if we can expand this into two fractions, do a little bit of partial fraction expansion, maybe we can come up with an expression that's a little bit easier to find the antiderivative of, and so I'm hoping that I can find an a and a b where a over n plus b over one minus n over k is equal to this business, is equal to one over n times one minus n over k. Let's see if we can find an a and a b. This is just partial fraction expansion. If this looks unfamiliar to you, I encourage you to review that part on Khan Academy. Do a search for partial fraction expansion on Khan Academy."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "This is kind of this messy thing, and maybe if we could break this out, if we can expand this into two fractions, do a little bit of partial fraction expansion, maybe we can come up with an expression that's a little bit easier to find the antiderivative of, and so I'm hoping that I can find an a and a b where a over n plus b over one minus n over k is equal to this business, is equal to one over n times one minus n over k. Let's see if we can find an a and a b. This is just partial fraction expansion. If this looks unfamiliar to you, I encourage you to review that part on Khan Academy. Do a search for partial fraction expansion on Khan Academy. So how do we do it? Well, let's just add these two right over here. So this, if we take the sum right over here, this is going to be a times this, which is a minus a over kn plus b times this, plus bn, over the product of these two."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Do a search for partial fraction expansion on Khan Academy. So how do we do it? Well, let's just add these two right over here. So this, if we take the sum right over here, this is going to be a times this, which is a minus a over kn plus b times this, plus bn, over the product of these two. So it's just going to be n times one minus n over k. One way to think about it, I just multiply the numerator and the denominator of this one by one minus n over k. So a times one minus n over k is that. N times one minus n over k is that. And I multiply the numerator and denominator of this both by n, bn, and n times that."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So this, if we take the sum right over here, this is going to be a times this, which is a minus a over kn plus b times this, plus bn, over the product of these two. So it's just going to be n times one minus n over k. One way to think about it, I just multiply the numerator and the denominator of this one by one minus n over k. So a times one minus n over k is that. N times one minus n over k is that. And I multiply the numerator and denominator of this both by n, bn, and n times that. And then I added the two now that I had the same denominator. So that's just adding fractions with unlike denominators. It's going to be equal to this, one over n times one minus n over k. And now we can try to think, well, what is a and b going to be equal to?"}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And I multiply the numerator and denominator of this both by n, bn, and n times that. And then I added the two now that I had the same denominator. So that's just adding fractions with unlike denominators. It's going to be equal to this, one over n times one minus n over k. And now we can try to think, well, what is a and b going to be equal to? Or what can they be equal to? Well, I have a constant term here. I don't have any n term here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "It's going to be equal to this, one over n times one minus n over k. And now we can try to think, well, what is a and b going to be equal to? Or what can they be equal to? Well, I have a constant term here. I don't have any n term here. I could say that maybe I have a zero times an n. And that actually helps things a little bit. Because maybe we could say that this thing, that this plus this, which are the coefficients on n, are going to sum up to zero. And that this, which is our a, is going to be equal to one."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "I don't have any n term here. I could say that maybe I have a zero times an n. And that actually helps things a little bit. Because maybe we could say that this thing, that this plus this, which are the coefficients on n, are going to sum up to zero. And that this, which is our a, is going to be equal to one. And that's pretty nice. So we could say that a is equal to one. a is equal to one."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And that this, which is our a, is going to be equal to one. And that's pretty nice. So we could say that a is equal to one. a is equal to one. And then if a is equal to one, we have negative one over k, negative one over k plus b, plus b is equal to zero. Well, what's b going to be? Well, b would be equal to one, one over k. So we can rewrite this as, we can rewrite it as one over n, one over n, one over n plus, one over n plus one over k, plus one over k over, over, all over, let me just do this, over one minus n over k. One minus n over k, over k, and then, and then times, times dn dt, dn dt is equal to, is equal to r. So I just did a little bit of partial fraction expansion."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "a is equal to one. And then if a is equal to one, we have negative one over k, negative one over k plus b, plus b is equal to zero. Well, what's b going to be? Well, b would be equal to one, one over k. So we can rewrite this as, we can rewrite it as one over n, one over n, one over n plus, one over n plus one over k, plus one over k over, over, all over, let me just do this, over one minus n over k. One minus n over k, over k, and then, and then times, times dn dt, dn dt is equal to, is equal to r. So I just did a little bit of partial fraction expansion. I don't wanna give myself a little bit of real estate. Let me clear this out. You can rewind the video and review that if you find it necessary."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Well, b would be equal to one, one over k. So we can rewrite this as, we can rewrite it as one over n, one over n, one over n plus, one over n plus one over k, plus one over k over, over, all over, let me just do this, over one minus n over k. One minus n over k, over k, and then, and then times, times dn dt, dn dt is equal to, is equal to r. So I just did a little bit of partial fraction expansion. I don't wanna give myself a little bit of real estate. Let me clear this out. You can rewind the video and review that if you find it necessary. So let's do that. And so how does this help us? Well, hey, you know, I kinda, you probably might be recognizing the antiderivative of one over n, and you might even see this."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "You can rewind the video and review that if you find it necessary. So let's do that. And so how does this help us? Well, hey, you know, I kinda, you probably might be recognizing the antiderivative of one over n, and you might even see this. So let's just think through this a little bit. We know the antiderivative of one over n is the natural log, I'm just in a new color, is the natural log of the absolute value of n. And we can see the derivative of that with respect to n is equal to one over n. And if we were to find the derivative of this with respect to t, derivative with respect to t of the natural log of the absolute value of n is equal to, this is going to be the derivative of this with respect to n times the derivative of n with respect to t, times dn dt. And we could do the same thing over here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Well, hey, you know, I kinda, you probably might be recognizing the antiderivative of one over n, and you might even see this. So let's just think through this a little bit. We know the antiderivative of one over n is the natural log, I'm just in a new color, is the natural log of the absolute value of n. And we can see the derivative of that with respect to n is equal to one over n. And if we were to find the derivative of this with respect to t, derivative with respect to t of the natural log of the absolute value of n is equal to, this is going to be the derivative of this with respect to n times the derivative of n with respect to t, times dn dt. And we could do the same thing over here. Notice, I have, I have an expression. What would be the derivative of this expression right over here? It would be negative one over k. If I'm talking about the derivative with respect to n. It would be negative one over k. I have a positive one over k here, and I can even make it negative."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And we could do the same thing over here. Notice, I have, I have an expression. What would be the derivative of this expression right over here? It would be negative one over k. If I'm talking about the derivative with respect to n. It would be negative one over k. I have a positive one over k here, and I can even make it negative. I can turn this, I can clear this out of the way. Instead of having a positive there, I could have a, I guess I could say a double negative. So a double negative, I haven't changed the value of it."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "It would be negative one over k. If I'm talking about the derivative with respect to n. It would be negative one over k. I have a positive one over k here, and I can even make it negative. I can turn this, I can clear this out of the way. Instead of having a positive there, I could have a, I guess I could say a double negative. So a double negative, I haven't changed the value of it. And notice, now this is the derivative of this, which makes it very nice for you substitution, or you might be already used to doing this thing in your head. And so we know that the derivative with respect to n, let me write this in a new color. We know that the derivative with respect to n of the natural log of one minus n over k, once again, this comes straight out of the chain rule, it's going to be the derivative of this with respect to n, which is negative one over k, times the derivative of this whole thing with respect to this thing, which is times one over one minus n over k, which is exactly what we have over here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So a double negative, I haven't changed the value of it. And notice, now this is the derivative of this, which makes it very nice for you substitution, or you might be already used to doing this thing in your head. And so we know that the derivative with respect to n, let me write this in a new color. We know that the derivative with respect to n of the natural log of one minus n over k, once again, this comes straight out of the chain rule, it's going to be the derivative of this with respect to n, which is negative one over k, times the derivative of this whole thing with respect to this thing, which is times one over one minus n over k, which is exactly what we have over here. But if I were to take the derivative of this with respect to t, so the derivative with respect to t of the natural log of one minus n over k, it's going to be the derivative of this with respect to n, which is that, which is what we just found. It's going to be this. So copy and paste."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "We know that the derivative with respect to n of the natural log of one minus n over k, once again, this comes straight out of the chain rule, it's going to be the derivative of this with respect to n, which is negative one over k, times the derivative of this whole thing with respect to this thing, which is times one over one minus n over k, which is exactly what we have over here. But if I were to take the derivative of this with respect to t, so the derivative with respect to t of the natural log of one minus n over k, it's going to be the derivative of this with respect to n, which is that, which is what we just found. It's going to be this. So copy and paste. It's going to be that. It's going to be that, times the derivative of n with respect to t, just straight out of the chain rule, dn dt. Well, notice, we have a dn dt."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So copy and paste. It's going to be that. It's going to be that, times the derivative of n with respect to t, just straight out of the chain rule, dn dt. Well, notice, we have a dn dt. We can multiply it times each of these things. And actually, why not? Let's just do that just to make it clear, because this is really, you know, this isn't so much differential equations, but sometimes some of the calculus that we learned not too long ago, and even frankly, some of the algebra, it's nice to not skip steps."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Well, notice, we have a dn dt. We can multiply it times each of these things. And actually, why not? Let's just do that just to make it clear, because this is really, you know, this isn't so much differential equations, but sometimes some of the calculus that we learned not too long ago, and even frankly, some of the algebra, it's nice to not skip steps. So if I distribute this, edit, paste, whoops. I wanted to copy and paste. So copy and paste."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Let's just do that just to make it clear, because this is really, you know, this isn't so much differential equations, but sometimes some of the calculus that we learned not too long ago, and even frankly, some of the algebra, it's nice to not skip steps. So if I distribute this, edit, paste, whoops. I wanted to copy and paste. So copy and paste. So I have that. And then I have, and then I have, I can copy and paste. So that, I'm just distributing, I'm just distributing this business right over here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So copy and paste. So I have that. And then I have, and then I have, I can copy and paste. So that, I'm just distributing, I'm just distributing this business right over here. So I get that. Now let me clean it up a little bit. So I have that right over here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So that, I'm just distributing, I'm just distributing this business right over here. So I get that. Now let me clean it up a little bit. So I have that right over here. And of course, we have this being equal to r. If I take the antiderivative of this with respect to t, well, I'm just gonna get, I'm just gonna get this. I am just going to get this. Notice the anti, I'm just gonna get this, and plus, or this minus, this minus this right over that."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So I have that right over here. And of course, we have this being equal to r. If I take the antiderivative of this with respect to t, well, I'm just gonna get, I'm just gonna get this. I am just going to get this. Notice the anti, I'm just gonna get this, and plus, or this minus, this minus this right over that. So let's do that. Let's take the antiderivative with respect to t. So the left-hand side, I'm gonna get the natural log, this, the antiderivative of that with respect to t is the natural log of the absolute value of n. And then minus the antiderivative of this with respect to t is the natural log, the natural log of the absolute value of one minus n over k is equal to, oh, and let's not forget, let's say it's, and we're gonna have some constant here. I wanna find a pretty general solution."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Notice the anti, I'm just gonna get this, and plus, or this minus, this minus this right over that. So let's do that. Let's take the antiderivative with respect to t. So the left-hand side, I'm gonna get the natural log, this, the antiderivative of that with respect to t is the natural log of the absolute value of n. And then minus the antiderivative of this with respect to t is the natural log, the natural log of the absolute value of one minus n over k is equal to, oh, and let's not forget, let's say it's, and we're gonna have some constant here. I wanna find a pretty general solution. Let's call it c one, is equal to the antiderivative of this with respect to t is r times t, maybe plus some other constant, plus some other constant, just like that. And now I'm going to assume that my n of t meets this assumption right over here. And so I'm going to assume, I'm going to, so assuming, assuming, assume, assume n of t is going to be less than k and greater than zero."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "I wanna find a pretty general solution. Let's call it c one, is equal to the antiderivative of this with respect to t is r times t, maybe plus some other constant, plus some other constant, just like that. And now I'm going to assume that my n of t meets this assumption right over here. And so I'm going to assume, I'm going to, so assuming, assuming, assume, assume n of t is going to be less than k and greater than zero. That means that this thing, this n is always going to be positive. And if n is always between zero and k, that means that this thing is always going to be positive. And so that helps me clear things up a little bit."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And so I'm going to assume, I'm going to, so assuming, assuming, assume, assume n of t is going to be less than k and greater than zero. That means that this thing, this n is always going to be positive. And if n is always between zero and k, that means that this thing is always going to be positive. And so that helps me clear things up a little bit. Actually, let me do it in this color. That helps me get rid of that, helps me get rid of, helps me get rid of that. I can instead throw some parentheses here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And so that helps me clear things up a little bit. Actually, let me do it in this color. That helps me get rid of that, helps me get rid of, helps me get rid of that. I can instead throw some parentheses here. And why don't I subtract the c one from both sides? So let me, that would get rid of it here. So edit, cut, and paste."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "I can instead throw some parentheses here. And why don't I subtract the c one from both sides? So let me, that would get rid of it here. So edit, cut, and paste. So I have the c one there, and I know I'm overwriting my own work, which is not making it look as clean as possible. I'm just trying to clean it up a little bit. And now, hey, I have an arbitrary constant that I haven't really solved for yet, minus another constant."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So edit, cut, and paste. So I have the c one there, and I know I'm overwriting my own work, which is not making it look as clean as possible. I'm just trying to clean it up a little bit. And now, hey, I have an arbitrary constant that I haven't really solved for yet, minus another constant. Well, let me just call this some other constant. Let me just call it a general c. So let me just call this, let me just call this c. So clear, and I'm just going to call that c. I am just going to call that c. And I could even simplify this a little bit. But actually, I just realized I'm 13 minutes into this video, which is longer than I like to do."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And now, hey, I have an arbitrary constant that I haven't really solved for yet, minus another constant. Well, let me just call this some other constant. Let me just call it a general c. So let me just call this, let me just call this c. So clear, and I'm just going to call that c. I am just going to call that c. And I could even simplify this a little bit. But actually, I just realized I'm 13 minutes into this video, which is longer than I like to do. So let's continue this in the next video. I was getting excited, because I'm so close, I'm so close to solving for an n of t that satisfies our logistic differential equation. Very, very exciting."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And if you have your initial conditions, you can solve for c1 and c2. But the question I'm asking is, what happens when you have two complex roots? Or essentially, when you're trying to solve the characteristic equation, when you're trying to solve that quadratic, the b squared minus 4ac, that's negative. So you get the two roots end up being complex conjugates. And we said, OK, let's say that our two roots are lambda plus or minus mu i. And we just did a bunch of algebra to, we said, well, if those are the roots and we substitute it back into this formula for the general solution, we get all of this. And we kept simplifying it all the way until we got here, where we said y is equal to e to the lambda x plus c1, et cetera, et cetera."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "So you get the two roots end up being complex conjugates. And we said, OK, let's say that our two roots are lambda plus or minus mu i. And we just did a bunch of algebra to, we said, well, if those are the roots and we substitute it back into this formula for the general solution, we get all of this. And we kept simplifying it all the way until we got here, where we said y is equal to e to the lambda x plus c1, et cetera, et cetera. And we said, can we simplify this further? And that's where we took out Euler's equation, or Euler's formula, or Euler's definition, depending on what you want, which I'm always in awe of every time I see it or use it. But we've talked a lot about that in the calculus playlist."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And we kept simplifying it all the way until we got here, where we said y is equal to e to the lambda x plus c1, et cetera, et cetera. And we said, can we simplify this further? And that's where we took out Euler's equation, or Euler's formula, or Euler's definition, depending on what you want, which I'm always in awe of every time I see it or use it. But we've talked a lot about that in the calculus playlist. We could use this to maybe further simplify it. So I wrote e to the mu xi as cosine mu x plus i sine mu x. And I wrote e to the minus mu xi as cosine minus mu x plus i sine minus mu x."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "But we've talked a lot about that in the calculus playlist. We could use this to maybe further simplify it. So I wrote e to the mu xi as cosine mu x plus i sine mu x. And I wrote e to the minus mu xi as cosine minus mu x plus i sine minus mu x. And now we could use a little bit about what we know about trigonometry. Cosine of minus theta is equal to cosine of theta. And we also know that sine of minus theta is equal to minus sine of theta."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And I wrote e to the minus mu xi as cosine minus mu x plus i sine minus mu x. And now we could use a little bit about what we know about trigonometry. Cosine of minus theta is equal to cosine of theta. And we also know that sine of minus theta is equal to minus sine of theta. So let's use these identities to simplify this a little bit more. So we get y is equal to e to the lambda x times, and we could actually distribute the c1 too, so times c1 cosine of mu x plus i times c1 sine of mu x plus, all of this is in this parentheses right here, plus c2, instead of cosine of negative mu x, we know this identity, so we could just write this as cosine of mu x as well, because cosine of minus x is the same thing as cosine of x. Plus c i times c2 sine of minus mu x, that's the same thing as minus sine of x."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And we also know that sine of minus theta is equal to minus sine of theta. So let's use these identities to simplify this a little bit more. So we get y is equal to e to the lambda x times, and we could actually distribute the c1 too, so times c1 cosine of mu x plus i times c1 sine of mu x plus, all of this is in this parentheses right here, plus c2, instead of cosine of negative mu x, we know this identity, so we could just write this as cosine of mu x as well, because cosine of minus x is the same thing as cosine of x. Plus c i times c2 sine of minus mu x, that's the same thing as minus sine of x. So actually, let's take the minus sign out there. So minus sine of mu x. And let's see, it seems like we're getting to a point that we can simplify it even more."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "Plus c i times c2 sine of minus mu x, that's the same thing as minus sine of x. So actually, let's take the minus sign out there. So minus sine of mu x. And let's see, it seems like we're getting to a point that we can simplify it even more. We could add the two cosine terms. So we get the general solution, and I know this is a high, this problem requires a lot of algebraic stamina, but as long as you don't make careless mistakes, you'll find it reasonably rewarding, because you'll see where things are coming from. So we get y, the general solution is y is equal to e to the lambda x times, let's add up the two cosine mu x terms."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And let's see, it seems like we're getting to a point that we can simplify it even more. We could add the two cosine terms. So we get the general solution, and I know this is a high, this problem requires a lot of algebraic stamina, but as long as you don't make careless mistakes, you'll find it reasonably rewarding, because you'll see where things are coming from. So we get y, the general solution is y is equal to e to the lambda x times, let's add up the two cosine mu x terms. So it's c1 plus c2 times cosine of mu x, and let's add the two sine of mu x terms. So plus i, we could call that c1i, minus c2i times sine of mu x, and we're almost done simplifying. And the last thing we can simplify is, well, you know, c1 and c2 are arbitrary constants."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "So we get y, the general solution is y is equal to e to the lambda x times, let's add up the two cosine mu x terms. So it's c1 plus c2 times cosine of mu x, and let's add the two sine of mu x terms. So plus i, we could call that c1i, minus c2i times sine of mu x, and we're almost done simplifying. And the last thing we can simplify is, well, you know, c1 and c2 are arbitrary constants. So let's just define this as another constant. I don't know, let's call it, I'll just call it c3, I mean, just to not confuse you by using c1 twice. I'll call this c3."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And the last thing we can simplify is, well, you know, c1 and c2 are arbitrary constants. So let's just define this as another constant. I don't know, let's call it, I'll just call it c3, I mean, just to not confuse you by using c1 twice. I'll call this c3. And now this might be a little bit of a stretch for you, but if you think about it, it really makes sense. This is still just a constant, right? Especially if I say, you know what, I'm not restricting the constants to the reals, c could be an imaginary number."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "I'll call this c3. And now this might be a little bit of a stretch for you, but if you think about it, it really makes sense. This is still just a constant, right? Especially if I say, you know what, I'm not restricting the constants to the reals, c could be an imaginary number. So if c is an imaginary number or some type of complex number, we don't even know whether this is necessarily an imaginary number. So we're not going to make any assumptions about it. Let's just say that this is some other arbitrary constant."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "Especially if I say, you know what, I'm not restricting the constants to the reals, c could be an imaginary number. So if c is an imaginary number or some type of complex number, we don't even know whether this is necessarily an imaginary number. So we're not going to make any assumptions about it. Let's just say that this is some other arbitrary constant. Call this c4, and we can worry about it when we're actually given the initial conditions. But what this gives us, if we make that simplification, we actually get a pretty straightforward general solution to our differential equation where the characteristic equation has complex roots. And that I'll do in a new color."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "Let's just say that this is some other arbitrary constant. Call this c4, and we can worry about it when we're actually given the initial conditions. But what this gives us, if we make that simplification, we actually get a pretty straightforward general solution to our differential equation where the characteristic equation has complex roots. And that I'll do in a new color. That is, y is equal to e to the lambda x times some constant, I'll call it c3, it could be c1, it could be c100, whatever, some constant times cosine of mu of x plus some other constant, I called it c4, it doesn't have to be c4, I just didn't want to confuse it with these, but some other constant times the sine of mu of x. So there's really two things I want you to realize. One is, we haven't done anything different."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And that I'll do in a new color. That is, y is equal to e to the lambda x times some constant, I'll call it c3, it could be c1, it could be c100, whatever, some constant times cosine of mu of x plus some other constant, I called it c4, it doesn't have to be c4, I just didn't want to confuse it with these, but some other constant times the sine of mu of x. So there's really two things I want you to realize. One is, we haven't done anything different. At the end of the day, we still just took the two roots and substituted it back into these equations for r1 and r2. The difference is, we just kept algebraically simplifying it so that we got rid of the i's. That's all we did."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "One is, we haven't done anything different. At the end of the day, we still just took the two roots and substituted it back into these equations for r1 and r2. The difference is, we just kept algebraically simplifying it so that we got rid of the i's. That's all we did. There was really nothing new here except for some algebra and the use of Euler's formula. But when r1 and r2 involved complex numbers, we got to this simplification. So in general, if you get the characteristic equation and your two roots are lambda plus or minus mu i, then the general solution is going to be this."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "That's all we did. There was really nothing new here except for some algebra and the use of Euler's formula. But when r1 and r2 involved complex numbers, we got to this simplification. So in general, if you get the characteristic equation and your two roots are lambda plus or minus mu i, then the general solution is going to be this. And it's not too, if you had to memorize it, although I don't want you to, you should be able to derive this on your own. But it's not too hard to think. And actually, if you ever forget it, just solve your characteristic equation, get your complex numbers, and just substitute it right back in this equation."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "So in general, if you get the characteristic equation and your two roots are lambda plus or minus mu i, then the general solution is going to be this. And it's not too, if you had to memorize it, although I don't want you to, you should be able to derive this on your own. But it's not too hard to think. And actually, if you ever forget it, just solve your characteristic equation, get your complex numbers, and just substitute it right back in this equation. And then with the real numbers, instead of the lambda and the mu, with the real numbers, just do the simplification we did and you'll get to the exact same point. But if you're taking an exam and you don't want to waste time and you want to be able to do something fairly quickly, you can just remember that if I have a complex root to my characteristic equation, lambda plus or minus mu i, then my general solution is e to the lambda x times some constant times cosine of mu x plus some constant times sine of mu x. And let's see if we can do a problem real fast that involves that."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And actually, if you ever forget it, just solve your characteristic equation, get your complex numbers, and just substitute it right back in this equation. And then with the real numbers, instead of the lambda and the mu, with the real numbers, just do the simplification we did and you'll get to the exact same point. But if you're taking an exam and you don't want to waste time and you want to be able to do something fairly quickly, you can just remember that if I have a complex root to my characteristic equation, lambda plus or minus mu i, then my general solution is e to the lambda x times some constant times cosine of mu x plus some constant times sine of mu x. And let's see if we can do a problem real fast that involves that. So let's say I had the differential equation y prime prime plus the first derivative plus y is equal to 0, so our characteristic equation is r squared plus r plus 1 is equal to 0. Let's break out the quadratic formula. So the roots are going to be negative b, so it's negative 1 plus or minus the square root of b squared, b squared is 1, minus 4 times ac."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "And let's see if we can do a problem real fast that involves that. So let's say I had the differential equation y prime prime plus the first derivative plus y is equal to 0, so our characteristic equation is r squared plus r plus 1 is equal to 0. Let's break out the quadratic formula. So the roots are going to be negative b, so it's negative 1 plus or minus the square root of b squared, b squared is 1, minus 4 times ac. Well, a and c are both 1, so it's just minus 4. All of that over 2, right? 2 times a, all of that over 2."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "So the roots are going to be negative b, so it's negative 1 plus or minus the square root of b squared, b squared is 1, minus 4 times ac. Well, a and c are both 1, so it's just minus 4. All of that over 2, right? 2 times a, all of that over 2. So the roots are going to be negative 1 plus or minus the square root of, this is negative 3, over 2. Or we could rewrite this as the roots are r is equal to negative 1 half plus or minus, well, we could rewrite this as i times the square root of 3, or square root of 3i, over 2. Or we could write this as square root of 3 over 2 times i, actually that's the best way to write it."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "2 times a, all of that over 2. So the roots are going to be negative 1 plus or minus the square root of, this is negative 3, over 2. Or we could rewrite this as the roots are r is equal to negative 1 half plus or minus, well, we could rewrite this as i times the square root of 3, or square root of 3i, over 2. Or we could write this as square root of 3 over 2 times i, actually that's the best way to write it. You just take the i out, so it takes the negative 1 out, and you have left with square root of 3 over 2. So these are the roots, and now if we want the general solution, we just have to throw this right back into that, and we'll have our general solution. So let me write that right down here."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "Or we could write this as square root of 3 over 2 times i, actually that's the best way to write it. You just take the i out, so it takes the negative 1 out, and you have left with square root of 3 over 2. So these are the roots, and now if we want the general solution, we just have to throw this right back into that, and we'll have our general solution. So let me write that right down here. So our general solution will be y is equal to e to the real part of our complex conjugate, so e to the minus 1 half times x, this is our lambda, times some constant, I'll write c1 now. c1 times cosine of the imaginary part without the i, so cosine of square root of 3 over 2x plus c2 times sine of square root of 3 over 2x. Not too bad."}, {"video_title": "Complex roots of the characteristic equations 2 Second order differential equations Khan Academy.mp3", "Sentence": "So let me write that right down here. So our general solution will be y is equal to e to the real part of our complex conjugate, so e to the minus 1 half times x, this is our lambda, times some constant, I'll write c1 now. c1 times cosine of the imaginary part without the i, so cosine of square root of 3 over 2x plus c2 times sine of square root of 3 over 2x. Not too bad. We had complex roots, and it really didn't take us any more time than when we had two real roots. You just have to realize this, and then you have to just use the quadratic equation to find the complex roots of the characteristic equation, and realize that this is lambda, this minus 1 half is lambda, and that the square root of 3 over 2 is equal to mu, and then substitute back into this solution that we got. Anyway, in the next video, I'll do another one of these problems, and we'll actually have initial conditions so we can solve for c1 and c2."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So let's say I have the differential equation, the derivative of y with respect to x is equal to two y squared. And let's say that the graph of a particular solution to this the graph of a particular solution passes through the point one comma negative one. So my question to you is, what is y, what is y when x is equal to three for this particular solution? So the particular solution to the differential equation that passes through the point one comma negative one, what is y when x is equal to three? And I encourage you to pause the video and try to work through it on your own. So I'm assuming you had a go at it and the key with a separable differential equation, and that's a big clue, that even calling it a separable differential equation is that you separate the x's from the y's, or all the x's and the dx's from the y's and dy's. So how do you do that here?"}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So the particular solution to the differential equation that passes through the point one comma negative one, what is y when x is equal to three? And I encourage you to pause the video and try to work through it on your own. So I'm assuming you had a go at it and the key with a separable differential equation, and that's a big clue, that even calling it a separable differential equation is that you separate the x's from the y's, or all the x's and the dx's from the y's and dy's. So how do you do that here? Well, what I could do, let me just rewrite it. So it's gonna be dy dx is equal to two y squared. Is equal to two y, equal to two y squared."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So how do you do that here? Well, what I could do, let me just rewrite it. So it's gonna be dy dx is equal to two y squared. Is equal to two y, equal to two y squared. So let's see, we can multiply both sides by dx. And let's see, so then we're gonna have, that cancels with that if we treat it as just a value or as a variable, we're gonna have dy is equal to two y squared dx. Well, we're not quite done yet."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "Is equal to two y, equal to two y squared. So let's see, we can multiply both sides by dx. And let's see, so then we're gonna have, that cancels with that if we treat it as just a value or as a variable, we're gonna have dy is equal to two y squared dx. Well, we're not quite done yet. We need to get this two y squared on the left-hand side. So we can divide both sides by two y squared. So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "Well, we're not quite done yet. We need to get this two y squared on the left-hand side. So we can divide both sides by two y squared. So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy. Let me, dy is equal to dx. And now we can integrate both sides. So we can integrate both sides."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy. Let me, dy is equal to dx. And now we can integrate both sides. So we can integrate both sides. Let me get myself a little bit more space. And so what is, what is this left-hand side going to be? Well, we increment the exponent and then divide by that value."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So we can integrate both sides. Let me get myself a little bit more space. And so what is, what is this left-hand side going to be? Well, we increment the exponent and then divide by that value. So y to the negative two, if you increment it to y to the negative one and then divide by negative one, so this is going to be negative 1 1 2 y to the negative one power. And we could do a plus c like we did in the previous video, but we're gonna have a plus c on both sides. And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "Well, we increment the exponent and then divide by that value. So y to the negative two, if you increment it to y to the negative one and then divide by negative one, so this is going to be negative 1 1 2 y to the negative one power. And we could do a plus c like we did in the previous video, but we're gonna have a plus c on both sides. And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other. So I'm just gonna write the constant only on one side. So you have that is equal to, well, if I integrate just dx, that's just going to give me x. That's just gonna give me x."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other. So I'm just gonna write the constant only on one side. So you have that is equal to, well, if I integrate just dx, that's just going to give me x. That's just gonna give me x. So this right over here is x, and of course I can have a plus c over there. And if I want, I can solve for y. If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "That's just gonna give me x. So this right over here is x, and of course I can have a plus c over there. And if I want, I can solve for y. If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y. Is equal to, if I multiply the right-hand side times negative two, I'm gonna have negative two times x plus, well, it's some arbitrary constant. It's still going to be negative two times this arbitrary constant, but I could still just call it some arbitrary constant. And then if we want, we can take the reciprocal of both sides."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y. Is equal to, if I multiply the right-hand side times negative two, I'm gonna have negative two times x plus, well, it's some arbitrary constant. It's still going to be negative two times this arbitrary constant, but I could still just call it some arbitrary constant. And then if we want, we can take the reciprocal of both sides. And so we will get y is equal to, is equal to one over negative two x plus c. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for c. So when x is negative one, so when x is negative one, or sorry, when x is one, when x is one, y is negative one. So we get negative one is equal to one over negative two plus c. Or we could say c minus two. We could multiply both sides times c minus two."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "And then if we want, we can take the reciprocal of both sides. And so we will get y is equal to, is equal to one over negative two x plus c. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for c. So when x is negative one, so when x is negative one, or sorry, when x is one, when x is one, y is negative one. So we get negative one is equal to one over negative two plus c. Or we could say c minus two. We could multiply both sides times c minus two. If then we will get, actually let me just scroll down a little bit. So if you multiply both sides times c minus two, negative one times c minus two is gonna be negative c plus two, or two minus c is equal to one. All I did is I multiplied c minus two times both sides."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "We could multiply both sides times c minus two. If then we will get, actually let me just scroll down a little bit. So if you multiply both sides times c minus two, negative one times c minus two is gonna be negative c plus two, or two minus c is equal to one. All I did is I multiplied c minus two times both sides. And then let's see, I can subtract two from both sides. So negative c is equal to negative one. And then if I multiply both sides by negative one, we get c is equal to one."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "All I did is I multiplied c minus two times both sides. And then let's see, I can subtract two from both sides. So negative c is equal to negative one. And then if I multiply both sides by negative one, we get c is equal to one. So our particular solution is y is equal to one over negative two x plus one. And we are almost done. They didn't just ask for, we didn't just ask for the particular solution."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "And then if I multiply both sides by negative one, we get c is equal to one. So our particular solution is y is equal to one over negative two x plus one. And we are almost done. They didn't just ask for, we didn't just ask for the particular solution. We asked what is y when x is equal to three. So y is going to be equal to one over, three times negative two is negative six plus one, which is equal to negative, is going to be equal to one over negative five, or negative one fifth. And we are done."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the easiest way to think about a slope field, if I needed to plot this slope field by hand, I would sample a bunch of x and y points and then I would figure out what the derivative would have to be at that point. And so what we can do here, since they've already drawn some candidate slope fields for us, is figure out what we think the slope field should be at some point and see which of these diagrams, these graphs or these slope fields, actually show that. So let's, let me make a little table here. So I'm gonna have, I'm gonna have x, y, and then the derivative of y with respect to x. And we can do it at a bunch of values, so let's think about it. Let's think about when, we're at this point right over here, when x is two and y is two. When x is two and y is two, the derivative of y with respect to x is going to be two minus two."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm gonna have, I'm gonna have x, y, and then the derivative of y with respect to x. And we can do it at a bunch of values, so let's think about it. Let's think about when, we're at this point right over here, when x is two and y is two. When x is two and y is two, the derivative of y with respect to x is going to be two minus two. It's going to be equal to zero. And just with that, let's see, here, this slope on this slope field does not look like it's zero. This looks like it's negative one."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "When x is two and y is two, the derivative of y with respect to x is going to be two minus two. It's going to be equal to zero. And just with that, let's see, here, this slope on this slope field does not look like it's zero. This looks like it's negative one. So already I could rule this one out. This slope right over here looks like it's positive one, so I'll rule that out. It's definitely not zero."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "This looks like it's negative one. So already I could rule this one out. This slope right over here looks like it's positive one, so I'll rule that out. It's definitely not zero. This slope also looks like positive one, so I can rule that one out. This slope at two comma two actually does look like zero, so I'm liking this one right over here. This slope at two comma two looks larger than one, so I could rule that out."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "It's definitely not zero. This slope also looks like positive one, so I can rule that one out. This slope at two comma two actually does look like zero, so I'm liking this one right over here. This slope at two comma two looks larger than one, so I could rule that out. So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one. But just for kicks, we could keep going to verify that this is indeed the slope field. So let's think about what happens when x is equal to, well, one, whenever x is equal to y, you're going to get the derivative equaling zero, and you see that here."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "This slope at two comma two looks larger than one, so I could rule that out. So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one. But just for kicks, we could keep going to verify that this is indeed the slope field. So let's think about what happens when x is equal to, well, one, whenever x is equal to y, you're going to get the derivative equaling zero, and you see that here. When you're at four, four, derivative equals zero. When it's six, six, derivative equals zero. At negative two, negative two, derivative equals zero."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what happens when x is equal to, well, one, whenever x is equal to y, you're going to get the derivative equaling zero, and you see that here. When you're at four, four, derivative equals zero. When it's six, six, derivative equals zero. At negative two, negative two, derivative equals zero. So that feels good that this is the right slope field. And then we could pick other arbitrary points. Let's say when x is four, y is two, then the derivative here should be four minus two, which is going to be two."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "At negative two, negative two, derivative equals zero. So that feels good that this is the right slope field. And then we could pick other arbitrary points. Let's say when x is four, y is two, then the derivative here should be four minus two, which is going to be two. So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here. And if it was the other way around, when x is, when x is, let's say, x is negative four, and y is negative two. So negative four, negative two."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say when x is four, y is two, then the derivative here should be four minus two, which is going to be two. So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here. And if it was the other way around, when x is, when x is, let's say, x is negative four, and y is negative two. So negative four, negative two. Well, negative four minus negative two is going to be negative two. And you can see that right over here. Negative four, negative two."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So negative four, negative two. Well, negative four minus negative two is going to be negative two. And you can see that right over here. Negative four, negative two. You can see the slope right over here. It's a little harder to see. Looks like negative two."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "A particle moves along a straight line. Its speed is inversely proportional to the square of the distance s it has traveled. Which equation describes this relationship? So I'm not gonna even look at these choices, and I'm just gonna try to parse this sentence up here and see if we can come up with an equation. So they tell us its speed is inversely proportional to what? To the square of the distance s it has traveled. So s is equal to distance."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm not gonna even look at these choices, and I'm just gonna try to parse this sentence up here and see if we can come up with an equation. So they tell us its speed is inversely proportional to what? To the square of the distance s it has traveled. So s is equal to distance. S is equal to distance. And how would we denote speed then, if s is distance? Well, speed is the rate of change of distance with respect to time."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So s is equal to distance. S is equal to distance. And how would we denote speed then, if s is distance? Well, speed is the rate of change of distance with respect to time. So our speed would be the rate of distance with respect to time. The rate of change of distance with respect to time. So this is going to be our speed."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Well, speed is the rate of change of distance with respect to time. So our speed would be the rate of distance with respect to time. The rate of change of distance with respect to time. So this is going to be our speed. So now that we got our notation, the s is the distance, the derivative of s with respect to time is speed, we can say the speed, which is d capital S, dt, is inversely proportional. So it's inversely proportional. I'll write a proportionality constant over what?"}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be our speed. So now that we got our notation, the s is the distance, the derivative of s with respect to time is speed, we can say the speed, which is d capital S, dt, is inversely proportional. So it's inversely proportional. I'll write a proportionality constant over what? It's inversely proportional to what? To the square of the distance. To the square of the distance it has traveled."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "I'll write a proportionality constant over what? It's inversely proportional to what? To the square of the distance. To the square of the distance it has traveled. So there you go. This is an equation that I think is describing a differential equation, really, that's describing what we have up here. Now let's see which of these choices match that."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "To the square of the distance it has traveled. So there you go. This is an equation that I think is describing a differential equation, really, that's describing what we have up here. Now let's see which of these choices match that. Well, actually, this one is exactly what we wrote. The speed, the rate of change of distance with respect to time is inversely proportional to the square of the distance. Now just to make sure we understand these other ones, let's just interpret them."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's see which of these choices match that. Well, actually, this one is exactly what we wrote. The speed, the rate of change of distance with respect to time is inversely proportional to the square of the distance. Now just to make sure we understand these other ones, let's just interpret them. This is saying that the distance, which is a function of time, is inversely proportional to the time squared. That's not what they told us. This is saying that the distance is inversely proportional to the distance squared."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Now just to make sure we understand these other ones, let's just interpret them. This is saying that the distance, which is a function of time, is inversely proportional to the time squared. That's not what they told us. This is saying that the distance is inversely proportional to the distance squared. That one is especially strange. And this is saying that the distance with respect to time, the change in distance with respect to time, the derivative of the distance with respect to time, ds dt, or the speed, is inversely proportional to time squared. Well, that's not what they said."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So now that we've spent some time thinking about what a differential equation is, and even visualizing solutions to a differential equation using things like slope field, let's start seeing if we can actually solve differential equations. And as we'll see, different types of differential equations might require different techniques, and some of them we might not be able to solve at all using analytic techniques. We'd have to resort to numeric techniques to estimate the solutions. But let's go to what I would argue is the simplest form of differential equation to solve, and that's what's called a separable, separable differential equation. And we will see in a second why it is called a separable differential equation. So let's say that we have the derivative of y with respect to x is equal to negative x over y e to the x squared. So we have this differential equation, and we want to find the particular solution that goes through the point, that goes through the point zero comma one."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "But let's go to what I would argue is the simplest form of differential equation to solve, and that's what's called a separable, separable differential equation. And we will see in a second why it is called a separable differential equation. So let's say that we have the derivative of y with respect to x is equal to negative x over y e to the x squared. So we have this differential equation, and we want to find the particular solution that goes through the point, that goes through the point zero comma one. And I encourage you to pause this video, and I'll give you a hint. If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it, don't worry, because we're about to work through it."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So we have this differential equation, and we want to find the particular solution that goes through the point, that goes through the point zero comma one. And I encourage you to pause this video, and I'll give you a hint. If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it, don't worry, because we're about to work through it. So as I said, let's use a little bit of algebra to get all the y's and d y's on one side, and all the x's and d x's on the other side. So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side. Well I can multiply both sides times y."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "Now if you can't do it, don't worry, because we're about to work through it. So as I said, let's use a little bit of algebra to get all the y's and d y's on one side, and all the x's and d x's on the other side. So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side. Well I can multiply both sides times y. So I can multiply both sides times y. That has the effect of putting the y's on the left hand side. And then I can multiply both sides times d x. I can multiply both sides times d x."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "Well I can multiply both sides times y. So I can multiply both sides times y. That has the effect of putting the y's on the left hand side. And then I can multiply both sides times d x. I can multiply both sides times d x. And we kind of treat, you can treat these differentials as you would treat a variable when you're manipulating it to essentially separate out the variables. And so this will cancel with that. And so we are left with, we are left with y d y, y d y is equal to negative x, and actually let me write it this way."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And then I can multiply both sides times d x. I can multiply both sides times d x. And we kind of treat, you can treat these differentials as you would treat a variable when you're manipulating it to essentially separate out the variables. And so this will cancel with that. And so we are left with, we are left with y d y, y d y is equal to negative x, and actually let me write it this way. Let me write it as negative x e, actually I might want a little more space. So negative x e to the negative x squared d x. D x."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And so we are left with, we are left with y d y, y d y is equal to negative x, and actually let me write it this way. Let me write it as negative x e, actually I might want a little more space. So negative x e to the negative x squared d x. D x. Now why is this interesting? Because we can integrate both sides. And now this also highlights why we call this separable."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "D x. Now why is this interesting? Because we can integrate both sides. And now this also highlights why we call this separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side. But this one we were able to."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And now this also highlights why we call this separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side. But this one we were able to. And so that's why this is called a separable differential equation. Differential, differential equation. And it's usually the first technique that you should try."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "But this one we were able to. And so that's why this is called a separable differential equation. Differential, differential equation. And it's usually the first technique that you should try. Hey, can I separate the y's and the x's? And as I said, this is not going to be true of many, if not most differential equations. But now that we did this, we can integrate both sides."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And it's usually the first technique that you should try. Hey, can I separate the y's and the x's? And as I said, this is not going to be true of many, if not most differential equations. But now that we did this, we can integrate both sides. So let's do that. So I'll find a nice color to integrate with. So I'm going to integrate, integrate both sides."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "But now that we did this, we can integrate both sides. So let's do that. So I'll find a nice color to integrate with. So I'm going to integrate, integrate both sides. Now if you integrate the left-hand side, what do you get? You get, and remember, we're integrating with respect to y here. So this is going to be y squared over two."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So I'm going to integrate, integrate both sides. Now if you integrate the left-hand side, what do you get? You get, and remember, we're integrating with respect to y here. So this is going to be y squared over two. And we could put some constant there. I could call that plus c one. And if you're integrating, now that's going to be equal to, now the right-hand side we're integrating with respect to x."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So this is going to be y squared over two. And we could put some constant there. I could call that plus c one. And if you're integrating, now that's going to be equal to, now the right-hand side we're integrating with respect to x. And let's see, you could do u substitution, or you could recognize that look, the derivative of negative x squared is going to be negative two x. So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there. So now you could either do u substitution explicitly, or you could do it in your head, where you said u is equal to negative x squared, and then du will be negative two x dx, or you can kind of do this in your head at this point."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And if you're integrating, now that's going to be equal to, now the right-hand side we're integrating with respect to x. And let's see, you could do u substitution, or you could recognize that look, the derivative of negative x squared is going to be negative two x. So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there. So now you could either do u substitution explicitly, or you could do it in your head, where you said u is equal to negative x squared, and then du will be negative two x dx, or you can kind of do this in your head at this point. So I have something and its derivative, so I really could just integrate with respect to that something, with respect to that u. So this is going to be 1 1 2, this 1 1 2 right over here. The antiderivative of this is e to the negative x squared, and then of course I might have some other constant."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So now you could either do u substitution explicitly, or you could do it in your head, where you said u is equal to negative x squared, and then du will be negative two x dx, or you can kind of do this in your head at this point. So I have something and its derivative, so I really could just integrate with respect to that something, with respect to that u. So this is going to be 1 1 2, this 1 1 2 right over here. The antiderivative of this is e to the negative x squared, and then of course I might have some other constant. I'll just call that c two. And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece. Now, what can I do here?"}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "The antiderivative of this is e to the negative x squared, and then of course I might have some other constant. I'll just call that c two. And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece. Now, what can I do here? Well I have a constant on the left-hand side, it's an arbitrary constant, we don't know what it is. I haven't used this initial condition yet, we could call it. So let me just subtract c one from both sides."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "Now, what can I do here? Well I have a constant on the left-hand side, it's an arbitrary constant, we don't know what it is. I haven't used this initial condition yet, we could call it. So let me just subtract c one from both sides. So if I just subtract c one from both sides, I have an arbitrary, so this is going to cancel, and I have c two, sorry, let me, so this is c one, so these are going to cancel, and c two minus c one, these are both constants, arbitrary constants, we don't know what they are yet, and so we could just rewrite this as, on the left-hand side we have y squared over two is equal to, on the right-hand side, I'll write one half e, let me write that in blue, just because I wrote it in blue before, one half, one half e to the negative x squared, and I'll just say c two minus c one, let's just call that c. So if you take the sum of those two things, let's just call that c. And so now, this is kind of a general solution, we don't know what this constant is, and we haven't explicitly solved for y yet, but even in this form, we can now find a particular solution using this initial condition. Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition. So it tells us when x is zero, y needs to be equal to one."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So let me just subtract c one from both sides. So if I just subtract c one from both sides, I have an arbitrary, so this is going to cancel, and I have c two, sorry, let me, so this is c one, so these are going to cancel, and c two minus c one, these are both constants, arbitrary constants, we don't know what they are yet, and so we could just rewrite this as, on the left-hand side we have y squared over two is equal to, on the right-hand side, I'll write one half e, let me write that in blue, just because I wrote it in blue before, one half, one half e to the negative x squared, and I'll just say c two minus c one, let's just call that c. So if you take the sum of those two things, let's just call that c. And so now, this is kind of a general solution, we don't know what this constant is, and we haven't explicitly solved for y yet, but even in this form, we can now find a particular solution using this initial condition. Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition. So it tells us when x is zero, y needs to be equal to one. So we would have one squared, which is just one, over two is equal to one half, e to the negative zero squared, well that's just going to be, e to the zero is just one, so it's going to be one half plus c, and just like that, we're able to figure out if you subtract one half from both sides, c is equal to zero. So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two. Now we can multiply both sides by two, and we're going to get y squared, y squared, let me do that, so we're going to get y squared is equal to, is equal to e to the negative x squared."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So it tells us when x is zero, y needs to be equal to one. So we would have one squared, which is just one, over two is equal to one half, e to the negative zero squared, well that's just going to be, e to the zero is just one, so it's going to be one half plus c, and just like that, we're able to figure out if you subtract one half from both sides, c is equal to zero. So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two. Now we can multiply both sides by two, and we're going to get y squared, y squared, let me do that, so we're going to get y squared is equal to, is equal to e to the negative x squared. Now we can take the square root of both sides, and you could say, well look, y squared is equal to this, so y could be equal to the plus or minus square root of e to the negative x squared, of e to the negative x squared, but they gave us an initial condition where y is actually positive. So we're finding the particular solution that goes through this point, that means y is going to be the positive square root. If this was the point zero, negative one, then we would say y is the negative square root, but we know that y is the positive square root, it's the principal root right over there."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "Now we can multiply both sides by two, and we're going to get y squared, y squared, let me do that, so we're going to get y squared is equal to, is equal to e to the negative x squared. Now we can take the square root of both sides, and you could say, well look, y squared is equal to this, so y could be equal to the plus or minus square root of e to the negative x squared, of e to the negative x squared, but they gave us an initial condition where y is actually positive. So we're finding the particular solution that goes through this point, that means y is going to be the positive square root. If this was the point zero, negative one, then we would say y is the negative square root, but we know that y is the positive square root, it's the principal root right over there. Actually let me do that a little bit neater, so we can get rid of, whoops, I thought I was writing in black. So we can get rid of this right over here. We're only going to be dealing with the positive square root, or we could, so we could write y is equal to e to the negative x squared to the 1 1 2 power, and that of course is equal to, is equal to e to the negative x squared over two."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We were in the midst of figuring out the Laplace transform of sine of at when I was running out of time. And so this is the definition of the Laplace transform of sine of at. I said that also equals y. This is going to be useful for us. It's going to be doing integration by parts twice. So I did integration by parts once, then integration by parts twice. I said, don't worry about the boundaries of the integral right now."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is going to be useful for us. It's going to be doing integration by parts twice. So I did integration by parts once, then integration by parts twice. I said, don't worry about the boundaries of the integral right now. Let's just worry about the indefinite integral. And then after we solve for y, let's just say y is the indefinite version of this, then we can evaluate the boundaries. And we got to this point, and we made the realization after doing two integration by parts and being very careful not to hopefully make any careless mistakes, we realized, wow, this is our original y."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I said, don't worry about the boundaries of the integral right now. Let's just worry about the indefinite integral. And then after we solve for y, let's just say y is the indefinite version of this, then we can evaluate the boundaries. And we got to this point, and we made the realization after doing two integration by parts and being very careful not to hopefully make any careless mistakes, we realized, wow, this is our original y. If I put the boundaries here, that's the same thing as the Laplace transform of sine of at. That's our original y. So now, and I'll switch colors just to avoid monotony, this is equal to y."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we got to this point, and we made the realization after doing two integration by parts and being very careful not to hopefully make any careless mistakes, we realized, wow, this is our original y. If I put the boundaries here, that's the same thing as the Laplace transform of sine of at. That's our original y. So now, and I'll switch colors just to avoid monotony, this is equal to y. That was our original definition. So let's add a squared over sine squared y to both sides of this. So this is equal to y plus, I'm just adding this whole term to both sides of this equation, plus a squared over s squared y is equal to, so this term is now gone, so it's equal to this stuff."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So now, and I'll switch colors just to avoid monotony, this is equal to y. That was our original definition. So let's add a squared over sine squared y to both sides of this. So this is equal to y plus, I'm just adding this whole term to both sides of this equation, plus a squared over s squared y is equal to, so this term is now gone, so it's equal to this stuff. And let's see if we can simplify this. So let's see if we can, let's factor out an e to the minus st. Actually, let's factor out a negative e to the minus st. So it's minus e to the minus st times sine of, well, actually, let's factor out it."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this is equal to y plus, I'm just adding this whole term to both sides of this equation, plus a squared over s squared y is equal to, so this term is now gone, so it's equal to this stuff. And let's see if we can simplify this. So let's see if we can, let's factor out an e to the minus st. Actually, let's factor out a negative e to the minus st. So it's minus e to the minus st times sine of, well, actually, let's factor out it. Well, let me just write. 1 over s sine of at minus 1 over s squared cosine of at. I really hope I haven't made any careless mistakes."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So it's minus e to the minus st times sine of, well, actually, let's factor out it. Well, let me just write. 1 over s sine of at minus 1 over s squared cosine of at. I really hope I haven't made any careless mistakes. And so this, we can add the coefficients. So we get 1 plus a squared over s squared times y. But that's the same thing as s squared over s squared plus a squared over s squared."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I really hope I haven't made any careless mistakes. And so this, we can add the coefficients. So we get 1 plus a squared over s squared times y. But that's the same thing as s squared over s squared plus a squared over s squared. So it's s squared plus a squared over s squared y is equal to e minus e to the minus st times this whole thing, sine of at minus 1 over s squared cosine of at. And now, this right here, since we're dealing everything with respect to dt, this is just a constant, right? So we could say a constant times the antiderivative is equal to this."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But that's the same thing as s squared over s squared plus a squared over s squared. So it's s squared plus a squared over s squared y is equal to e minus e to the minus st times this whole thing, sine of at minus 1 over s squared cosine of at. And now, this right here, since we're dealing everything with respect to dt, this is just a constant, right? So we could say a constant times the antiderivative is equal to this. This is as good a time as any to evaluate the boundaries. If this had a t here, I would have to somehow get them back on the other side, because the t's are involved in evaluating the boundaries, since we're doing our definite integral or improper integral. So let's evaluate the boundaries now."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we could say a constant times the antiderivative is equal to this. This is as good a time as any to evaluate the boundaries. If this had a t here, I would have to somehow get them back on the other side, because the t's are involved in evaluating the boundaries, since we're doing our definite integral or improper integral. So let's evaluate the boundaries now. And we could have kept them along with us the whole time, right, and just factored out this term right here. But anyway, so let's evaluate this from 0 to infinity, and this should simplify things. So the right-hand side of this equation, when I evaluated at infinity, what is e to the minus infinity?"}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's evaluate the boundaries now. And we could have kept them along with us the whole time, right, and just factored out this term right here. But anyway, so let's evaluate this from 0 to infinity, and this should simplify things. So the right-hand side of this equation, when I evaluated at infinity, what is e to the minus infinity? Well, that is 0. We've established that multiple times. And now it approaches 0 from the negative side, but it's still going to be 0, or it approaches 0."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So the right-hand side of this equation, when I evaluated at infinity, what is e to the minus infinity? Well, that is 0. We've established that multiple times. And now it approaches 0 from the negative side, but it's still going to be 0, or it approaches 0. And then that times, well, what's sine of infinity? Well, sine just keeps oscillating, right, between negative 1 and plus 1, and so does cosine, right? So this is bounded."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And now it approaches 0 from the negative side, but it's still going to be 0, or it approaches 0. And then that times, well, what's sine of infinity? Well, sine just keeps oscillating, right, between negative 1 and plus 1, and so does cosine, right? So this is bounded. So this thing is going to overpower these. And if you're curious, you can graph it. This kind of forms an envelope around these oscillations."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this is bounded. So this thing is going to overpower these. And if you're curious, you can graph it. This kind of forms an envelope around these oscillations. So the limit as this approaches infinity is going to be equal to 0. And that makes sense, right? These are bounded between 0 and negative 1, and this approaches 0 very quickly."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This kind of forms an envelope around these oscillations. So the limit as this approaches infinity is going to be equal to 0. And that makes sense, right? These are bounded between 0 and negative 1, and this approaches 0 very quickly. So it's 0 times something bounded between 1 and negative 1. Another way to view it is the largest value this could equal is 1 times whatever coefficient's on it, and this is going to 0, so it's like 0 times 1. Anyway, I don't want to focus too much on that."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "These are bounded between 0 and negative 1, and this approaches 0 very quickly. So it's 0 times something bounded between 1 and negative 1. Another way to view it is the largest value this could equal is 1 times whatever coefficient's on it, and this is going to 0, so it's like 0 times 1. Anyway, I don't want to focus too much on that. You can play around with that if you like. Minus this whole thing evaluated at 0. So what's e to the minus 0?"}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Anyway, I don't want to focus too much on that. You can play around with that if you like. Minus this whole thing evaluated at 0. So what's e to the minus 0? Well, that is e to the minus 0 is 1, right? That's e to the 0. We have a minus 1, so it becomes plus 1 times, now, sine of 0 is 0 minus 1 over s squared cosine of 0."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So what's e to the minus 0? Well, that is e to the minus 0 is 1, right? That's e to the 0. We have a minus 1, so it becomes plus 1 times, now, sine of 0 is 0 minus 1 over s squared cosine of 0. So what is 1 over cosine of 0? It's 1. So we have minus 1 over s squared times 1."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We have a minus 1, so it becomes plus 1 times, now, sine of 0 is 0 minus 1 over s squared cosine of 0. So what is 1 over cosine of 0? It's 1. So we have minus 1 over s squared times 1. So that is equal to minus 1 over s squared. And I think I made a mistake, because I shouldn't be having a negative number here. So let's backtrack and let's see where that mistake might be."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we have minus 1 over s squared times 1. So that is equal to minus 1 over s squared. And I think I made a mistake, because I shouldn't be having a negative number here. So let's backtrack and let's see where that mistake might be. Maybe this isn't a negative number. Let's see. Infinity, right?"}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's backtrack and let's see where that mistake might be. Maybe this isn't a negative number. Let's see. Infinity, right? This whole thing is 0. Minus, let's see, when you put 0 here, this becomes a minus 1. Yeah, I'm getting a, let me see where my, so either this is a plus or this is a plus."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Infinity, right? This whole thing is 0. Minus, let's see, when you put 0 here, this becomes a minus 1. Yeah, I'm getting a, let me see where my, so either this is a plus or this is a plus. Let's see where I made my mistake. e to the minus st. Oh, I see where my mistake is, right? Up here, where I factored out a minus e to the minus st, fair enough."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Yeah, I'm getting a, let me see where my, so either this is a plus or this is a plus. Let's see where I made my mistake. e to the minus st. Oh, I see where my mistake is, right? Up here, where I factored out a minus e to the minus st, fair enough. So that makes this 1 over s sine of at. But if I factor out a minus e to the minus st, this becomes a plus, right? There was a minus here, but I'm factoring out a minus e to the minus st."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Up here, where I factored out a minus e to the minus st, fair enough. So that makes this 1 over s sine of at. But if I factor out a minus e to the minus st, this becomes a plus, right? There was a minus here, but I'm factoring out a minus e to the minus st. So that's a plus. This is a plus. Boy, I'm glad that was not too difficult to find."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "There was a minus here, but I'm factoring out a minus e to the minus st. So that's a plus. This is a plus. Boy, I'm glad that was not too difficult to find. So then this becomes a plus, and then this becomes a plus. Thank God. It would have been sad if I wasted two videos and ended up with a careless negative number."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Boy, I'm glad that was not too difficult to find. So then this becomes a plus, and then this becomes a plus. Thank God. It would have been sad if I wasted two videos and ended up with a careless negative number. Anyway, so now we have s squared plus a squared over s squared times y is equal to this. Multiply both sides times s squared over s squared plus a squared, divide both sides by this, and we get y is equal to 1 over s squared. Actually, let me make sure that that is right."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It would have been sad if I wasted two videos and ended up with a careless negative number. Anyway, so now we have s squared plus a squared over s squared times y is equal to this. Multiply both sides times s squared over s squared plus a squared, divide both sides by this, and we get y is equal to 1 over s squared. Actually, let me make sure that that is right. It's 1 over s squared. y is equal to 1 over s squared times s squared over s squared plus a squared, and then these cancel out. And let me make sure that I haven't made another careless mistake, because I have a feeling I have."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Actually, let me make sure that that is right. It's 1 over s squared. y is equal to 1 over s squared times s squared over s squared plus a squared, and then these cancel out. And let me make sure that I haven't made another careless mistake, because I have a feeling I have. I have a feeling I have. Yep, there. I see the careless mistake."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And let me make sure that I haven't made another careless mistake, because I have a feeling I have. I have a feeling I have. Yep, there. I see the careless mistake. And it was all in this term. And I hope you don't mind my careless mistakes, but I want you to see that I'm doing these things in real time, and I'm human, in case you haven't realized already. Anyway, so I made the same careless mistake."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I see the careless mistake. And it was all in this term. And I hope you don't mind my careless mistakes, but I want you to see that I'm doing these things in real time, and I'm human, in case you haven't realized already. Anyway, so I made the same careless mistake. So I factored out e to the minus st here. So it's plus, but it was a over s squared. So this is an a."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Anyway, so I made the same careless mistake. So I factored out e to the minus st here. So it's plus, but it was a over s squared. So this is an a. That's an a. And so this is an a. And so this is an a."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this is an a. That's an a. And so this is an a. And so this is an a. And so this is an a. This was an a. And so we're left with, and this is the correct answer, a over s squared plus a squared."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And so this is an a. And so this is an a. This was an a. And so we're left with, and this is the correct answer, a over s squared plus a squared. So I hope those careless mistakes didn't throw you off too much. These things happen when you do integration by parts twice with a bunch of variables. But anyway, now we are ready to add a significant entry into our table of Laplace transforms."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And so we're left with, and this is the correct answer, a over s squared plus a squared. So I hope those careless mistakes didn't throw you off too much. These things happen when you do integration by parts twice with a bunch of variables. But anyway, now we are ready to add a significant entry into our table of Laplace transforms. And that is that the Laplace transform of sine of at is equal to a over s squared plus a squared. And that's a significant entry. And maybe a good exercise for you, just to see how fun it is to do these integration by parts problems twice, is to figure out the Laplace transform of cosine of at."}, {"video_title": "Part 2 of the transform of the sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But anyway, now we are ready to add a significant entry into our table of Laplace transforms. And that is that the Laplace transform of sine of at is equal to a over s squared plus a squared. And that's a significant entry. And maybe a good exercise for you, just to see how fun it is to do these integration by parts problems twice, is to figure out the Laplace transform of cosine of at. And I'll give you a hint. It's s over s squared plus a squared. And it's nice that there's that symmetry there."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "I've spoken a lot about second order linear homogenous differential equations in abstract terms and how if g is a solution, then some constant times g is also a solution. Or if g and h are solutions, then g plus h is also a solution. Let's actually do problems because I think that will actually help you learn as opposed to help you get confused. So let's say I have this differential equation. The second derivative of y with respect to x plus 5 times the first derivative of y with respect to x plus 6 times y is equal to 0. So we need to find a y where 1 times its second derivative plus 5 times its first derivative plus 6 times itself is equal to 0. And now let's just do a little bit of, take a step back and think about what kind of function."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "So let's say I have this differential equation. The second derivative of y with respect to x plus 5 times the first derivative of y with respect to x plus 6 times y is equal to 0. So we need to find a y where 1 times its second derivative plus 5 times its first derivative plus 6 times itself is equal to 0. And now let's just do a little bit of, take a step back and think about what kind of function. Most functions, if I have the function and I take its derivative and then I take its second derivative, most times I get something completely different. Like if I have y was x squared, then y prime would be 2x and y prime prime would be 2. And then to add them together, you'd say, well, how would my x terms cancel out so that you get 0 in the end?"}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And now let's just do a little bit of, take a step back and think about what kind of function. Most functions, if I have the function and I take its derivative and then I take its second derivative, most times I get something completely different. Like if I have y was x squared, then y prime would be 2x and y prime prime would be 2. And then to add them together, you'd say, well, how would my x terms cancel out so that you get 0 in the end? So draw back into your brain and think, is there some function that when I take its first and second derivatives and third and fourth derivatives, it essentially becomes the same function. Maybe the constant in front of the function changes as I take the derivative. And if you've listened to a lot of my videos, you'd realize that it probably is what I consider to be the most amazing function in mathematics."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And then to add them together, you'd say, well, how would my x terms cancel out so that you get 0 in the end? So draw back into your brain and think, is there some function that when I take its first and second derivatives and third and fourth derivatives, it essentially becomes the same function. Maybe the constant in front of the function changes as I take the derivative. And if you've listened to a lot of my videos, you'd realize that it probably is what I consider to be the most amazing function in mathematics. And that is the function e to the x. And in particular, maybe e to the x won't work here. You could even try it out."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And if you've listened to a lot of my videos, you'd realize that it probably is what I consider to be the most amazing function in mathematics. And that is the function e to the x. And in particular, maybe e to the x won't work here. You could even try it out. If you did e to the x, it won't satisfy this equation. e to the x, you'd get e to the x plus 5e to the x plus 6e to the x, that would not equal to 0. But maybe y is equal to e to some constant r times x."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "You could even try it out. If you did e to the x, it won't satisfy this equation. e to the x, you'd get e to the x plus 5e to the x plus 6e to the x, that would not equal to 0. But maybe y is equal to e to some constant r times x. Let's just make the assumption that y is equal to some constant r times x, substitute it back into this, and then see if we can actually solve for an r that makes this equation true. And if we can, we've found the solution. Or maybe we found several solutions."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "But maybe y is equal to e to some constant r times x. Let's just make the assumption that y is equal to some constant r times x, substitute it back into this, and then see if we can actually solve for an r that makes this equation true. And if we can, we've found the solution. Or maybe we found several solutions. So let's try it out. Let's try y is equal to e to the rx into this differential equation. So what is the first derivative of it, first of all?"}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "Or maybe we found several solutions. So let's try it out. Let's try y is equal to e to the rx into this differential equation. So what is the first derivative of it, first of all? It's always useful to. So y prime is equal to what? Derivative chain rule."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "So what is the first derivative of it, first of all? It's always useful to. So y prime is equal to what? Derivative chain rule. Derivative of the inside is r, and then derivative of the outside is still just e to the rx. And what's the second derivative? y prime prime is equal to derivative."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "Derivative chain rule. Derivative of the inside is r, and then derivative of the outside is still just e to the rx. And what's the second derivative? y prime prime is equal to derivative. r is just a constant, so derivative of the inside is r times r on the outside, that's r squared times e to the rx. And now we're ready to substitute back in. And I will switch colors."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "y prime prime is equal to derivative. r is just a constant, so derivative of the inside is r times r on the outside, that's r squared times e to the rx. And now we're ready to substitute back in. And I will switch colors. So the second derivative, that's r squared times e to the rx plus 5 times the first derivative, so that's 5r e to the rx plus 6 times our function, 6 times e to the rx is equal to 0. And something might already be surfacing to you as something we can do to this equation to solve for r. All of these terms on the left all have an e to the rx. So let's factor that out."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And I will switch colors. So the second derivative, that's r squared times e to the rx plus 5 times the first derivative, so that's 5r e to the rx plus 6 times our function, 6 times e to the rx is equal to 0. And something might already be surfacing to you as something we can do to this equation to solve for r. All of these terms on the left all have an e to the rx. So let's factor that out. So this is equal to e to the rx times r squared plus 5r plus 6 is equal to 0. And our goal, remember, was to solve for the r or the r's that will make this true. And in order for this side of the equation to be 0, what do we know?"}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "So let's factor that out. So this is equal to e to the rx times r squared plus 5r plus 6 is equal to 0. And our goal, remember, was to solve for the r or the r's that will make this true. And in order for this side of the equation to be 0, what do we know? Can e to the rx ever equal 0? Can you ever get something to some exponent and get 0? Well, we'll know."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And in order for this side of the equation to be 0, what do we know? Can e to the rx ever equal 0? Can you ever get something to some exponent and get 0? Well, we'll know. So this cannot equal 0. So in order for this left-hand side of the equation to be 0, this term, this expression right here has to be 0. And I'll do that in a different color."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "Well, we'll know. So this cannot equal 0. So in order for this left-hand side of the equation to be 0, this term, this expression right here has to be 0. And I'll do that in a different color. So we know if we want to solve for r that this, r squared plus 5r plus 6, that has to be 0. And this is called the characteristic equation. This or this, the r squared plus 5r plus 6 is called the characteristic equation."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And I'll do that in a different color. So we know if we want to solve for r that this, r squared plus 5r plus 6, that has to be 0. And this is called the characteristic equation. This or this, the r squared plus 5r plus 6 is called the characteristic equation. And it should be obvious to you that now this is no longer calculus. This is just factoring a quadratic. And this one actually is fairly straightforward to factor."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "This or this, the r squared plus 5r plus 6 is called the characteristic equation. And it should be obvious to you that now this is no longer calculus. This is just factoring a quadratic. And this one actually is fairly straightforward to factor. So what is this? This is r plus 2 times r plus 3 is equal to 0. And so the solutions of the characteristic equation, or actually the solutions to this original equation, are r is equal to negative 2 and r is equal to minus 3."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And this one actually is fairly straightforward to factor. So what is this? This is r plus 2 times r plus 3 is equal to 0. And so the solutions of the characteristic equation, or actually the solutions to this original equation, are r is equal to negative 2 and r is equal to minus 3. So you say, hey, we found two solutions, because we found two suitable r's that make this equation true, this differential equation true. And what are those? Well, the first one is y is equal to e to the minus 2x."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And so the solutions of the characteristic equation, or actually the solutions to this original equation, are r is equal to negative 2 and r is equal to minus 3. So you say, hey, we found two solutions, because we found two suitable r's that make this equation true, this differential equation true. And what are those? Well, the first one is y is equal to e to the minus 2x. r is minus 2. And then we could call that y1. And then the second solution we found, y2, is e to the, what is this, r is minus 3x."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "Well, the first one is y is equal to e to the minus 2x. r is minus 2. And then we could call that y1. And then the second solution we found, y2, is e to the, what is this, r is minus 3x. Now my question to you is, is this the most general solution? Well, in the last video, in kind of our introductory video, we learned that a constant times a solution is still a solution. So if y1 is a solution, we also know that we can multiply y1 times any constant."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And then the second solution we found, y2, is e to the, what is this, r is minus 3x. Now my question to you is, is this the most general solution? Well, in the last video, in kind of our introductory video, we learned that a constant times a solution is still a solution. So if y1 is a solution, we also know that we can multiply y1 times any constant. Let's do that. Let's multiply it by c1. That's a c1 there."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "So if y1 is a solution, we also know that we can multiply y1 times any constant. Let's do that. Let's multiply it by c1. That's a c1 there. This is also going to be a solution. Now it's a little bit more general, right? It's a whole class of functions."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "That's a c1 there. This is also going to be a solution. Now it's a little bit more general, right? It's a whole class of functions. The c doesn't have to just be a 1. It can be any constant. And then when you use your initial values, you actually can figure out what that constant is."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "It's a whole class of functions. The c doesn't have to just be a 1. It can be any constant. And then when you use your initial values, you actually can figure out what that constant is. And same for y2. y2 doesn't have to be 1 times e to the minus 3x. It has to be some, any constant."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And then when you use your initial values, you actually can figure out what that constant is. And same for y2. y2 doesn't have to be 1 times e to the minus 3x. It has to be some, any constant. And we learned that in the last video, that if something's a solution, some constant times that is also a solution. And we also learned that if we have two different solutions, that if you add them together, you also get a solution. So the most general solution to this differential equation is y of x."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "It has to be some, any constant. And we learned that in the last video, that if something's a solution, some constant times that is also a solution. And we also learned that if we have two different solutions, that if you add them together, you also get a solution. So the most general solution to this differential equation is y of x. Just to hit it home that this is definitely a function of x. y of x is equal to c1 e to the minus 2x plus c2 e to the minus 3x. And this is the general solution of this differential equation. And I won't prove it, because the proof is fairly involved."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "So the most general solution to this differential equation is y of x. Just to hit it home that this is definitely a function of x. y of x is equal to c1 e to the minus 2x plus c2 e to the minus 3x. And this is the general solution of this differential equation. And I won't prove it, because the proof is fairly involved. I mean, we just tried out e to the rx. Maybe there's some other wacko function that would have worked here. But I'll tell you now, and you kind of have to take it as a leap of faith, that this is the only general solution."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And I won't prove it, because the proof is fairly involved. I mean, we just tried out e to the rx. Maybe there's some other wacko function that would have worked here. But I'll tell you now, and you kind of have to take it as a leap of faith, that this is the only general solution. There isn't some crazy outside function there that would have also worked. And so the other question that might be popping in your brain is, Sal, in previous, when we did first order differential equations, we only had one constant. And that was OK, because we had one set of initial conditions, and we solved for our constants."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "But I'll tell you now, and you kind of have to take it as a leap of faith, that this is the only general solution. There isn't some crazy outside function there that would have also worked. And so the other question that might be popping in your brain is, Sal, in previous, when we did first order differential equations, we only had one constant. And that was OK, because we had one set of initial conditions, and we solved for our constants. But here I have two constants. So if I wanted a particular solution, how can I solve for two variables if I'm only given one initial condition? And if that's what you actually thought, your intuition would be correct."}, {"video_title": "2nd order linear homogeneous differential equations 2 Khan Academy.mp3", "Sentence": "And that was OK, because we had one set of initial conditions, and we solved for our constants. But here I have two constants. So if I wanted a particular solution, how can I solve for two variables if I'm only given one initial condition? And if that's what you actually thought, your intuition would be correct. You actually need two initial conditions to solve this differential equation. You would need to know at a given value of x what y is equal to, and maybe at a given value of x what the first derivative is. And that is what we will do in the next video."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And now I'll show you, at least in the context of differential equations. And I've gotten, actually, a bunch of letters on the Laplace transform, what does it really mean, and all of that, and those are excellent questions and you should strive for that. It's hard to really have an intuition of the Laplace transform in the differential equations context, other than it being a very useful tool that converts differential or integral problems into algebra problems. But I'll give you a hint, and if you want a path to learn it in, you should learn about Fourier series and Fourier transforms, which are very similar to Laplace transforms, and that'll actually build up the intuition on what the frequency domain is all about. Anyway, let's actually use a Laplace transform to solve a differential equation, and this is one we've seen before. So let me see, there it is. OK, so let's say the differential equation is y prime prime plus 5 times the first derivative plus 6y is equal to 0."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But I'll give you a hint, and if you want a path to learn it in, you should learn about Fourier series and Fourier transforms, which are very similar to Laplace transforms, and that'll actually build up the intuition on what the frequency domain is all about. Anyway, let's actually use a Laplace transform to solve a differential equation, and this is one we've seen before. So let me see, there it is. OK, so let's say the differential equation is y prime prime plus 5 times the first derivative plus 6y is equal to 0. And you know how to solve this one, but I just want to show you, with a fairly straightforward differential equation, that you could solve it with a Laplace transform, and actually you end up kind of having a characteristic equation. And the initial conditions are y of 0 is equal to 2, and y prime of 0 is equal to 3. Now, to use a Laplace transform here, we essentially just take the Laplace transform of both sides of this equation."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "OK, so let's say the differential equation is y prime prime plus 5 times the first derivative plus 6y is equal to 0. And you know how to solve this one, but I just want to show you, with a fairly straightforward differential equation, that you could solve it with a Laplace transform, and actually you end up kind of having a characteristic equation. And the initial conditions are y of 0 is equal to 2, and y prime of 0 is equal to 3. Now, to use a Laplace transform here, we essentially just take the Laplace transform of both sides of this equation. So we get, let me use a more vibrant color, so we get the Laplace transform of y, the second derivative, plus, well we could say Laplace transform of 5 times y prime, but that's the same thing as 5 times the Laplace transform of y prime plus 6 times Laplace transform of y. Let me ask you a question. What's the Laplace transform of 0?"}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, to use a Laplace transform here, we essentially just take the Laplace transform of both sides of this equation. So we get, let me use a more vibrant color, so we get the Laplace transform of y, the second derivative, plus, well we could say Laplace transform of 5 times y prime, but that's the same thing as 5 times the Laplace transform of y prime plus 6 times Laplace transform of y. Let me ask you a question. What's the Laplace transform of 0? Let me do that in a, so the Laplace transform of 0 would be the integral from 0 to infinity of 0 times e to the minus st dt. So this is a 0 in here, so this is equal to 0. So Laplace transform of 0 is 0, and that's good, because I didn't have space to draw another curly L. So what are the Laplace transforms of these things?"}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What's the Laplace transform of 0? Let me do that in a, so the Laplace transform of 0 would be the integral from 0 to infinity of 0 times e to the minus st dt. So this is a 0 in here, so this is equal to 0. So Laplace transform of 0 is 0, and that's good, because I didn't have space to draw another curly L. So what are the Laplace transforms of these things? Well, this is where we break out one of the useful properties that we learned, and that useful property, let me write it over here, because I think that's going to be some, I'll need as much real estate as possible. Let me erase this. So we learned that the Laplace transform, I'll do it here, actually I'll do it down here."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So Laplace transform of 0 is 0, and that's good, because I didn't have space to draw another curly L. So what are the Laplace transforms of these things? Well, this is where we break out one of the useful properties that we learned, and that useful property, let me write it over here, because I think that's going to be some, I'll need as much real estate as possible. Let me erase this. So we learned that the Laplace transform, I'll do it here, actually I'll do it down here. The Laplace transform of f prime, or we could even say y prime, is equal to s times the Laplace transform of y minus y of 0. We proved that to you, and this is extremely important to know. So let's see if we can apply that."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we learned that the Laplace transform, I'll do it here, actually I'll do it down here. The Laplace transform of f prime, or we could even say y prime, is equal to s times the Laplace transform of y minus y of 0. We proved that to you, and this is extremely important to know. So let's see if we can apply that. So the Laplace transform of y prime prime, if we apply that, that's equal to s times the Laplace transform of, well, to go from y prime to y, you're just taking the antiderivative. So if we're taking the antiderivative of the second derivative, we just end up with the first derivative. Minus the first derivative at 0."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's see if we can apply that. So the Laplace transform of y prime prime, if we apply that, that's equal to s times the Laplace transform of, well, to go from y prime to y, you're just taking the antiderivative. So if we're taking the antiderivative of the second derivative, we just end up with the first derivative. Minus the first derivative at 0. Notice, we're already using our initial conditions. I won't substitute it just yet. And then we end up with plus 5 times, I'll write it every time, just so we can, so plus 5 times the Laplace transform of y prime plus 6 times the Laplace transform of y."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Minus the first derivative at 0. Notice, we're already using our initial conditions. I won't substitute it just yet. And then we end up with plus 5 times, I'll write it every time, just so we can, so plus 5 times the Laplace transform of y prime plus 6 times the Laplace transform of y. All of that is equal to 0. So just to be clear, all I did is I expanded this into this. Using this."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then we end up with plus 5 times, I'll write it every time, just so we can, so plus 5 times the Laplace transform of y prime plus 6 times the Laplace transform of y. All of that is equal to 0. So just to be clear, all I did is I expanded this into this. Using this. So how can we rewrite the Laplace transform of y prime? Well, we could use this once again. So let's do that."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Using this. So how can we rewrite the Laplace transform of y prime? Well, we could use this once again. So let's do that. So this over here, I'll do it in magenta, this is equal to s times what? s times the Laplace transform of y prime. Well, that's s times the Laplace transform of y minus y of 0, right?"}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's do that. So this over here, I'll do it in magenta, this is equal to s times what? s times the Laplace transform of y prime. Well, that's s times the Laplace transform of y minus y of 0, right? I took this part and replaced it with what I have in parentheses. So minus y prime of 0. And now I'll switch colors."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, that's s times the Laplace transform of y minus y of 0, right? I took this part and replaced it with what I have in parentheses. So minus y prime of 0. And now I'll switch colors. Plus 5 times, once again, the Laplace transform of y prime. Well, we could use this again. So 5 times s times the Laplace transform of y minus y of 0 plus 6 times the Laplace transform of y."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And now I'll switch colors. Plus 5 times, once again, the Laplace transform of y prime. Well, we could use this again. So 5 times s times the Laplace transform of y minus y of 0 plus 6 times the Laplace transform of y. All of that is equal to 0. I know this looks really confusing, but we'll simplify right now. We could get rid of this right here, because we've used it as much as we need to."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So 5 times s times the Laplace transform of y minus y of 0 plus 6 times the Laplace transform of y. All of that is equal to 0. I know this looks really confusing, but we'll simplify right now. We could get rid of this right here, because we've used it as much as we need to. So now we just simplify. And notice, using the Laplace transform, we didn't have to guess at a general solution or anything like that. Even when we did characteristic equation, we guessed what the original general solution was."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We could get rid of this right here, because we've used it as much as we need to. So now we just simplify. And notice, using the Laplace transform, we didn't have to guess at a general solution or anything like that. Even when we did characteristic equation, we guessed what the original general solution was. Now we're just taking Laplace transforms. And let's see where this gets us. So simplifying, actually I just want to make it clear, because I know it's very confusing."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Even when we did characteristic equation, we guessed what the original general solution was. Now we're just taking Laplace transforms. And let's see where this gets us. So simplifying, actually I just want to make it clear, because I know it's very confusing. So I rewrote this part as this, and I rewrote this thing as this. And everything else is the same. But now let's simplify the math."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So simplifying, actually I just want to make it clear, because I know it's very confusing. So I rewrote this part as this, and I rewrote this thing as this. And everything else is the same. But now let's simplify the math. So we get s squared times the Laplace transform of y. I'm going to write smaller. I've learned my lesson. Minus s times y of 0."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But now let's simplify the math. So we get s squared times the Laplace transform of y. I'm going to write smaller. I've learned my lesson. Minus s times y of 0. Let's substitute y of 0 here. y of 0 is 2. So s times y of 0 is 2 times s. So 2s."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Minus s times y of 0. Let's substitute y of 0 here. y of 0 is 2. So s times y of 0 is 2 times s. So 2s. Minus y prime of 0. y prime of 0 is 3. So minus 3. So we have 5 times s times the Laplace transform of y."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So s times y of 0 is 2 times s. So 2s. Minus y prime of 0. y prime of 0 is 3. So minus 3. So we have 5 times s times the Laplace transform of y. So plus 5s times the Laplace transform of y. Minus 5 times y of 0. y of 0 is 2. So minus 10."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we have 5 times s times the Laplace transform of y. So plus 5s times the Laplace transform of y. Minus 5 times y of 0. y of 0 is 2. So minus 10. 5 times, this is 2 right here. So 5 times 2. Plus 6 times the Laplace transform of y."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So minus 10. 5 times, this is 2 right here. So 5 times 2. Plus 6 times the Laplace transform of y. All of that is equal to 0. Now let's group our Laplace transform of y terms and our constant terms. And we should be hopefully getting someplace."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Plus 6 times the Laplace transform of y. All of that is equal to 0. Now let's group our Laplace transform of y terms and our constant terms. And we should be hopefully getting someplace. So let's see. My Laplace transform of y terms. I have this one."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And we should be hopefully getting someplace. So let's see. My Laplace transform of y terms. I have this one. I have this one. And I have that one. So what am I left with?"}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I have this one. I have this one. And I have that one. So what am I left with? Well let me factor out the Laplace transform of y part. So I get the Laplace transform of y. And that's good because it's a pain to keep writing it over and over, times s squared plus 5s plus 6."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So what am I left with? Well let me factor out the Laplace transform of y part. So I get the Laplace transform of y. And that's good because it's a pain to keep writing it over and over, times s squared plus 5s plus 6. So those are all my Laplace transform terms. And then I have my constant terms. So minus 2s minus 3 minus 10 is equal to 0."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And that's good because it's a pain to keep writing it over and over, times s squared plus 5s plus 6. So those are all my Laplace transform terms. And then I have my constant terms. So minus 2s minus 3 minus 10 is equal to 0. And what can we do here? Well this is interesting, first of all. Notice that the coefficients on the Laplace transform of y terms, that those are that characteristic equation that we dealt with so much."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So minus 2s minus 3 minus 10 is equal to 0. And what can we do here? Well this is interesting, first of all. Notice that the coefficients on the Laplace transform of y terms, that those are that characteristic equation that we dealt with so much. And that is hopefully, to some degree, second nature to you. So that's a little bit of a clue. And just if you want some very tenuous connections, well that makes a lot of sense."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Notice that the coefficients on the Laplace transform of y terms, that those are that characteristic equation that we dealt with so much. And that is hopefully, to some degree, second nature to you. So that's a little bit of a clue. And just if you want some very tenuous connections, well that makes a lot of sense. Because the characteristic equation, to get that we substituted e to the rt. And the Laplace transform involves a very similar function. But anyway, let's go back to the problem."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And just if you want some very tenuous connections, well that makes a lot of sense. Because the characteristic equation, to get that we substituted e to the rt. And the Laplace transform involves a very similar function. But anyway, let's go back to the problem. So how do we solve this? So actually, let me just give you the big picture here, because this is a good point. What I'm going to do is I'm going to solve this."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But anyway, let's go back to the problem. So how do we solve this? So actually, let me just give you the big picture here, because this is a good point. What I'm going to do is I'm going to solve this. I'm going to say the Laplace transform of y is equal to something. And then I'm going to say, boy, what functions Laplace transform is at something? And then I'll have the solution."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What I'm going to do is I'm going to solve this. I'm going to say the Laplace transform of y is equal to something. And then I'm going to say, boy, what functions Laplace transform is at something? And then I'll have the solution. If that confuses you, just wait and hopefully it'll make some sense. From here until that point, it's just some fairly hairy algebra. So let's scroll down a little bit, just so we have some breathing room."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then I'll have the solution. If that confuses you, just wait and hopefully it'll make some sense. From here until that point, it's just some fairly hairy algebra. So let's scroll down a little bit, just so we have some breathing room. And so I get the Laplace transform of y times s squared plus 5s plus 6 is equal to, let's add these terms to both sides of this equation, is equal to 2s plus 3 plus 10. Oh, that's silly, plus 13. This is minus 13 here."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's scroll down a little bit, just so we have some breathing room. And so I get the Laplace transform of y times s squared plus 5s plus 6 is equal to, let's add these terms to both sides of this equation, is equal to 2s plus 3 plus 10. Oh, that's silly, plus 13. This is minus 13 here. A phone call. Who's calling? I think it's some kind of marketing phone call."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is minus 13 here. A phone call. Who's calling? I think it's some kind of marketing phone call. Anyway, 2s plus 13. And now what can I do? Well, let's divide both sides by this s squared plus 5s plus 6."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I think it's some kind of marketing phone call. Anyway, 2s plus 13. And now what can I do? Well, let's divide both sides by this s squared plus 5s plus 6. So I get the Laplace transform of y is equal to 2s plus 13 over s squared plus 5s plus 6. Now we're almost done. Everything here is just a little bit of algebra."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, let's divide both sides by this s squared plus 5s plus 6. So I get the Laplace transform of y is equal to 2s plus 13 over s squared plus 5s plus 6. Now we're almost done. Everything here is just a little bit of algebra. So now we're almost done. We haven't solved for y yet, but we know that the Laplace transform of y is equal to this. Now if we just had this in our table of our Laplace transforms, we would immediately know what y was."}, {"video_title": "Laplace transform to solve an equation Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Everything here is just a little bit of algebra. So now we're almost done. We haven't solved for y yet, but we know that the Laplace transform of y is equal to this. Now if we just had this in our table of our Laplace transforms, we would immediately know what y was. But I don't see something, or I don't remember anything we did in our table, that looks like this expression of s. I'm essentially out of time. So in the next video, we're going to figure out what function's Laplace transform is this. And it actually turns out it's a sum of things we already know, and we just have to manipulate this a little bit algebraically."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "Derivative of y with respect to x is equal to y cosine of x divided by 1 plus 2y squared. And they give us an initial condition that y of 0 is equal to 1, or when x is equal to 0, y is equal to 1. And I know we did a couple already, but another way to think about separable differential equations is really all you're doing is implicit differentiation in reverse. Or another way to think about it is, whenever you took an implicit derivative, the end product was a separable differential equation. And so that just hopefully forms a little bit of a connection. But anyway, let's just do this. We have to separate the y's from the x's."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "Or another way to think about it is, whenever you took an implicit derivative, the end product was a separable differential equation. And so that just hopefully forms a little bit of a connection. But anyway, let's just do this. We have to separate the y's from the x's. Let's multiply both sides times 1 plus 2y squared. We get 1 plus 2y squared times dy dx is equal to y cosine of x. We still haven't fully separated the y's and the x's."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "We have to separate the y's from the x's. Let's multiply both sides times 1 plus 2y squared. We get 1 plus 2y squared times dy dx is equal to y cosine of x. We still haven't fully separated the y's and the x's. Let's divide both sides of this by y. And then let's see. We get 1 over y plus 2y squared divided by y."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "We still haven't fully separated the y's and the x's. Let's divide both sides of this by y. And then let's see. We get 1 over y plus 2y squared divided by y. That's just 2y times dy dx is equal to cosine of x. I can just multiply both sides by dx. 1 over y plus 2y times dy is equal to cosine of x dx. And now we can integrate both sides."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "We get 1 over y plus 2y squared divided by y. That's just 2y times dy dx is equal to cosine of x. I can just multiply both sides by dx. 1 over y plus 2y times dy is equal to cosine of x dx. And now we can integrate both sides. Let's integrate both sides. So what's the integral of 1 over y with respect to y? I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function than that whose derivative is actually 1 over y."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "And now we can integrate both sides. Let's integrate both sides. So what's the integral of 1 over y with respect to y? I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function than that whose derivative is actually 1 over y. And that's the natural log of the absolute value of y. And this is just a slightly broader function because its domain includes positive and negative numbers. It just excludes 0, while natural log of y only includes numbers larger than 0."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function than that whose derivative is actually 1 over y. And that's the natural log of the absolute value of y. And this is just a slightly broader function because its domain includes positive and negative numbers. It just excludes 0, while natural log of y only includes numbers larger than 0. So natural log of absolute value of y is nice. And it's actually true that at all points other than 0, its derivative is 1 over y. So it's just a slightly broader function."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "It just excludes 0, while natural log of y only includes numbers larger than 0. So natural log of absolute value of y is nice. And it's actually true that at all points other than 0, its derivative is 1 over y. So it's just a slightly broader function. So that's the antiderivative of 1 over y. We proved that, or at least we proved that the derivative of natural log of y is 1 over y. Plus, what's the antiderivative of 2y with respect to y?"}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "So it's just a slightly broader function. So that's the antiderivative of 1 over y. We proved that, or at least we proved that the derivative of natural log of y is 1 over y. Plus, what's the antiderivative of 2y with respect to y? Well, it's y squared. So 2y is equal to, I'll do the plus c on this side, whose derivative is cosine of x? What's sine of x?"}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "Plus, what's the antiderivative of 2y with respect to y? Well, it's y squared. So 2y is equal to, I'll do the plus c on this side, whose derivative is cosine of x? What's sine of x? And then we could add the plus c. We can add that plus c there. And what was our initial condition? y of 0 is equal to 1."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "What's sine of x? And then we could add the plus c. We can add that plus c there. And what was our initial condition? y of 0 is equal to 1. So when x is equal to 0, y is equal to 1. So when x is equal to 0, y is equal to 1. So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of 1, e to the what power is 1?"}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "y of 0 is equal to 1. So when x is equal to 0, y is equal to 1. So when x is equal to 0, y is equal to 1. So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of 1, e to the what power is 1? Well, 0. Plus 1. Sine of 0 is 0."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of 1, e to the what power is 1? Well, 0. Plus 1. Sine of 0 is 0. It's equal to c. So we get c is equal to 1. So the solution to this differential equation up here is, I don't even have to rewrite it. We figured out c is equal to 1."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "Sine of 0 is 0. It's equal to c. So we get c is equal to 1. So the solution to this differential equation up here is, I don't even have to rewrite it. We figured out c is equal to 1. So we could just scratch this out. We could put a 1. The natural log of the absolute value of y plus y squared is equal to sine of x plus 1."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "We figured out c is equal to 1. So we could just scratch this out. We could put a 1. The natural log of the absolute value of y plus y squared is equal to sine of x plus 1. And actually, if you were to graph this, you would see that y never actually dips below or even hits the x-axis. So you could actually get rid of that absolute value function there. But anyway, that's just a little technicality."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "The natural log of the absolute value of y plus y squared is equal to sine of x plus 1. And actually, if you were to graph this, you would see that y never actually dips below or even hits the x-axis. So you could actually get rid of that absolute value function there. But anyway, that's just a little technicality. But this is the implicit form of the solution to this differential equation. That makes sense, because these separable differential equations are really just implicit derivatives backwards. And in general, one thing that's kind of fun about differential equations, but kind of not as satisfying about differential equations, is it really is just a whole hodgepodge of tools to solve different types of equations."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "But anyway, that's just a little technicality. But this is the implicit form of the solution to this differential equation. That makes sense, because these separable differential equations are really just implicit derivatives backwards. And in general, one thing that's kind of fun about differential equations, but kind of not as satisfying about differential equations, is it really is just a whole hodgepodge of tools to solve different types of equations. There isn't just one tool or one theory that will solve all differential equations. There are a few that will solve a certain class of differential equations. But there's not just one consistent way to solve all of them."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "And in general, one thing that's kind of fun about differential equations, but kind of not as satisfying about differential equations, is it really is just a whole hodgepodge of tools to solve different types of equations. There isn't just one tool or one theory that will solve all differential equations. There are a few that will solve a certain class of differential equations. But there's not just one consistent way to solve all of them. And even today, there are unsolved differential equations, where the only way that we know how to get solutions is using a computer and numerically. And one day I'll do videos on that. And actually, you'll find that in most applications, that's what you end up doing anyway, because most differential equations you encounter in science or in any kind of science, whether it's economics or physics or engineering, they often are unsolvable, because they're going to have a second or third derivatives involved, or they're going to multiply."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "But there's not just one consistent way to solve all of them. And even today, there are unsolved differential equations, where the only way that we know how to get solutions is using a computer and numerically. And one day I'll do videos on that. And actually, you'll find that in most applications, that's what you end up doing anyway, because most differential equations you encounter in science or in any kind of science, whether it's economics or physics or engineering, they often are unsolvable, because they're going to have a second or third derivatives involved, or they're going to multiply. I mean, they're just going to be really complicated, very hard to solve analytically. And you actually are going to solve them numerically, which is often much easier. But anyway, hopefully at this point you have a pretty good sense of separable equations."}, {"video_title": "Old separable differential equations example First order differential equations Khan Academy.mp3", "Sentence": "And actually, you'll find that in most applications, that's what you end up doing anyway, because most differential equations you encounter in science or in any kind of science, whether it's economics or physics or engineering, they often are unsolvable, because they're going to have a second or third derivatives involved, or they're going to multiply. I mean, they're just going to be really complicated, very hard to solve analytically. And you actually are going to solve them numerically, which is often much easier. But anyway, hopefully at this point you have a pretty good sense of separable equations. There's just implicit differentiation backwards, and it's really nothing new. Our next thing we'll learn is exact differential equations. And then we'll go off into more and more methods."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "We'll now move from the world of first order differential equations to the world of second order differential equations. What does that mean? That means that we're now going to start involving the second derivative. And the first class that I'm going to show you, and this is probably the most useful class when you're studying classical physics, are linear second order differential equations. So what is a linear second order differential equation? So I think I touched on it a little bit in our very first intro video, but it's something that looks like this. If I have a of x, so some function only of x, times the second derivative of y with respect to x, plus b of x times the first derivative of y with respect to x, plus c of x times y is equal to some function that's only a function of x."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "And the first class that I'm going to show you, and this is probably the most useful class when you're studying classical physics, are linear second order differential equations. So what is a linear second order differential equation? So I think I touched on it a little bit in our very first intro video, but it's something that looks like this. If I have a of x, so some function only of x, times the second derivative of y with respect to x, plus b of x times the first derivative of y with respect to x, plus c of x times y is equal to some function that's only a function of x. So just to review our terminology, y is the second order because the highest derivative here is the second derivative, so that makes it second order. And what makes it linear? Well, all of the coefficients on, and I want to be careful with the term coefficients because traditionally we view coefficients as always being constants, but here we have functions of x as coefficients."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "If I have a of x, so some function only of x, times the second derivative of y with respect to x, plus b of x times the first derivative of y with respect to x, plus c of x times y is equal to some function that's only a function of x. So just to review our terminology, y is the second order because the highest derivative here is the second derivative, so that makes it second order. And what makes it linear? Well, all of the coefficients on, and I want to be careful with the term coefficients because traditionally we view coefficients as always being constants, but here we have functions of x as coefficients. So in order for this to be a linear differential equation, a of x, b of x, c of x, and d of x, they all have to be functions only of x, as I've drawn it here. And now before we start trying to solve this generally, we'll do a special case of this where a, b, c are constants and d is 0. So what will that look like?"}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "Well, all of the coefficients on, and I want to be careful with the term coefficients because traditionally we view coefficients as always being constants, but here we have functions of x as coefficients. So in order for this to be a linear differential equation, a of x, b of x, c of x, and d of x, they all have to be functions only of x, as I've drawn it here. And now before we start trying to solve this generally, we'll do a special case of this where a, b, c are constants and d is 0. So what will that look like? So I could just rewrite that as a, so now a is not a function anymore, it's just a number. a times the second derivative of y with respect to x plus b times the first derivative plus c times y. And instead of having just a fourth constant, instead of d of x, I'm just going to set that equal to 0."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "So what will that look like? So I could just rewrite that as a, so now a is not a function anymore, it's just a number. a times the second derivative of y with respect to x plus b times the first derivative plus c times y. And instead of having just a fourth constant, instead of d of x, I'm just going to set that equal to 0. And by setting this equal to 0, I have now introduced you to the other form of homogenous differential equation, and this one is called homogenous. And I haven't made the connection yet on how these second order differential equations are related to the first order ones that I just introduced, those other homogenous differential equations I introduced you to. I think they just happen to have the same name, even though they're not that related."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "And instead of having just a fourth constant, instead of d of x, I'm just going to set that equal to 0. And by setting this equal to 0, I have now introduced you to the other form of homogenous differential equation, and this one is called homogenous. And I haven't made the connection yet on how these second order differential equations are related to the first order ones that I just introduced, those other homogenous differential equations I introduced you to. I think they just happen to have the same name, even though they're not that related. So the reason why this one is called homogenous is because you have it equal to 0. So this is what makes it homogenous. And actually, I do see more of a connection between this type of equation and milk where all the fat is spread out."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "I think they just happen to have the same name, even though they're not that related. So the reason why this one is called homogenous is because you have it equal to 0. So this is what makes it homogenous. And actually, I do see more of a connection between this type of equation and milk where all the fat is spread out. Because if you think about it, no matter the solution for all homogenous equations, or when you kind of apply, when you kind of solve the equation, they always equal 0. So they're homogenized, I guess, is the best way that I can draw any kind of parallel. So we could call this a second order linear, because a, b, and c definitely are functions just of, well, they're not even functions of x or y, they're just constants."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "And actually, I do see more of a connection between this type of equation and milk where all the fat is spread out. Because if you think about it, no matter the solution for all homogenous equations, or when you kind of apply, when you kind of solve the equation, they always equal 0. So they're homogenized, I guess, is the best way that I can draw any kind of parallel. So we could call this a second order linear, because a, b, and c definitely are functions just of, well, they're not even functions of x or y, they're just constants. So second order linear, homogenous, because they equal 0, differential equations. And I think you'll see that these are, in some ways, the most fun differential equations to solve. And actually, often the most useful, because a lot of the applications in classical mechanics, this is all you need to solve."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "So we could call this a second order linear, because a, b, and c definitely are functions just of, well, they're not even functions of x or y, they're just constants. So second order linear, homogenous, because they equal 0, differential equations. And I think you'll see that these are, in some ways, the most fun differential equations to solve. And actually, often the most useful, because a lot of the applications in classical mechanics, this is all you need to solve. But they're the most fun to solve because they all boil down to algebra 2 problems. And I'll touch on that in a second. But let's just think about this a little bit."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "And actually, often the most useful, because a lot of the applications in classical mechanics, this is all you need to solve. But they're the most fun to solve because they all boil down to algebra 2 problems. And I'll touch on that in a second. But let's just think about this a little bit. Think about what the properties of these solutions might be. Let me just throw out something. Let's say that g of x is a solution."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "But let's just think about this a little bit. Think about what the properties of these solutions might be. Let me just throw out something. Let's say that g of x is a solution. So that means that a times g prime prime plus b times g prime plus c times g is equal to 0. That means the same thing. Now my question to you is, what if I have some constant times g?"}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "Let's say that g of x is a solution. So that means that a times g prime prime plus b times g prime plus c times g is equal to 0. That means the same thing. Now my question to you is, what if I have some constant times g? Is that still a solution? So let's say, so my question is, let's say some constant, c1 gx, c1 times g. Is this a solution? Let's try it out."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "Now my question to you is, what if I have some constant times g? Is that still a solution? So let's say, so my question is, let's say some constant, c1 gx, c1 times g. Is this a solution? Let's try it out. Let's substitute this into our original equation. So a times the second derivative of this would just be, and I'll switch colors here. I'll switch maybe, let me switch to brown."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "Let's try it out. Let's substitute this into our original equation. So a times the second derivative of this would just be, and I'll switch colors here. I'll switch maybe, let me switch to brown. So a times the second derivative of this would be the constant, every time you take a derivative, the constant just carries over, right? So that'll just be a times c1 g prime prime plus the same thing for the first derivative, b times c1 g prime plus c times, and this c is different than this c1 c, times g. And let's see whether this is equal to 0. So we could factor out that c1 constant and we get c1 times a g prime prime plus b g prime plus cg."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "I'll switch maybe, let me switch to brown. So a times the second derivative of this would be the constant, every time you take a derivative, the constant just carries over, right? So that'll just be a times c1 g prime prime plus the same thing for the first derivative, b times c1 g prime plus c times, and this c is different than this c1 c, times g. And let's see whether this is equal to 0. So we could factor out that c1 constant and we get c1 times a g prime prime plus b g prime plus cg. And lo and behold, we already know, because we know that g of x is a solution, we know that this is true, so this is going to be equal to 0, right? Because g is a solution. So if this is 0, c1 times 0 is going to be equal to 0."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "So we could factor out that c1 constant and we get c1 times a g prime prime plus b g prime plus cg. And lo and behold, we already know, because we know that g of x is a solution, we know that this is true, so this is going to be equal to 0, right? Because g is a solution. So if this is 0, c1 times 0 is going to be equal to 0. So this expression up here is also equal to 0. Or another way to view it is that if g is a solution to this second order linear homogenous differential equation, then some constant times g is also a solution. So this is also a solution to the differential equation."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "So if this is 0, c1 times 0 is going to be equal to 0. So this expression up here is also equal to 0. Or another way to view it is that if g is a solution to this second order linear homogenous differential equation, then some constant times g is also a solution. So this is also a solution to the differential equation. And then the next property I want to show you, and this is all going someplace, don't worry. The next question I want to ask you is, OK, we know that g of x is a solution to the differential equation. What if I were to also tell you that h of x is also a solution?"}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "So this is also a solution to the differential equation. And then the next property I want to show you, and this is all going someplace, don't worry. The next question I want to ask you is, OK, we know that g of x is a solution to the differential equation. What if I were to also tell you that h of x is also a solution? So my question to you is, is g of x plus h of x a solution? If you add these two functions that are both solutions, if you add them together, is that still a solution of our original differential equation? Well, let's substitute this whole thing into our original differential equation, right?"}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "What if I were to also tell you that h of x is also a solution? So my question to you is, is g of x plus h of x a solution? If you add these two functions that are both solutions, if you add them together, is that still a solution of our original differential equation? Well, let's substitute this whole thing into our original differential equation, right? So we'll have g of x plus h of x. a times the second derivative of this thing. Well, that's straightforward enough. That's just g prime prime plus h prime prime."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "Well, let's substitute this whole thing into our original differential equation, right? So we'll have g of x plus h of x. a times the second derivative of this thing. Well, that's straightforward enough. That's just g prime prime plus h prime prime. Plus b times the first derivative of this thing, g prime plus h prime. Plus c times this function, g plus h. And now what can we do? Let's distribute all of these constants."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "That's just g prime prime plus h prime prime. Plus b times the first derivative of this thing, g prime plus h prime. Plus c times this function, g plus h. And now what can we do? Let's distribute all of these constants. We get a times g prime prime plus a times h prime prime plus b times the first derivative of g plus b times the first derivative of h plus c times g plus c times h. And now we can rearrange them. We get a, let's take this one, let's take all the g terms, a times the second derivative of g plus b times the first derivative plus c times g. That's these three terms. Plus a times the second derivative of h plus b times first derivative plus c times h. And now we know that both g and h are solutions of the original differential equation."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "Let's distribute all of these constants. We get a times g prime prime plus a times h prime prime plus b times the first derivative of g plus b times the first derivative of h plus c times g plus c times h. And now we can rearrange them. We get a, let's take this one, let's take all the g terms, a times the second derivative of g plus b times the first derivative plus c times g. That's these three terms. Plus a times the second derivative of h plus b times first derivative plus c times h. And now we know that both g and h are solutions of the original differential equation. So by definition, if g is a solution of the original differential equation, this was the left-hand side of that differential equation, this is going to be equal to 0. This is going to be equal to 0. And so is this going to be equal to 0."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "Plus a times the second derivative of h plus b times first derivative plus c times h. And now we know that both g and h are solutions of the original differential equation. So by definition, if g is a solution of the original differential equation, this was the left-hand side of that differential equation, this is going to be equal to 0. This is going to be equal to 0. And so is this going to be equal to 0. So we've shown that this whole expression is equal to 0. So if g is a solution of the differential equation, of this second-order linear homogenous differential equation, and h is also a solution, then if you were to add them together, the sum of them is also a solution. So in general, if we show that g is a solution and h is a solution, you can add them."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "And so is this going to be equal to 0. So we've shown that this whole expression is equal to 0. So if g is a solution of the differential equation, of this second-order linear homogenous differential equation, and h is also a solution, then if you were to add them together, the sum of them is also a solution. So in general, if we show that g is a solution and h is a solution, you can add them. And we showed before that any constant times them is also a solution. So you could also say that some constant times g of x plus some constant times h of x is also a solution. And maybe the constant in one of the cases is 0 or something."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "So in general, if we show that g is a solution and h is a solution, you can add them. And we showed before that any constant times them is also a solution. So you could also say that some constant times g of x plus some constant times h of x is also a solution. And maybe the constant in one of the cases is 0 or something. I don't know. But anyway, these are useful properties to maybe internalize for second-order homogenous linear differential equations. And in the next video, we're actually going to apply these properties to figure out the solutions for these."}, {"video_title": "2nd order linear homogeneous differential equations 1 Khan Academy.mp3", "Sentence": "And maybe the constant in one of the cases is 0 or something. I don't know. But anyway, these are useful properties to maybe internalize for second-order homogenous linear differential equations. And in the next video, we're actually going to apply these properties to figure out the solutions for these. And you'll see that they're actually straightforward. I would say a lot easier than what we did with the previous first-order homogenous differential equations or the exact equations. This is much, much easier."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Let's say we don't know how to find the solutions to this, but we at least want to get a sense of what the solutions might look like. And to do that, what we could do is we could look at a coordinate plane. So, let me draw some axes here. So, let me draw a relatively straight line. Alright, so that's my y-axis, and this is my x-axis. Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two. And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could visualize the slope, if a solution goes to that point, what the slope needs to be there, and I can visualize that with a line segment, a little small line segment that has the same slope as the slope in question."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, let me draw a relatively straight line. Alright, so that's my y-axis, and this is my x-axis. Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two. And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could visualize the slope, if a solution goes to that point, what the slope needs to be there, and I can visualize that with a line segment, a little small line segment that has the same slope as the slope in question. So, let's actually do that. Let me set up a little table here. Let me do a little table here to do a bunch of x and y values."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could visualize the slope, if a solution goes to that point, what the slope needs to be there, and I can visualize that with a line segment, a little small line segment that has the same slope as the slope in question. So, let's actually do that. Let me set up a little table here. Let me do a little table here to do a bunch of x and y values. Once again, I'm just sampling some points on the coordinate plane to be able to visualize. So, x, y, and this is dy, dx. So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x?"}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Let me do a little table here to do a bunch of x and y values. Once again, I'm just sampling some points on the coordinate plane to be able to visualize. So, x, y, and this is dy, dx. So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative zero over one, so it's just going to be zero. And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero. And so, we can visualize that by doing a little horizontal line segment right there."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative zero over one, so it's just going to be zero. And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero. And so, we can visualize that by doing a little horizontal line segment right there. So, let's keep going. What about when x is one and y is one? Well then, dy, dx, the derivative of y with respect to x is negative one over one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And so, we can visualize that by doing a little horizontal line segment right there. So, let's keep going. What about when x is one and y is one? Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one. So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one. And so, I draw a little line segment that has a slope of negative one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one. So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one. And so, I draw a little line segment that has a slope of negative one. What about when x is, let me do this in a new color, what about when x is one and y is zero? Well then, it's negative one over zero. So, this is actually undefined."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And so, I draw a little line segment that has a slope of negative one. What about when x is, let me do this in a new color, what about when x is one and y is zero? Well then, it's negative one over zero. So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical. So, I'll put that as a question mark. Vertical there."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical. So, I'll put that as a question mark. Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it. But, let's not draw that just yet, but let's try some other points. Let's say that we had, let's try the point negative one, negative one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it. But, let's not draw that just yet, but let's try some other points. Let's say that we had, let's try the point negative one, negative one. So, now we have negative negative one, which is one over negative one. Well, you would have a slope of negative one here. So, negative one, negative one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Let's say that we had, let's try the point negative one, negative one. So, now we have negative negative one, which is one over negative one. Well, you would have a slope of negative one here. So, negative one, negative one. You would have a slope of, you would have a slope of negative one. What about if you had one negative one? Well, now it's negative one over negative one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, negative one, negative one. You would have a slope of, you would have a slope of negative one. What about if you had one negative one? Well, now it's negative one over negative one. Your slope is now one. So, one negative one. If your solution, if a solution goes through this, its slope would look like that."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Well, now it's negative one over negative one. Your slope is now one. So, one negative one. If your solution, if a solution goes through this, its slope would look like that. And we could keep going. We could even do two negative two. That's going to have a slope of one as well."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "If your solution, if a solution goes through this, its slope would look like that. And we could keep going. We could even do two negative two. That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here. And so, we could do a bunch of points."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here. And so, we could do a bunch of points. Just keep going. I'm now just doing them in my head. I'm not going on the table."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And so, we could do a bunch of points. Just keep going. I'm now just doing them in my head. I'm not going on the table. But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one. It's going to have a slope of one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "I'm not going on the table. But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one. It's going to have a slope of one. So, at this point, your slope, negative one, one. So, negative negative one is one over one. So, you're going to have a slope like that."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "It's going to have a slope of one. So, at this point, your slope, negative one, one. So, negative negative one is one over one. So, you're going to have a slope like that. At negative two, two, same exact idea. It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do?"}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, you're going to have a slope like that. At negative two, two, same exact idea. It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something like this. This would be a solution. So, maybe it would have to do something like this."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something like this. This would be a solution. So, maybe it would have to do something like this. Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this. Or, if the function started like here, based on what we've seen so far, maybe it would have to do something, maybe it would have to do something like this. Or, if it was, if this were a point on the function over here, it would have to do something like this."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, maybe it would have to do something like this. Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this. Or, if the function started like here, based on what we've seen so far, maybe it would have to do something, maybe it would have to do something like this. Or, if it was, if this were a point on the function over here, it would have to do something like this. And once again, I'm doing this based on what the slope field is telling me. So, this field that I'm creating, where I'm taking, I'm sampling a bunch of points and I'm visualizing the slope with the line segment, once again, this is called a slope, slope field. So, hopefully that gives you kind of the basic idea of what a slope field is."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "And so here we have a differential equation, and we haven't started exploring how we find the solutions for differential equations yet, but let's just say you saw this and someone just walked up to you on the street and says, hey, I will give you a clue, that there's a solution to this differential equation that is essentially a linear function, where y is equal to mx plus b, and you just need to figure out the m's and the b's, or maybe the m and the b, that makes this linear function satisfy this differential equation. And what I now encourage you to do is pause the video and see if you can do it. So I'm assuming you have paused it and had a go at it, so let's think this through together. So if we know that this kind of a solution can be described in this way, we have to figure out some m's and b's here, this is telling us that if we were to take the derivative of this with respect to x, if we take the derivative of mx plus b with respect to x, that that should be equal to negative two times x plus three times y, well, we know y is this thing, minus five, and that should be true for all x's. In order for this to be a solution to this differential equation, remember, a solution to a differential equation is not a value or a set of values, it is a function or a set of functions. So in order for this to be, to satisfy this differential equation, it needs to be true for all of these x's here. So let's work through it."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So if we know that this kind of a solution can be described in this way, we have to figure out some m's and b's here, this is telling us that if we were to take the derivative of this with respect to x, if we take the derivative of mx plus b with respect to x, that that should be equal to negative two times x plus three times y, well, we know y is this thing, minus five, and that should be true for all x's. In order for this to be a solution to this differential equation, remember, a solution to a differential equation is not a value or a set of values, it is a function or a set of functions. So in order for this to be, to satisfy this differential equation, it needs to be true for all of these x's here. So let's work through it. Let's figure out first what our dy dx is. So dy dx, we'll just take the derivative here with respect to x, dy dx is derivative of mx with respect to x is just going to be m, and of course derivative of b with respect to x is just a constant, so it's just going to be zero. So dy dx is m. So we could write m is equal to negative two x, is equal to negative two x plus three times, instead of putting y there, I could write mx plus b."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So let's work through it. Let's figure out first what our dy dx is. So dy dx, we'll just take the derivative here with respect to x, dy dx is derivative of mx with respect to x is just going to be m, and of course derivative of b with respect to x is just a constant, so it's just going to be zero. So dy dx is m. So we could write m is equal to negative two x, is equal to negative two x plus three times, instead of putting y there, I could write mx plus b. Remember, y is equal to mx plus b. And just as a repeated reminder, this has to be true for all x's. mx plus b, and then of course we have the minus five."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So dy dx is m. So we could write m is equal to negative two x, is equal to negative two x plus three times, instead of putting y there, I could write mx plus b. Remember, y is equal to mx plus b. And just as a repeated reminder, this has to be true for all x's. mx plus b, and then of course we have the minus five. And so if you weren't able to solve it the first time, I encourage you to start from here and now figure out what m and b needs to be in order for this equation right over here, in order for this to be true for all x's. In order for this to be true for all x's. So I'm assuming you have paused again and had a go at it, so let's just keep algebraically manipulating this."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "mx plus b, and then of course we have the minus five. And so if you weren't able to solve it the first time, I encourage you to start from here and now figure out what m and b needs to be in order for this equation right over here, in order for this to be true for all x's. In order for this to be true for all x's. So I'm assuming you have paused again and had a go at it, so let's just keep algebraically manipulating this. And I'll just switch to one color here. So we have m, m is equal to negative two x plus, if we distribute this three, we're going to have three mx plus three b, and then of course we're going to have minus five. And now we can group the x terms."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So I'm assuming you have paused again and had a go at it, so let's just keep algebraically manipulating this. And I'll just switch to one color here. So we have m, m is equal to negative two x plus, if we distribute this three, we're going to have three mx plus three b, and then of course we're going to have minus five. And now we can group the x terms. So if we were to group, if we were to group, let me find a new color here, maybe this blue. So if we were to take these two and add them together, that's going to be negative two plus three m times x, or we could write this as three m minus two times x, and then you have your constant terms. So you have these terms right over here."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "And now we can group the x terms. So if we were to group, if we were to group, let me find a new color here, maybe this blue. So if we were to take these two and add them together, that's going to be negative two plus three m times x, or we could write this as three m minus two times x, and then you have your constant terms. So you have these terms right over here. So plus three b minus five, and of course that's all going to be equal to m. That's going to be equal to m. Now remember, this needs to be true, this needs to be true for all x's. So notice, over here I have some, I have some coefficient times x on the right-hand side, but on the left-hand side I have no x's. So somehow this thing must disappear."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So you have these terms right over here. So plus three b minus five, and of course that's all going to be equal to m. That's going to be equal to m. Now remember, this needs to be true, this needs to be true for all x's. So notice, over here I have some, I have some coefficient times x on the right-hand side, but on the left-hand side I have no x's. So somehow this thing must disappear. This is a constant, so it's completely reasonable, it's completely reasonable that this constant could be equal to m, but the only way that I get these x's to disappear, so all I'm left with is an m, is if this thing is equal to zero. Let me say that again, because I think it might maybe be a little bit counterintuitive what I'm about to do. We're saying that m, some constant value, is equal to some coefficient times x plus some other constant value."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So somehow this thing must disappear. This is a constant, so it's completely reasonable, it's completely reasonable that this constant could be equal to m, but the only way that I get these x's to disappear, so all I'm left with is an m, is if this thing is equal to zero. Let me say that again, because I think it might maybe be a little bit counterintuitive what I'm about to do. We're saying that m, some constant value, is equal to some coefficient times x plus some other constant value. Well, in order for a constant value to be equal to a coefficient times x plus some other constant value, the coefficient on x must be equal to zero. Another way to think about it is, this should be, you could rewrite the left-hand side here as zero x plus m. And so you see, you kind of match the coefficients. So zero must be equal to three m minus two, and m is equal to three b minus five."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "We're saying that m, some constant value, is equal to some coefficient times x plus some other constant value. Well, in order for a constant value to be equal to a coefficient times x plus some other constant value, the coefficient on x must be equal to zero. Another way to think about it is, this should be, you could rewrite the left-hand side here as zero x plus m. And so you see, you kind of match the coefficients. So zero must be equal to three m minus two, and m is equal to three b minus five. m is equal to three b minus five. So let's use that knowledge, that information, to solve for m and b. So we could use this first one."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So zero must be equal to three m minus two, and m is equal to three b minus five. m is equal to three b minus five. So let's use that knowledge, that information, to solve for m and b. So we could use this first one. So three m minus two must be equal to zero. So let's write that. Three m minus two is equal to zero, or three m is equal to two, or m is equal to 2 3rds."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "So we could use this first one. So three m minus two must be equal to zero. So let's write that. Three m minus two is equal to zero, or three m is equal to two, or m is equal to 2 3rds. So we figured out what m is. And then we could use that information because we know that m is equal to three b minus five. m is 2 3rds, so we get 2 3rds is equal to three b minus five."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "Three m minus two is equal to zero, or three m is equal to two, or m is equal to 2 3rds. So we figured out what m is. And then we could use that information because we know that m is equal to three b minus five. m is 2 3rds, so we get 2 3rds is equal to three b minus five. We could add five to both sides, which is the same thing as adding 15 3rds to both sides. Did I do that right? Yeah, adding five to both sides is the same thing as adding 15 3rds to both sides."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "m is 2 3rds, so we get 2 3rds is equal to three b minus five. We could add five to both sides, which is the same thing as adding 15 3rds to both sides. Did I do that right? Yeah, adding five to both sides is the same thing as adding 15 3rds to both sides. So let's do that, 15 over three plus 15 over three. These cancel out, that's just five right over there on the left-hand side. We have 17 over three is equal to three b."}, {"video_title": "Finding particular linear solution to differential equation Khan Academy.mp3", "Sentence": "Yeah, adding five to both sides is the same thing as adding 15 3rds to both sides. So let's do that, 15 over three plus 15 over three. These cancel out, that's just five right over there on the left-hand side. We have 17 over three is equal to three b. Or if you divide both sides by three, you get b is equal to 17 over nine. And we're done. We just found a particular solution for this differential equation."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Now, I'll introduce you to the concept of exact equations. And it's just another method for solving a certain type of differential equations. Let me write that down. Exact equations. Before I show you what an exact equation is, I'm just going to give you a little bit of the building blocks, just so that when I later prove it, or at least give you the intuition behind it, it doesn't seem like it's coming out of the blue. So let's say I had some function of x and y, and we'll call it psi, just because that's what people tend to use for these exact equations. So psi is a function of x and y."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Exact equations. Before I show you what an exact equation is, I'm just going to give you a little bit of the building blocks, just so that when I later prove it, or at least give you the intuition behind it, it doesn't seem like it's coming out of the blue. So let's say I had some function of x and y, and we'll call it psi, just because that's what people tend to use for these exact equations. So psi is a function of x and y. So you're probably not familiar with taking the chain rule onto partial derivatives, but I'll show it to you now. And I'll give you a little intuition, although I won't prove it. So if I were to take the derivative of this with respect to x, where y is also a function of x."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So psi is a function of x and y. So you're probably not familiar with taking the chain rule onto partial derivatives, but I'll show it to you now. And I'll give you a little intuition, although I won't prove it. So if I were to take the derivative of this with respect to x, where y is also a function of x. So maybe I could also write this as y, sorry, not y, psi. So I could also write this as psi as x and y, which is a function of x. I could write it just like that. These are just two different ways of writing the same thing."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So if I were to take the derivative of this with respect to x, where y is also a function of x. So maybe I could also write this as y, sorry, not y, psi. So I could also write this as psi as x and y, which is a function of x. I could write it just like that. These are just two different ways of writing the same thing. Now, if I were to take the derivative of psi with respect to x, and these are just the building blocks. If I were to take the derivative of psi with respect to x, it is equal to, and this is the chain rule using partial derivatives, and I won't prove it, but I'll give you the intuition right here. So this is going to be equal to the partial derivative of psi with respect to x plus the partial derivative of psi with respect to y times dy dx."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "These are just two different ways of writing the same thing. Now, if I were to take the derivative of psi with respect to x, and these are just the building blocks. If I were to take the derivative of psi with respect to x, it is equal to, and this is the chain rule using partial derivatives, and I won't prove it, but I'll give you the intuition right here. So this is going to be equal to the partial derivative of psi with respect to x plus the partial derivative of psi with respect to y times dy dx. And this should make a little bit of intuition. I'm kind of taking the derivative with respect to x, and then if you could say, and I know you can't, because this partial with respect to y and the dy, they're two different things, but if these canceled out, then you'd kind of have another partial with respect to x. And if you were to kind of add them up, then you would get the full derivative with respect to x."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So this is going to be equal to the partial derivative of psi with respect to x plus the partial derivative of psi with respect to y times dy dx. And this should make a little bit of intuition. I'm kind of taking the derivative with respect to x, and then if you could say, and I know you can't, because this partial with respect to y and the dy, they're two different things, but if these canceled out, then you'd kind of have another partial with respect to x. And if you were to kind of add them up, then you would get the full derivative with respect to x. That's not even in the intuition. That's just to kind of show you that even this should make a little bit of intuitive sense. Now the intuition here, let's just say, and I'm not, let's just say psi, and psi doesn't always have to take this form, but you could use the same methodology to take psi to kind of more complex notations."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And if you were to kind of add them up, then you would get the full derivative with respect to x. That's not even in the intuition. That's just to kind of show you that even this should make a little bit of intuitive sense. Now the intuition here, let's just say, and I'm not, let's just say psi, and psi doesn't always have to take this form, but you could use the same methodology to take psi to kind of more complex notations. But let's say that psi, and I won't write this, it's a function of x and y. We know it's a function of x and y. Let's say it's equal to some function of x, we'll call that f1 of x, times some function of y."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Now the intuition here, let's just say, and I'm not, let's just say psi, and psi doesn't always have to take this form, but you could use the same methodology to take psi to kind of more complex notations. But let's say that psi, and I won't write this, it's a function of x and y. We know it's a function of x and y. Let's say it's equal to some function of x, we'll call that f1 of x, times some function of y. And let's say there's a bunch of terms like this. So there's n terms like this. Plus all the way, and the n-th term is the n-th function of x times the n-th function of y. I just define psi like this, just so I can give you the intuition that when I use implicit differentiation on this, when I take the derivative of this with respect to x, I actually get something that looks just like that."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Let's say it's equal to some function of x, we'll call that f1 of x, times some function of y. And let's say there's a bunch of terms like this. So there's n terms like this. Plus all the way, and the n-th term is the n-th function of x times the n-th function of y. I just define psi like this, just so I can give you the intuition that when I use implicit differentiation on this, when I take the derivative of this with respect to x, I actually get something that looks just like that. So what's the derivative of psi with respect to x? The derivative of psi with respect to x, and this is just the implicit differentiation that you learned in your first, or that you hopefully learned in your first semester calculus course. That's equal, we just do the product rule, right?"}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Plus all the way, and the n-th term is the n-th function of x times the n-th function of y. I just define psi like this, just so I can give you the intuition that when I use implicit differentiation on this, when I take the derivative of this with respect to x, I actually get something that looks just like that. So what's the derivative of psi with respect to x? The derivative of psi with respect to x, and this is just the implicit differentiation that you learned in your first, or that you hopefully learned in your first semester calculus course. That's equal, we just do the product rule, right? So the first expression, you take the derivative of that with respect to x, well that's just going to be f1 prime of x times the second function, well that's just g1 of y. Now you add that to the derivative of the second function times the first function. So plus f1 of x, that's just the first function, times the derivative of the second function."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "That's equal, we just do the product rule, right? So the first expression, you take the derivative of that with respect to x, well that's just going to be f1 prime of x times the second function, well that's just g1 of y. Now you add that to the derivative of the second function times the first function. So plus f1 of x, that's just the first function, times the derivative of the second function. Now the derivative of the second function, it's going to be this function with respect to y, so you could write that as g1 prime of y. But of course we're doing the chain rule, so it's that times dy dx, and you might want to review the implicit differentiation videos if this seems a little bit foreign. But this right here, what I just did, this expression right here, this is the derivative with respect to x of this, right?"}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So plus f1 of x, that's just the first function, times the derivative of the second function. Now the derivative of the second function, it's going to be this function with respect to y, so you could write that as g1 prime of y. But of course we're doing the chain rule, so it's that times dy dx, and you might want to review the implicit differentiation videos if this seems a little bit foreign. But this right here, what I just did, this expression right here, this is the derivative with respect to x of this, right? And we have n terms like that, so if we keep adding them, I'll do them vertically down. So plus, and then you'd have a bunch of them, and then the last one's going to look the same, it's just the nth function of x. So fn prime of x times the second function, gn of y, plus the first function, fn of x, times the derivative of the second function."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "But this right here, what I just did, this expression right here, this is the derivative with respect to x of this, right? And we have n terms like that, so if we keep adding them, I'll do them vertically down. So plus, and then you'd have a bunch of them, and then the last one's going to look the same, it's just the nth function of x. So fn prime of x times the second function, gn of y, plus the first function, fn of x, times the derivative of the second function. The derivative of the second function with respect to y is just g prime of y times dy dx. That's just the chain rule. dy dx."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So fn prime of x times the second function, gn of y, plus the first function, fn of x, times the derivative of the second function. The derivative of the second function with respect to y is just g prime of y times dy dx. That's just the chain rule. dy dx. Now we have two n terms. We had n terms here, right? Where each term was f of x times g of y, or f1 of x times g1 of y, and then all the way to fn of x times gn of y."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "dy dx. Now we have two n terms. We had n terms here, right? Where each term was f of x times g of y, or f1 of x times g1 of y, and then all the way to fn of x times gn of y. Now we have, for each of those, we got two of them when we did the product rule. If we group the terms, so if we group all the terms that don't have a dy dx on them, what do we get? If we add all of these, I guess you could call them on the left-hand side, you get, I'm just rearranging, it all equals f1 prime of x times g1 of y plus f2, g2, all the way to fn prime of x, gn of y."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Where each term was f of x times g of y, or f1 of x times g1 of y, and then all the way to fn of x times gn of y. Now we have, for each of those, we got two of them when we did the product rule. If we group the terms, so if we group all the terms that don't have a dy dx on them, what do we get? If we add all of these, I guess you could call them on the left-hand side, you get, I'm just rearranging, it all equals f1 prime of x times g1 of y plus f2, g2, all the way to fn prime of x, gn of y. That's just all of these added up. Plus all of these added up. All of the terms that have the dy dx in them."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "If we add all of these, I guess you could call them on the left-hand side, you get, I'm just rearranging, it all equals f1 prime of x times g1 of y plus f2, g2, all the way to fn prime of x, gn of y. That's just all of these added up. Plus all of these added up. All of the terms that have the dy dx in them. So those are, and I'll do them in a different color. So all of these terms, which are going to be the different color, I'll do it in different parentheses, plus f1 of x g1 prime of y, and I'll do the dy dx later. I'll distribute it out."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "All of the terms that have the dy dx in them. So those are, and I'll do them in a different color. So all of these terms, which are going to be the different color, I'll do it in different parentheses, plus f1 of x g1 prime of y, and I'll do the dy dx later. I'll distribute it out. Plus, and we have n terms, plus fn of x gn prime of y. And then all of these terms are multiplied by dy dx. Now something looks interesting here."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "I'll distribute it out. Plus, and we have n terms, plus fn of x gn prime of y. And then all of these terms are multiplied by dy dx. Now something looks interesting here. We originally defined our xi up here as this right here. But what is this green term? Well, what we did is we took all of these individual terms and these green terms here are just taking the derivative with respect to just x on each of these terms."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Now something looks interesting here. We originally defined our xi up here as this right here. But what is this green term? Well, what we did is we took all of these individual terms and these green terms here are just taking the derivative with respect to just x on each of these terms. Because if you take the derivative just with respect to x of this, then the function of y is just a constant, right? If you were to take just the partial derivative with respect to x. So if you take the partial derivative with respect to x of this term, you treat a function of y as a constant."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "Well, what we did is we took all of these individual terms and these green terms here are just taking the derivative with respect to just x on each of these terms. Because if you take the derivative just with respect to x of this, then the function of y is just a constant, right? If you were to take just the partial derivative with respect to x. So if you take the partial derivative with respect to x of this term, you treat a function of y as a constant. So the derivative of this would just be f prime of x g1 of y, right? Because g1 of y is just a constant. And so forth and so on."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So if you take the partial derivative with respect to x of this term, you treat a function of y as a constant. So the derivative of this would just be f prime of x g1 of y, right? Because g1 of y is just a constant. And so forth and so on. All of these green terms you can view as the partial derivative of xi with respect to x. We just pretended like y is a constant. And that same logic, if you just look at this part right here, what is this?"}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And so forth and so on. All of these green terms you can view as the partial derivative of xi with respect to x. We just pretended like y is a constant. And that same logic, if you just look at this part right here, what is this? We took xi up here. We treated the functions of x as a constant. And we just took the partial derivative with respect to y."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And that same logic, if you just look at this part right here, what is this? We took xi up here. We treated the functions of x as a constant. And we just took the partial derivative with respect to y. And that's why the primes are on all the g's. And then we multiply that times dy dx. So you could write this."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And we just took the partial derivative with respect to y. And that's why the primes are on all the g's. And then we multiply that times dy dx. So you could write this. This is equal to, I'll do this green. This green is the same thing as the partial of xi with respect to x plus, what's this purple? This part of the purple."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "So you could write this. This is equal to, I'll do this green. This green is the same thing as the partial of xi with respect to x plus, what's this purple? This part of the purple. Let me do it in a different color, in a magenta. This right here is the partial of xi with respect to y. And then times dy dx."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "This part of the purple. Let me do it in a different color, in a magenta. This right here is the partial of xi with respect to y. And then times dy dx. So that's essentially all I wanted to show you right now in this video, because I realize I'm almost running out of time, that the chain rule, when you're taking, with respect to one of the variables, but the second variable of the function is also a function of x, the chain rule is this. If xi is a function of x and y, and I take not a partial derivative, I take the full derivative of xi with respect to x is equal to the partial of xi with respect to x plus the partial of xi with respect to y times dy dx. If y wasn't a function of x, or if y in no way was independent of x, then dy dx would be 0."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "And then times dy dx. So that's essentially all I wanted to show you right now in this video, because I realize I'm almost running out of time, that the chain rule, when you're taking, with respect to one of the variables, but the second variable of the function is also a function of x, the chain rule is this. If xi is a function of x and y, and I take not a partial derivative, I take the full derivative of xi with respect to x is equal to the partial of xi with respect to x plus the partial of xi with respect to y times dy dx. If y wasn't a function of x, or if y in no way was independent of x, then dy dx would be 0. And this term would be 0. And then the derivative of xi with respect to x would be just the partial of xi with respect to x. But anyway, I want you to just keep this in mind."}, {"video_title": "Exact equations intuition 1 (proofy) First order differential equations Khan Academy.mp3", "Sentence": "If y wasn't a function of x, or if y in no way was independent of x, then dy dx would be 0. And this term would be 0. And then the derivative of xi with respect to x would be just the partial of xi with respect to x. But anyway, I want you to just keep this in mind. And in this video, I didn't prove it, but I hopefully gave you a little intuition if I didn't confuse you. And we're going to use this property in the next series of videos to understand exact equations a little bit more. I realize in this video I just got as far as kind of giving you an intuition here."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "We're now ready to solve non-homogenous, second-order, linear differential equations with constant coefficients. So what does all that mean? Well, it means an equation that looks like this. a times the second derivative plus b times the first derivative plus c times the function is equal to g of x. Before I show you an actual example, I want to show you something interesting. That the general solution of this non-homogenous equation is actually the general solution of the homogenous equation plus a particular solution. I'll explain what that means in a second."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "a times the second derivative plus b times the first derivative plus c times the function is equal to g of x. Before I show you an actual example, I want to show you something interesting. That the general solution of this non-homogenous equation is actually the general solution of the homogenous equation plus a particular solution. I'll explain what that means in a second. So let's say that h is a solution of the homogenous equation. h is homogenous. And that worked out well, because h for homogenous."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "I'll explain what that means in a second. So let's say that h is a solution of the homogenous equation. h is homogenous. And that worked out well, because h for homogenous. h is solution for homogenous. There should be some shorthand notation for homogenous. So what does that mean?"}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "And that worked out well, because h for homogenous. h is solution for homogenous. There should be some shorthand notation for homogenous. So what does that mean? That means that a times the second derivative of h plus b times h prime plus c times h is equal to 0. That's what I mean when I say that h is a solution. And actually, let's just say that h is the general solution for this homogenous equation."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So what does that mean? That means that a times the second derivative of h plus b times h prime plus c times h is equal to 0. That's what I mean when I say that h is a solution. And actually, let's just say that h is the general solution for this homogenous equation. And we know how to solve that. Take the characteristic equation, depending on how many roots it has and whether they're real or complex. You can figure out a general equation, a general solution."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "And actually, let's just say that h is the general solution for this homogenous equation. And we know how to solve that. Take the characteristic equation, depending on how many roots it has and whether they're real or complex. You can figure out a general equation, a general solution. And then if you have initial conditions, you can substitute them and get the values of the constants. Fair enough. Now let's say that I were to say that g is a solution."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "You can figure out a general equation, a general solution. And then if you have initial conditions, you can substitute them and get the values of the constants. Fair enough. Now let's say that I were to say that g is a solution. Let's say it's a. Well, I already used g up here. Let's say, what am I?"}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Now let's say that I were to say that g is a solution. Let's say it's a. Well, I already used g up here. Let's say, what am I? Well, I don't like using vowels. Let's say j. Let's say j is a particular solution to this differential equation."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Let's say, what am I? Well, I don't like using vowels. Let's say j. Let's say j is a particular solution to this differential equation. So what does that mean? That means that a times j prime prime plus b times j prime plus c times j is equal to g of x. So that means we're just defining j of x to be a particular solution."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Let's say j is a particular solution to this differential equation. So what does that mean? That means that a times j prime prime plus b times j prime plus c times j is equal to g of x. So that means we're just defining j of x to be a particular solution. Now what I want to show you is that j of x plus h of x is also going to be a solution to this original equation. And that it's the general solution for this non- homogenous equation. And before I just do it mathematically, what's the intuition?"}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So that means we're just defining j of x to be a particular solution. Now what I want to show you is that j of x plus h of x is also going to be a solution to this original equation. And that it's the general solution for this non- homogenous equation. And before I just do it mathematically, what's the intuition? Well, when you substitute h here, you get 0. When you substitute j here, you get g of x. So when you add them together, you're going to get 0 plus g of x here."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "And before I just do it mathematically, what's the intuition? Well, when you substitute h here, you get 0. When you substitute j here, you get g of x. So when you add them together, you're going to get 0 plus g of x here. So you're going to get g of x here. And I'll show you that right now. So let's say I wanted to substitute h plus j here."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So when you add them together, you're going to get 0 plus g of x here. So you're going to get g of x here. And I'll show you that right now. So let's say I wanted to substitute h plus j here. So what is, and I'll do it in a different color, a. So the second derivative of the sum of those two functions is going to be the second derivative of both of them summed up. Plus b times the first derivative of the sum."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So let's say I wanted to substitute h plus j here. So what is, and I'll do it in a different color, a. So the second derivative of the sum of those two functions is going to be the second derivative of both of them summed up. Plus b times the first derivative of the sum. Plus c times the sum of the functions. And my goal is to show that this is equal to g of x. So what does this simplify to?"}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Plus b times the first derivative of the sum. Plus c times the sum of the functions. And my goal is to show that this is equal to g of x. So what does this simplify to? Well, if we take all the h terms, we get a h prime prime plus b h prime plus c h plus, let's do all the j terms, a j prime prime plus b j prime plus c j. Well, by definition of how we defined h and j, what is this equal to? Well, we said that h is a solution for the homogeneous equation or that this expression is equal to 0."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So what does this simplify to? Well, if we take all the h terms, we get a h prime prime plus b h prime plus c h plus, let's do all the j terms, a j prime prime plus b j prime plus c j. Well, by definition of how we defined h and j, what is this equal to? Well, we said that h is a solution for the homogeneous equation or that this expression is equal to 0. So that equals 0. And by our definition for j, what does this equal? Well, we said j is a particular solution for the non-homogeneous equation or that this expression is equal to g of x."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Well, we said that h is a solution for the homogeneous equation or that this expression is equal to 0. So that equals 0. And by our definition for j, what does this equal? Well, we said j is a particular solution for the non-homogeneous equation or that this expression is equal to g of x. So this is equal to g of x. So when you substitute h plus j into this differential equation, on the left-hand side, on the right-hand side, true enough, you get g of x. So we've just shown that if you define h and j this way, that the function, I don't know, we'll call it k of x is equal to h of x plus j of x. I'm running out of space."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Well, we said j is a particular solution for the non-homogeneous equation or that this expression is equal to g of x. So this is equal to g of x. So when you substitute h plus j into this differential equation, on the left-hand side, on the right-hand side, true enough, you get g of x. So we've just shown that if you define h and j this way, that the function, I don't know, we'll call it k of x is equal to h of x plus j of x. I'm running out of space. That is the general solution. I haven't proven that it is the most general solution, but I think you have the intuition, right? Because the general solution on the homogeneous one, that was the general solution, the most general solution."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So we've just shown that if you define h and j this way, that the function, I don't know, we'll call it k of x is equal to h of x plus j of x. I'm running out of space. That is the general solution. I haven't proven that it is the most general solution, but I think you have the intuition, right? Because the general solution on the homogeneous one, that was the general solution, the most general solution. And now we're adding a particular solution that gets you the g of x on the right-hand side. That might be very confusing to you, so let's actually try to do it with some real numbers. I think it'll make a lot more sense."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Because the general solution on the homogeneous one, that was the general solution, the most general solution. And now we're adding a particular solution that gets you the g of x on the right-hand side. That might be very confusing to you, so let's actually try to do it with some real numbers. I think it'll make a lot more sense. So let's say we have the differential equation, and I'm going to teach you a technique now for figuring out that g or that j in that last example. So how do you figure out that particular solution? So let's say I have the differential equation, the second derivative of y minus 3 times the first derivative minus 4 times y is equal to 3e to the 2x."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "I think it'll make a lot more sense. So let's say we have the differential equation, and I'm going to teach you a technique now for figuring out that g or that j in that last example. So how do you figure out that particular solution? So let's say I have the differential equation, the second derivative of y minus 3 times the first derivative minus 4 times y is equal to 3e to the 2x. So the first step is we want the general solution of the homogeneous equation. So first we can get, and in that example I just did, that would have been our h of x. So we want the solution of y prime prime minus 3, y prime minus 4y is equal to 0."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So let's say I have the differential equation, the second derivative of y minus 3 times the first derivative minus 4 times y is equal to 3e to the 2x. So the first step is we want the general solution of the homogeneous equation. So first we can get, and in that example I just did, that would have been our h of x. So we want the solution of y prime prime minus 3, y prime minus 4y is equal to 0. Take the characteristic equation. 4 is equal to 0. r, what is this? r minus 4 times r plus 1 is equal to 0."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So we want the solution of y prime prime minus 3, y prime minus 4y is equal to 0. Take the characteristic equation. 4 is equal to 0. r, what is this? r minus 4 times r plus 1 is equal to 0. 2 roots, r could be 4 or negative 1. And so our general solution, I'll call that h, or let's call that y general, y sub g. So our general solution is equal to, and we've done this many times, c1 e to the 4x plus c2 e to the minus 1x, or minus x. Fair enough."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "r minus 4 times r plus 1 is equal to 0. 2 roots, r could be 4 or negative 1. And so our general solution, I'll call that h, or let's call that y general, y sub g. So our general solution is equal to, and we've done this many times, c1 e to the 4x plus c2 e to the minus 1x, or minus x. Fair enough. So we solved the homogeneous equation. So how do we get, in that last example, a j of x that'll give us a particular solution so on the right-hand side we get this? Well here we just have to think a little bit."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Fair enough. So we solved the homogeneous equation. So how do we get, in that last example, a j of x that'll give us a particular solution so on the right-hand side we get this? Well here we just have to think a little bit. And this method is called the method of undetermined coefficients. And you have to say, well, if I want some function where I take its second derivative and add that or subtract it, some multiple of its first derivative minus some multiple of the function, I get e to the 2x. That function and its derivative and its second derivatives must be something of the form something times e to the 2x."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Well here we just have to think a little bit. And this method is called the method of undetermined coefficients. And you have to say, well, if I want some function where I take its second derivative and add that or subtract it, some multiple of its first derivative minus some multiple of the function, I get e to the 2x. That function and its derivative and its second derivatives must be something of the form something times e to the 2x. So essentially we take a guess. We say, well, what does it look like when we take the derivatives and the various derivatives and the functions and we multiply multiples of it plus each other and all of that, that we get e to the 2x, or some multiple of e to the 2x. Well, a good guess could just be that j, which is our, well, I'll call it y particular."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "That function and its derivative and its second derivatives must be something of the form something times e to the 2x. So essentially we take a guess. We say, well, what does it look like when we take the derivatives and the various derivatives and the functions and we multiply multiples of it plus each other and all of that, that we get e to the 2x, or some multiple of e to the 2x. Well, a good guess could just be that j, which is our, well, I'll call it y particular. Our particular solution here could be that, and particular solution I'm using a little different than the particular solution when we had initial conditions. Here we can view this as a particular solution, a solution that gives us this on the right-hand side. So let's say that the one I pick is some constant a times e to the 2x."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Well, a good guess could just be that j, which is our, well, I'll call it y particular. Our particular solution here could be that, and particular solution I'm using a little different than the particular solution when we had initial conditions. Here we can view this as a particular solution, a solution that gives us this on the right-hand side. So let's say that the one I pick is some constant a times e to the 2x. If that's my guess, then the derivative of that is equal to 2ae to the 2x. And the second derivative of that, of my particular solution, is equal to 4ae to the 2x. And now I can substitute in here, and let's see if I can solve for a, and then I'll have my particular solution."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So let's say that the one I pick is some constant a times e to the 2x. If that's my guess, then the derivative of that is equal to 2ae to the 2x. And the second derivative of that, of my particular solution, is equal to 4ae to the 2x. And now I can substitute in here, and let's see if I can solve for a, and then I'll have my particular solution. So the second derivative, that's this. So I get 4ae to the 2x minus 3 times the first derivative. So minus 3 times this."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "And now I can substitute in here, and let's see if I can solve for a, and then I'll have my particular solution. So the second derivative, that's this. So I get 4ae to the 2x minus 3 times the first derivative. So minus 3 times this. So that's minus 6ae to the 2x minus 4 times the function. So minus 4ae to the 2x. And all of that is going to be equal to 3e to the 2x."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "So minus 3 times this. So that's minus 6ae to the 2x minus 4 times the function. So minus 4ae to the 2x. And all of that is going to be equal to 3e to the 2x. Well, we know e to the 2x doesn't equal 0, so we can divide both sides by that, just factor it out, really. Get rid of all the e's of the 2x. And on the left-hand side, we have a 4a and a minus 4a."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "And all of that is going to be equal to 3e to the 2x. Well, we know e to the 2x doesn't equal 0, so we can divide both sides by that, just factor it out, really. Get rid of all the e's of the 2x. And on the left-hand side, we have a 4a and a minus 4a. Well, those cancel out. And then, lo and behold, we have minus 6a is equal to 3. Divide both sides by 6, you get a is equal to minus 1.5."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "And on the left-hand side, we have a 4a and a minus 4a. Well, those cancel out. And then, lo and behold, we have minus 6a is equal to 3. Divide both sides by 6, you get a is equal to minus 1.5. So there, we have our particular solution. It is equal to minus 1.5e to the 2x. And now, like I just showed you before I cleared the screen, our general solution of this non-homogeneous equation is going to be our particular solution plus the general solution to the homogenous equation."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "Divide both sides by 6, you get a is equal to minus 1.5. So there, we have our particular solution. It is equal to minus 1.5e to the 2x. And now, like I just showed you before I cleared the screen, our general solution of this non-homogeneous equation is going to be our particular solution plus the general solution to the homogenous equation. So we could call this the most general solution. Or I don't know, I'll just call it y. It is our general solution, c1e to the 4x plus c2e to the minus x plus our particular solution we found."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "And now, like I just showed you before I cleared the screen, our general solution of this non-homogeneous equation is going to be our particular solution plus the general solution to the homogenous equation. So we could call this the most general solution. Or I don't know, I'll just call it y. It is our general solution, c1e to the 4x plus c2e to the minus x plus our particular solution we found. So that's minus 1.5e to the 2x. Pretty neat. Anyway, well, I'll do a couple more examples of this."}, {"video_title": "Undetermined coefficients 1 Second order differential equations Khan Academy.mp3", "Sentence": "It is our general solution, c1e to the 4x plus c2e to the minus x plus our particular solution we found. So that's minus 1.5e to the 2x. Pretty neat. Anyway, well, I'll do a couple more examples of this. I think you'll get the hang of it. In the next example, we'll do something other than an e to the 2x or an e function here. We'll try to do stuff with polynomials and trig functions as well."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Instead of just taking Laplace transforms and taking their inverse, let's actually solve a problem. So let's say that I had the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function, 0 up until 2 pi of t, times sine of t minus 2 pi. Let's solve this differential equation, an interpretation of it. I actually do a whole playlist on interpretations of differential equations and how do you model it. But you can kind of view this as a forcing function. It's a weird forcing function of this being applied to some weight with, this is the acceleration term, right? The second derivative with respect to time is the acceleration."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "I actually do a whole playlist on interpretations of differential equations and how do you model it. But you can kind of view this as a forcing function. It's a weird forcing function of this being applied to some weight with, this is the acceleration term, right? The second derivative with respect to time is the acceleration. So the mass would be 1, whatever units. And then just the function of its position, this is probably some type of spring constant. Anyway, I won't go there."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "The second derivative with respect to time is the acceleration. So the mass would be 1, whatever units. And then just the function of its position, this is probably some type of spring constant. Anyway, I won't go there. I don't want to waste your time with the interpretation of it. But let's solve it. We can do more about interpretations later."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Anyway, I won't go there. I don't want to waste your time with the interpretation of it. But let's solve it. We can do more about interpretations later. So we're going to take the Laplace transform of both sides of this equation. So what's the Laplace transform of the left-hand side? So the Laplace transform of the second derivative of y is just s squared."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "We can do more about interpretations later. So we're going to take the Laplace transform of both sides of this equation. So what's the Laplace transform of the left-hand side? So the Laplace transform of the second derivative of y is just s squared. So now I'm taking the Laplace transform of just that. The Laplace transform of s squared times the Laplace transform of y minus, lower the degree there once, minus s times y of 0 minus y prime of 0. So clearly I must have to give you some initial conditions in order to do this properly."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So the Laplace transform of the second derivative of y is just s squared. So now I'm taking the Laplace transform of just that. The Laplace transform of s squared times the Laplace transform of y minus, lower the degree there once, minus s times y of 0 minus y prime of 0. So clearly I must have to give you some initial conditions in order to do this properly. And then plus 4 times the Laplace transform of y is equal to, what's the Laplace transform of sine of t? That should be second nature by now. It's just 1 over s squared plus 1."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So clearly I must have to give you some initial conditions in order to do this properly. And then plus 4 times the Laplace transform of y is equal to, what's the Laplace transform of sine of t? That should be second nature by now. It's just 1 over s squared plus 1. And then we have minus the Laplace transform of this thing, and I'll do a little side note here to figure out the Laplace transform of this thing right here. And we know it. I mean, I showed it to you a couple of videos ago."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "It's just 1 over s squared plus 1. And then we have minus the Laplace transform of this thing, and I'll do a little side note here to figure out the Laplace transform of this thing right here. And we know it. I mean, I showed it to you a couple of videos ago. We showed that the Laplace transform, actually I could just write it out here. This is going to be the same thing as the Laplace transform of sine of t, but we're going to have to multiply it by e to the minus, if you remember that last formula, e to the minus cs, where c is 2 pi. Let me actually write that down."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "I mean, I showed it to you a couple of videos ago. We showed that the Laplace transform, actually I could just write it out here. This is going to be the same thing as the Laplace transform of sine of t, but we're going to have to multiply it by e to the minus, if you remember that last formula, e to the minus cs, where c is 2 pi. Let me actually write that down. Let me, I decided to write it down, and then I decided, oh no, I don't want to do this. But let me write the Laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the Laplace transform of just the original function, times the Laplace transform of f of t. So if we're taking the Laplace transform of this thing, our c is 2 pi, our f of t is just sine of t. So then this is just going to be equal to, if we just do this piece right here, it's going to be equal to e to the minus cs, our c is 2 pi, e to the minus 2 pi s times the Laplace transform of f of t. f of t is just sine of t, right, before we shift it. This is f of t minus 2 pi."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Let me actually write that down. Let me, I decided to write it down, and then I decided, oh no, I don't want to do this. But let me write the Laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the Laplace transform of just the original function, times the Laplace transform of f of t. So if we're taking the Laplace transform of this thing, our c is 2 pi, our f of t is just sine of t. So then this is just going to be equal to, if we just do this piece right here, it's going to be equal to e to the minus cs, our c is 2 pi, e to the minus 2 pi s times the Laplace transform of f of t. f of t is just sine of t, right, before we shift it. This is f of t minus 2 pi. So f of t is just going to be sine of t. So it's going to be times 1 over s squared plus 1. This is a Laplace transform of sine of t. So let's go back to where we had left off. So we've taken the Laplace transform of both sides of this equation."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "This is f of t minus 2 pi. So f of t is just going to be sine of t. So it's going to be times 1 over s squared plus 1. This is a Laplace transform of sine of t. So let's go back to where we had left off. So we've taken the Laplace transform of both sides of this equation. And clearly I have some initial conditions here. So the problem must have given me some, and I just forgot to write them down. So let's see, the initial conditions I'm given, they are written kind of in the margin here."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So we've taken the Laplace transform of both sides of this equation. And clearly I have some initial conditions here. So the problem must have given me some, and I just forgot to write them down. So let's see, the initial conditions I'm given, they are written kind of in the margin here. They tell us, I'll do it in orange, they tell us that y of 0 is equal to 0, and y prime of 0 is equal to 0. That makes the math easy. That's 0, and that's 0."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So let's see, the initial conditions I'm given, they are written kind of in the margin here. They tell us, I'll do it in orange, they tell us that y of 0 is equal to 0, and y prime of 0 is equal to 0. That makes the math easy. That's 0, and that's 0. So let's see if I can simplify my equation. So the left-hand side, let's factor out the Laplace transform. So let's factor out this term and that term."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "That's 0, and that's 0. So let's see if I can simplify my equation. So the left-hand side, let's factor out the Laplace transform. So let's factor out this term and that term. So we get the Laplace transform of y times this plus this, times s squared plus 4, is equal to the right-hand side, and what's the right-hand side? It is, we get, we could simplify this. Well, I'll just write it out."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So let's factor out this term and that term. So we get the Laplace transform of y times this plus this, times s squared plus 4, is equal to the right-hand side, and what's the right-hand side? It is, we get, we could simplify this. Well, I'll just write it out. I don't want to do too many steps at once. It's 1 over s squared plus 1, and then plus, or minus actually, this is a minus, minus Laplace transform of this thing, which was e to the minus 2 pi s over s squared plus 1. And so we can write this whole thing as, so if we divide both sides of this equation by the s squared plus 4, then we get the Laplace transform of y is equal to, and actually I can just merge these two."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Well, I'll just write it out. I don't want to do too many steps at once. It's 1 over s squared plus 1, and then plus, or minus actually, this is a minus, minus Laplace transform of this thing, which was e to the minus 2 pi s over s squared plus 1. And so we can write this whole thing as, so if we divide both sides of this equation by the s squared plus 4, then we get the Laplace transform of y is equal to, and actually I can just merge these two. They have the same denominator. So before I even divide by s squared plus 4, that right-hand side will look like this. It will look like, with a denominator of s squared plus 1, and you have a numerator of 1 minus e to the minus 2 pi s. And of course, we're dividing both sides of this equation by s squared plus 4, so we're going to have to stick that s squared plus 4 over here."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "And so we can write this whole thing as, so if we divide both sides of this equation by the s squared plus 4, then we get the Laplace transform of y is equal to, and actually I can just merge these two. They have the same denominator. So before I even divide by s squared plus 4, that right-hand side will look like this. It will look like, with a denominator of s squared plus 1, and you have a numerator of 1 minus e to the minus 2 pi s. And of course, we're dividing both sides of this equation by s squared plus 4, so we're going to have to stick that s squared plus 4 over here. Now we're at the hard part. In order to figure out y, we have to take the inverse Laplace transform of this thing. So how do we take the inverse Laplace transform of this thing, that's where the hard part is always, it's kind of solving the differential equation is easy if you know the Laplace transforms."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "It will look like, with a denominator of s squared plus 1, and you have a numerator of 1 minus e to the minus 2 pi s. And of course, we're dividing both sides of this equation by s squared plus 4, so we're going to have to stick that s squared plus 4 over here. Now we're at the hard part. In order to figure out y, we have to take the inverse Laplace transform of this thing. So how do we take the inverse Laplace transform of this thing, that's where the hard part is always, it's kind of solving the differential equation is easy if you know the Laplace transforms. So it looks like we're going to have to do some partial fraction expansion. So let's see if we can do that. So we can rewrite this equation right here."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So how do we take the inverse Laplace transform of this thing, that's where the hard part is always, it's kind of solving the differential equation is easy if you know the Laplace transforms. So it looks like we're going to have to do some partial fraction expansion. So let's see if we can do that. So we can rewrite this equation right here. We can rewrite this equation. Let's write it as this, because this will kind of simplify our work. Let's factor this whole thing out."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So we can rewrite this equation right here. We can rewrite this equation. Let's write it as this, because this will kind of simplify our work. Let's factor this whole thing out. So we're going to write it as 1 minus e to the minus 2 pi s. All of that times 1 over s squared plus 1 times s squared plus 4. Now we need to do some partial fraction expansion to simplify this thing right here. So we're going to say, we're going to do this on the side."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Let's factor this whole thing out. So we're going to write it as 1 minus e to the minus 2 pi s. All of that times 1 over s squared plus 1 times s squared plus 4. Now we need to do some partial fraction expansion to simplify this thing right here. So we're going to say, we're going to do this on the side. Maybe I should do this over on the right right here. This thing, let me rewrite it, 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions, s squared plus 1 and s squared plus 4, with the numerators. This one would be a s plus b."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So we're going to say, we're going to do this on the side. Maybe I should do this over on the right right here. This thing, let me rewrite it, 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions, s squared plus 1 and s squared plus 4, with the numerators. This one would be a s plus b. It's going to have to have degree 1, because this is degree 2 here. And then we'd have c s plus d. And so when you add these two things up, you get a s plus b times s squared plus 4 plus c s plus d times s squared plus 1. All of that over the common denominator."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "This one would be a s plus b. It's going to have to have degree 1, because this is degree 2 here. And then we'd have c s plus d. And so when you add these two things up, you get a s plus b times s squared plus 4 plus c s plus d times s squared plus 1. All of that over the common denominator. And we've seen this story before. We just have to do some algebra here. As you can tell, these differential equations problems, they require a lot of stamina."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "All of that over the common denominator. And we've seen this story before. We just have to do some algebra here. As you can tell, these differential equations problems, they require a lot of stamina. You kind of just have to say, I will keep moving forward and do the algebra that I need to do in order to get the answer. And you kind of have to get excited about that notion, that you have all of this algebra to do. So let's figure it out."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "As you can tell, these differential equations problems, they require a lot of stamina. You kind of just have to say, I will keep moving forward and do the algebra that I need to do in order to get the answer. And you kind of have to get excited about that notion, that you have all of this algebra to do. So let's figure it out. So this top can be simplified to a s to the third plus b s squared plus 4 a s plus 4 b. And then this one is, you end up with c s to the third plus d s squared plus c s plus d. So when you add all of these up together, you get, and this is all the algebra that we have to do for better or for worse, a plus c over s to the third plus b plus d times s squared plus 4 a plus c times s. Scroll over a little bit. Plus 4 b plus d. And now we just have to say, OK, so do we see any, all of this is equal to this thing up here."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So let's figure it out. So this top can be simplified to a s to the third plus b s squared plus 4 a s plus 4 b. And then this one is, you end up with c s to the third plus d s squared plus c s plus d. So when you add all of these up together, you get, and this is all the algebra that we have to do for better or for worse, a plus c over s to the third plus b plus d times s squared plus 4 a plus c times s. Scroll over a little bit. Plus 4 b plus d. And now we just have to say, OK, so do we see any, all of this is equal to this thing up here. This is the numerator. We just simplified the numerator. This is the numerator."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Plus 4 b plus d. And now we just have to say, OK, so do we see any, all of this is equal to this thing up here. This is the numerator. We just simplified the numerator. This is the numerator. That's the numerator right there. And all of this is going to be over your original s squared plus 1 times your s squared plus 4. And we established that this thing should be, let me just write this, that 1 over s squared plus 1 times s squared plus 4 should equal this thing."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "This is the numerator. That's the numerator right there. And all of this is going to be over your original s squared plus 1 times your s squared plus 4. And we established that this thing should be, let me just write this, that 1 over s squared plus 1 times s squared plus 4 should equal this thing. And then you just pattern match on the coefficients. This is all just intense partial fraction expansion. And you say, look, a plus c is the coefficient of the s cubed terms."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "And we established that this thing should be, let me just write this, that 1 over s squared plus 1 times s squared plus 4 should equal this thing. And then you just pattern match on the coefficients. This is all just intense partial fraction expansion. And you say, look, a plus c is the coefficient of the s cubed terms. I don't see any s cubed terms here. So a plus c must be equal to 0. And then you see, OK, b plus d is the coefficient on the s squared terms."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "And you say, look, a plus c is the coefficient of the s cubed terms. I don't see any s cubed terms here. So a plus c must be equal to 0. And then you see, OK, b plus d is the coefficient on the s squared terms. Don't see any s squared terms there. So b plus d must be equal to 0. 4 a plus c, coefficient on the s terms."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "And then you see, OK, b plus d is the coefficient on the s squared terms. Don't see any s squared terms there. So b plus d must be equal to 0. 4 a plus c, coefficient on the s terms. Don't see any s terms over here. So 4 a plus c must be equal to 0. And then we're almost done."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "4 a plus c, coefficient on the s terms. Don't see any s terms over here. So 4 a plus c must be equal to 0. And then we're almost done. 4 b plus d must be the constant terms. There is a constant term there. So 4 b plus d is equal to 1."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "And then we're almost done. 4 b plus d must be the constant terms. There is a constant term there. So 4 b plus d is equal to 1. So let's see if we can do anything here. If we subtract this from that, we get minus 3 a is equal to 0, or a is equal to 0. If a is equal to 0, then c is equal to 0."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So 4 b plus d is equal to 1. So let's see if we can do anything here. If we subtract this from that, we get minus 3 a is equal to 0, or a is equal to 0. If a is equal to 0, then c is equal to 0. And let's see what we can get here. If we subtract this from that, we get minus 3 b, the d's cancel out, is equal to minus 1. Or b is equal to 1 third."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "If a is equal to 0, then c is equal to 0. And let's see what we can get here. If we subtract this from that, we get minus 3 b, the d's cancel out, is equal to minus 1. Or b is equal to 1 third. And then of course we have d is equal to minus b. If we subtract b from both sides, so d is equal to 1 third. So all of that work, and we actually have a pretty simple result."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Or b is equal to 1 third. And then of course we have d is equal to minus b. If we subtract b from both sides, so d is equal to 1 third. So all of that work, and we actually have a pretty simple result. Our equation, this thing here, can be rewritten as the a disappeared, it's 1 third over s squared plus 1. b was the coefficient on the, let me make it very clear, b was a coefficient on the, or it was a term on top of the s squared plus 1. So that's why I'm using b there. And then d is minus b."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So all of that work, and we actually have a pretty simple result. Our equation, this thing here, can be rewritten as the a disappeared, it's 1 third over s squared plus 1. b was the coefficient on the, let me make it very clear, b was a coefficient on the, or it was a term on top of the s squared plus 1. So that's why I'm using b there. And then d is minus b. So d is minus 1 third. Let me make sure I have that. b is 1 third."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "And then d is minus b. So d is minus 1 third. Let me make sure I have that. b is 1 third. Let me make sure I get that right. d is 1 third, so, sorry, b as in boy is 1 third. So d is minus 1 third."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "b is 1 third. Let me make sure I get that right. d is 1 third, so, sorry, b as in boy is 1 third. So d is minus 1 third. So b is a term on top of the s squared plus 1. And then you have minus d over the minus 1 third over s squared plus 4. This takes a lot of stamina to record this video."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So d is minus 1 third. So b is a term on top of the s squared plus 1. And then you have minus d over the minus 1 third over s squared plus 4. This takes a lot of stamina to record this video. I hope you appreciate it. OK, so let me rewrite everything, just so we can get back to the problem. Because when you take that partial fraction detour, it kind of, you forget, I mean, not even to speak of the problem, you forget what day it is."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "This takes a lot of stamina to record this video. I hope you appreciate it. OK, so let me rewrite everything, just so we can get back to the problem. Because when you take that partial fraction detour, it kind of, you forget, I mean, not even to speak of the problem, you forget what day it is. Let's see, so you get the Laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for. Times, and I'll write it like this, 1 third times 1 over s squared plus 1 minus 1 third times, actually, let me write it this way, because I have this s squared plus 4. So I really want to have a 2 there, right?"}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Because when you take that partial fraction detour, it kind of, you forget, I mean, not even to speak of the problem, you forget what day it is. Let's see, so you get the Laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for. Times, and I'll write it like this, 1 third times 1 over s squared plus 1 minus 1 third times, actually, let me write it this way, because I have this s squared plus 4. So I really want to have a 2 there, right? So I want to have a 2 in the numerator. So you want to have a 2 over s squared plus 4. So if I put a 2 in the numerator, then I have to divide this by 2 as well."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So I really want to have a 2 there, right? So I want to have a 2 in the numerator. So you want to have a 2 over s squared plus 4. So if I put a 2 in the numerator, then I have to divide this by 2 as well. So times, so let me change this to a 6. Minus 1 sixth times 2 is minus 1 third. So I did that just so I get this in the form of the Laplace transform of sine of t. Now let's see if there's anything that I can do from here."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So if I put a 2 in the numerator, then I have to divide this by 2 as well. So times, so let me change this to a 6. Minus 1 sixth times 2 is minus 1 third. So I did that just so I get this in the form of the Laplace transform of sine of t. Now let's see if there's anything that I can do from here. This is an epic problem. I'll be amazed if I don't make a careless mistake while I do this. So we can rewrite everything."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So I did that just so I get this in the form of the Laplace transform of sine of t. Now let's see if there's anything that I can do from here. This is an epic problem. I'll be amazed if I don't make a careless mistake while I do this. So we can rewrite everything. Let's see if we can simplify this. And by simplifying it, I'm just going to make it longer. We can write the Laplace transform of y is equal to, I'm just going to multiply the 1 out, and then I'm going to multiply the e to the minus 2 pi s out."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So we can rewrite everything. Let's see if we can simplify this. And by simplifying it, I'm just going to make it longer. We can write the Laplace transform of y is equal to, I'm just going to multiply the 1 out, and then I'm going to multiply the e to the minus 2 pi s out. So if you multiply the 1 out, you get 1 third times 1 over s squared plus 1. I'm just multiplying the 1 out. Minus 1 sixth, these are all the 1's times the 1, times 2 over s squared plus 4."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "We can write the Laplace transform of y is equal to, I'm just going to multiply the 1 out, and then I'm going to multiply the e to the minus 2 pi s out. So if you multiply the 1 out, you get 1 third times 1 over s squared plus 1. I'm just multiplying the 1 out. Minus 1 sixth, these are all the 1's times the 1, times 2 over s squared plus 4. And I'm going to multiply the minus e. So this, and then you get, let me switch colors, do the minus e, so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1. And then the minus and the minus cancel out, so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4. Now, taking the inverse Laplace transform of these things are pretty straightforward."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Minus 1 sixth, these are all the 1's times the 1, times 2 over s squared plus 4. And I'm going to multiply the minus e. So this, and then you get, let me switch colors, do the minus e, so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1. And then the minus and the minus cancel out, so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4. Now, taking the inverse Laplace transform of these things are pretty straightforward. So let's do that. Let's take the inverse Laplace transform of the whole thing. And we get y is equal to, inverse Laplace transform of this guy right here is just 1 third sine of t. 1 third, I don't have to write a parentheses there."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Now, taking the inverse Laplace transform of these things are pretty straightforward. So let's do that. Let's take the inverse Laplace transform of the whole thing. And we get y is equal to, inverse Laplace transform of this guy right here is just 1 third sine of t. 1 third, I don't have to write a parentheses there. Sine of t, and then this is minus 1 sixth times, this is the Laplace transform of sine of 2t. That's that term right there. Now, these are almost the same, but we have this little pesky character over here."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "And we get y is equal to, inverse Laplace transform of this guy right here is just 1 third sine of t. 1 third, I don't have to write a parentheses there. Sine of t, and then this is minus 1 sixth times, this is the Laplace transform of sine of 2t. That's that term right there. Now, these are almost the same, but we have this little pesky character over here. We have this e to the minus 2 pi s. And there we just have to remind ourselves, I'll write it here in the bottom, we just have to remind ourselves that the Laplace transform of the unit step function, I'll put the pi there, just 2 pi times f of t minus 2 pi, I should put the step function of t, is equal to e to the minus 2 pi s times the Laplace transform of just, or let me just write it this way, times the Laplace transform of f of t. So if we view f of t as just sine of t or sine of 2t, then we can kind of backwards pattern match and we'll have to shift it and multiply it by the unit step function. So I want to make that clear, right? If you didn't have this guy here, the inverse Laplace transform of this guy would be the same thing as this guy."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Now, these are almost the same, but we have this little pesky character over here. We have this e to the minus 2 pi s. And there we just have to remind ourselves, I'll write it here in the bottom, we just have to remind ourselves that the Laplace transform of the unit step function, I'll put the pi there, just 2 pi times f of t minus 2 pi, I should put the step function of t, is equal to e to the minus 2 pi s times the Laplace transform of just, or let me just write it this way, times the Laplace transform of f of t. So if we view f of t as just sine of t or sine of 2t, then we can kind of backwards pattern match and we'll have to shift it and multiply it by the unit step function. So I want to make that clear, right? If you didn't have this guy here, the inverse Laplace transform of this guy would be the same thing as this guy. It would just be sine of t. Inverse Laplace transform of this guy would be sine of 2t. But we have this pesky character here, which essentially, instead of having the inverse Laplace transform just being our f of t, it's going to be our f of t shifted by 2 pi times the unit step function where it steps up at 2 pi. So this is going to be minus 1 third times the unit step function where c is 2 pi of t times, instead of sine of t, sine of t minus 2 pi."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "If you didn't have this guy here, the inverse Laplace transform of this guy would be the same thing as this guy. It would just be sine of t. Inverse Laplace transform of this guy would be sine of 2t. But we have this pesky character here, which essentially, instead of having the inverse Laplace transform just being our f of t, it's going to be our f of t shifted by 2 pi times the unit step function where it steps up at 2 pi. So this is going to be minus 1 third times the unit step function where c is 2 pi of t times, instead of sine of t, sine of t minus 2 pi. And then we're almost done. I'll do it in magenta to celebrate it. Plus this very last term, which is 1 sixth times the unit step function, 2 pi of t. The unit step function that steps up at 2 pi."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So this is going to be minus 1 third times the unit step function where c is 2 pi of t times, instead of sine of t, sine of t minus 2 pi. And then we're almost done. I'll do it in magenta to celebrate it. Plus this very last term, which is 1 sixth times the unit step function, 2 pi of t. The unit step function that steps up at 2 pi. Times sine of, and we have to be careful here. Instead of, wherever we had a t before, we're going to replace it with a t minus 2 pi. So sine of, instead of 2t, it's going to be 2 times t minus 2 pi."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "Plus this very last term, which is 1 sixth times the unit step function, 2 pi of t. The unit step function that steps up at 2 pi. Times sine of, and we have to be careful here. Instead of, wherever we had a t before, we're going to replace it with a t minus 2 pi. So sine of, instead of 2t, it's going to be 2 times t minus 2 pi. And there you have it. We finally have solved our very hairy problem. We could take some time, if we want, to simplify this a little bit."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "So sine of, instead of 2t, it's going to be 2 times t minus 2 pi. And there you have it. We finally have solved our very hairy problem. We could take some time, if we want, to simplify this a little bit. In fact, we might as well. At the risk of making a careless mistake at the last moment, let me see if I can make any simplifications here. Actually, it's not obvious that there's any, without kind of resorting to some type of, well, we could factor out this guy right here."}, {"video_title": "Laplace step function differential equation Laplace transform Khan Academy.mp3", "Sentence": "We could take some time, if we want, to simplify this a little bit. In fact, we might as well. At the risk of making a careless mistake at the last moment, let me see if I can make any simplifications here. Actually, it's not obvious that there's any, without kind of resorting to some type of, well, we could factor out this guy right here. But other than that, that seems about as simple as we can get. So this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here. This looked fairly straightforward, but we got this big mess to actually satisfy that equation, given those initial conditions that we had initially."}, {"video_title": "Worked example range of solution curve from slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So, let's see, zero comma six, so this is part of the solution, and we want to know the range of the solution curve. So the solution curve, you can eyeball a little bit by looking at the slope field. So as x, remember, x is gonna be greater than or equal to zero, so it's going to include this point right over here, and as x increases, you can tell from the slope, okay, y is gonna decrease, but it's gonna keep decreasing at a slower and slower rate, and it looks like it's asymptoting towards the line y is equal to four. So it's gonna get really, as x gets larger and larger and larger, it's gonna get infinitely close to y is equal to four, but it's not quite gonna get there. So the range, the y values that this is going to take on, y is going to be greater than four. It's not ever gonna be equal to four, so I'll do, it's going to be greater than four, that's gonna be the bottom end of my range, and at the top end of my range, I will be equal to six. Six is the largest value that I am going to take on."}, {"video_title": "Worked example range of solution curve from slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So it's gonna get really, as x gets larger and larger and larger, it's gonna get infinitely close to y is equal to four, but it's not quite gonna get there. So the range, the y values that this is going to take on, y is going to be greater than four. It's not ever gonna be equal to four, so I'll do, it's going to be greater than four, that's gonna be the bottom end of my range, and at the top end of my range, I will be equal to six. Six is the largest value that I am going to take on. Another way I could have written this is four is less than y, is less than or equal to six. Either way, this is a way of describing the range, the y values that the solution will take on for x being greater than or equal to zero. If they said for all x's, well, then you might have been able to go back this way and keep going, but they're saying the range of the solution curve for x is greater than or equal to zero so we won't consider those values of x less than zero."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Let's now actually apply Newton's law of cooling. So just to remind ourselves, if capital T is the temperature of something in Celsius degrees and lowercase t is time in minutes, we can say that the rate of change, the rate of change of our temperature with respect to time is going to be proportional, and I'll write a negative k right over here. If we assumed our constant k is positive, then a negative k is going to be proportional to the difference between the temperature of our thing and the ambient temperature in the room. And once again, why did I have a negative there? Well, because if the temperature of our thing is larger than the temperature of the room, we would expect that we would be decreasing in temperature, that we would have a negative rate of change. That temperature should be decreasing with time. If, on the other hand, our temperature is lower than the ambient temperature of the room, then this thing is going to be negative, but we would want a positive rate of change of temperature."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And once again, why did I have a negative there? Well, because if the temperature of our thing is larger than the temperature of the room, we would expect that we would be decreasing in temperature, that we would have a negative rate of change. That temperature should be decreasing with time. If, on the other hand, our temperature is lower than the ambient temperature of the room, then this thing is going to be negative, but we would want a positive rate of change of temperature. Things would be warming up, and so that's why a negative of a negative would give you the positive. So this right over here, this differential equation, we already saw it in the previous video on Newton's Law of Cooling, and we even saw a general solution to that. And the general solution that I care about, because we're now going to deal with the scenario where we're putting something warm, or we're gonna put a warm bowl of oatmeal in a room temperature room."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "If, on the other hand, our temperature is lower than the ambient temperature of the room, then this thing is going to be negative, but we would want a positive rate of change of temperature. Things would be warming up, and so that's why a negative of a negative would give you the positive. So this right over here, this differential equation, we already saw it in the previous video on Newton's Law of Cooling, and we even saw a general solution to that. And the general solution that I care about, because we're now going to deal with the scenario where we're putting something warm, or we're gonna put a warm bowl of oatmeal in a room temperature room. So given that, we're gonna assume the case that we saw in the last video, where our temperature is greater than or equal to the ambient temperature, and in that situation, our general solution boiled down to, it boiled down to temperature as a function of time is equal to some constant times e to the negative kt, e to the negative kt, negative kt, plus our ambient temperature, plus our ambient temperature. Once again, we figured this out in the last video. Now let's actually apply it."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And the general solution that I care about, because we're now going to deal with the scenario where we're putting something warm, or we're gonna put a warm bowl of oatmeal in a room temperature room. So given that, we're gonna assume the case that we saw in the last video, where our temperature is greater than or equal to the ambient temperature, and in that situation, our general solution boiled down to, it boiled down to temperature as a function of time is equal to some constant times e to the negative kt, e to the negative kt, negative kt, plus our ambient temperature, plus our ambient temperature. Once again, we figured this out in the last video. Now let's actually apply it. So I said we were dealing with the scenario where our temperature is greater than or equal to the ambient temperature, so let's assume we're in a scenario, let's assume a scenario where our ambient temperature is 20 degrees Celsius, and we assume that doesn't change, that the room is just large enough that even if something is warmer is put into it, that the ambient temperature does not change. And let's say that the thing that we put in it, our warm bowl of oatmeal, let's say it starts off, the moment we put it in the room at time equals zero is 80 degrees Celsius. And let's say we also know, just from previous tests, that after two minutes, after two minutes, it gets to 60 degrees Celsius."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Now let's actually apply it. So I said we were dealing with the scenario where our temperature is greater than or equal to the ambient temperature, so let's assume we're in a scenario, let's assume a scenario where our ambient temperature is 20 degrees Celsius, and we assume that doesn't change, that the room is just large enough that even if something is warmer is put into it, that the ambient temperature does not change. And let's say that the thing that we put in it, our warm bowl of oatmeal, let's say it starts off, the moment we put it in the room at time equals zero is 80 degrees Celsius. And let's say we also know, just from previous tests, that after two minutes, after two minutes, it gets to 60 degrees Celsius. So we also know that T of two is 60 degrees Celsius. And given all of this information right over here, using Newton's Law of Cooling and using all of this information we know about how bowls of oatmeal that started at this temperature have cooled in the past, we wanna know how long will it take, how many minutes will have to pass when you put an 80 degree bowl of oatmeal in the room, how many minutes have to pass in order for it to get to 40 degrees using this model. So let me write that down."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And let's say we also know, just from previous tests, that after two minutes, after two minutes, it gets to 60 degrees Celsius. So we also know that T of two is 60 degrees Celsius. And given all of this information right over here, using Newton's Law of Cooling and using all of this information we know about how bowls of oatmeal that started at this temperature have cooled in the past, we wanna know how long will it take, how many minutes will have to pass when you put an 80 degree bowl of oatmeal in the room, how many minutes have to pass in order for it to get to 40 degrees using this model. So let me write that down. So how long, how many minutes, minutes for, or let me say to cool, to cool, to cool to 40 degrees Celsius. And so I encourage you to pause the video now and try to figure it out. So I'm assuming you have paused the video and you have had your go at it."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So let me write that down. So how long, how many minutes, minutes for, or let me say to cool, to cool, to cool to 40 degrees Celsius. And so I encourage you to pause the video now and try to figure it out. So I'm assuming you have paused the video and you have had your go at it. And the key is to use all of this information right over here to solve for the constants C and K. And then once you know that, you essentially have described your model and then you can apply it to solve for the time that gets you to a temperature of 40 degrees Celsius. So let's do that. So the first thing we know is the ambient temperature is 20 degrees Celsius."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So I'm assuming you have paused the video and you have had your go at it. And the key is to use all of this information right over here to solve for the constants C and K. And then once you know that, you essentially have described your model and then you can apply it to solve for the time that gets you to a temperature of 40 degrees Celsius. So let's do that. So the first thing we know is the ambient temperature is 20 degrees Celsius. So this right over here, this right over here is 20 degrees. And so the most obvious thing to solve for or to apply is what happens with T of zero? And what's neat about T of zero, when T equals zero, this exponent is zero, E to the zero power is one."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So the first thing we know is the ambient temperature is 20 degrees Celsius. So this right over here, this right over here is 20 degrees. And so the most obvious thing to solve for or to apply is what happens with T of zero? And what's neat about T of zero, when T equals zero, this exponent is zero, E to the zero power is one. And so T of zero is essentially going to simplify to C plus 20 degrees. Let me actually write that down. So T of zero, which we already know is 80 degrees, we already know is 80 degrees Celsius."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And what's neat about T of zero, when T equals zero, this exponent is zero, E to the zero power is one. And so T of zero is essentially going to simplify to C plus 20 degrees. Let me actually write that down. So T of zero, which we already know is 80 degrees, we already know is 80 degrees Celsius. And I'm just gonna write 80. I'm just gonna write 80. We'll assume it's in degrees Celsius."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So T of zero, which we already know is 80 degrees, we already know is 80 degrees Celsius. And I'm just gonna write 80. I'm just gonna write 80. We'll assume it's in degrees Celsius. That is going to be equal to, that is going to be equal to, when T equals zero, this, the E to the negative, or E to the zero is just gonna be one. So it's going to be equal to C plus 20. And so if you wanna solve for C, you just subtract 20 from both sides of this equation."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "We'll assume it's in degrees Celsius. That is going to be equal to, that is going to be equal to, when T equals zero, this, the E to the negative, or E to the zero is just gonna be one. So it's going to be equal to C plus 20. And so if you wanna solve for C, you just subtract 20 from both sides of this equation. And we are left with, we are left with 80 minus 20 is 60, is equal to C. 60 is equal to C. So we were able to figure out C. So let's figure out, let's write what we know right now. So we know that T, let me do that in that magenta color. We know that T of T, that's confusing, uppercase T of lowercase T, temperature as a function of time, is going to be equal to, is going to be equal to, let me do that same color, 60 E, 60 E to the negative KT."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And so if you wanna solve for C, you just subtract 20 from both sides of this equation. And we are left with, we are left with 80 minus 20 is 60, is equal to C. 60 is equal to C. So we were able to figure out C. So let's figure out, let's write what we know right now. So we know that T, let me do that in that magenta color. We know that T of T, that's confusing, uppercase T of lowercase T, temperature as a function of time, is going to be equal to, is going to be equal to, let me do that same color, 60 E, 60 E to the negative KT. Negative KT plus 20, plus our ambient temperature. Plus our ambient temperature. Now we need to solve for K. And we could use this information right over here to solve for K. T of two is equal to 60 degrees."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "We know that T of T, that's confusing, uppercase T of lowercase T, temperature as a function of time, is going to be equal to, is going to be equal to, let me do that same color, 60 E, 60 E to the negative KT. Negative KT plus 20, plus our ambient temperature. Plus our ambient temperature. Now we need to solve for K. And we could use this information right over here to solve for K. T of two is equal to 60 degrees. So if we make T is equal to two, this thing is going to be 60 degrees. So let me write that down. So we could, let me write it over here so I have some space."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Now we need to solve for K. And we could use this information right over here to solve for K. T of two is equal to 60 degrees. So if we make T is equal to two, this thing is going to be 60 degrees. So let me write that down. So we could, let me write it over here so I have some space. So we have 60 is equal to, 60 is equal to 60, is equal to 60 E to the negative KT. All this color switching takes time. E to the negative KT plus, oh, let me be careful, that's at time two."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So we could, let me write it over here so I have some space. So we have 60 is equal to, 60 is equal to 60, is equal to 60 E to the negative KT. All this color switching takes time. E to the negative KT plus, oh, let me be careful, that's at time two. So that's E to the negative K times two. That's at time equals two. Or I could write that E to the negative two K. E to the negative two K. And then of course we have our plus 20."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "E to the negative KT plus, oh, let me be careful, that's at time two. So that's E to the negative K times two. That's at time equals two. Or I could write that E to the negative two K. E to the negative two K. And then of course we have our plus 20. And then we have our plus 20. And now we just have to solve for K. And once again, at any point, if you feel inspired to do so, I encourage you to try to solve it on your own. All right, so let's do this."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Or I could write that E to the negative two K. E to the negative two K. And then of course we have our plus 20. And then we have our plus 20. And now we just have to solve for K. And once again, at any point, if you feel inspired to do so, I encourage you to try to solve it on your own. All right, so let's do this. So if we subtract, if we subtract 20 from both sides, we get 40 is equal to, is equal to 60 E, E to the negative two K, negative two K. Now divide both sides by 60. You are left with, you are left with two thirds. 40 divided by 60 is two thirds, is equal to E to the negative two K. E to the negative two K. Negative two K. All this color changing takes work."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "All right, so let's do this. So if we subtract, if we subtract 20 from both sides, we get 40 is equal to, is equal to 60 E, E to the negative two K, negative two K. Now divide both sides by 60. You are left with, you are left with two thirds. 40 divided by 60 is two thirds, is equal to E to the negative two K. E to the negative two K. Negative two K. All this color changing takes work. Let me know if y'all want me to keep changing. I enjoy changing colors. It just keeps it interesting on the screen."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "40 divided by 60 is two thirds, is equal to E to the negative two K. E to the negative two K. Negative two K. All this color changing takes work. Let me know if y'all want me to keep changing. I enjoy changing colors. It just keeps it interesting on the screen. But anyway. So E to the negative two K. Actually, let me scroll down a little bit so I have some more real estate to work with. And now I could take, let's see, I could take the natural log of both sides."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "It just keeps it interesting on the screen. But anyway. So E to the negative two K. Actually, let me scroll down a little bit so I have some more real estate to work with. And now I could take, let's see, I could take the natural log of both sides. And so I'll have the natural log, natural log of two thirds, is equal to the natural log of E to the negative two K is just going to be negative two K. That's the whole reason why I took the natural log of both sides. And then to solve for K, I divide both sides by negative two. I get K is equal to negative one half, negative one half, natural log, natural log of two thirds."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And now I could take, let's see, I could take the natural log of both sides. And so I'll have the natural log, natural log of two thirds, is equal to the natural log of E to the negative two K is just going to be negative two K. That's the whole reason why I took the natural log of both sides. And then to solve for K, I divide both sides by negative two. I get K is equal to negative one half, negative one half, natural log, natural log of two thirds. I just swap sides. Natural log of two thirds. Let me do that since I kept the colors going so long."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "I get K is equal to negative one half, negative one half, natural log, natural log of two thirds. I just swap sides. Natural log of two thirds. Let me do that since I kept the colors going so long. Let me keep it that way. Natural log of two thirds. And so we have solved for all of the constants."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Let me do that since I kept the colors going so long. Let me keep it that way. Natural log of two thirds. And so we have solved for all of the constants. So now we can rewrite this thing right over here. We can rewrite it as, so we decided for a mini drum roll here. We're not completely done yet."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And so we have solved for all of the constants. So now we can rewrite this thing right over here. We can rewrite it as, so we decided for a mini drum roll here. We're not completely done yet. We get T of T is equal to, is equal to 60 E, E to the negative K. Well negative K, the negative and negative is going to be positive. It's going to be one half natural log of two thirds. So one half natural log of two thirds."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "We're not completely done yet. We get T of T is equal to, is equal to 60 E, E to the negative K. Well negative K, the negative and negative is going to be positive. It's going to be one half natural log of two thirds. So one half natural log of two thirds. Negative K, so negative of a negative. So one half natural log of two thirds. Which actually will be a negative value."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So one half natural log of two thirds. Negative K, so negative of a negative. So one half natural log of two thirds. Which actually will be a negative value. Two thirds is less than E. So you're going to have a natural log of it is going to be negative. So it makes you feel good that the temperature is going to be going down over time. And so plus, oh times T, times T plus 20."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Which actually will be a negative value. Two thirds is less than E. So you're going to have a natural log of it is going to be negative. So it makes you feel good that the temperature is going to be going down over time. And so plus, oh times T, times T plus 20. Plus 20. And so now all we have to do is we have to figure out what T gets us to a temperature of 40 degrees Celsius. So let's solve for that."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And so plus, oh times T, times T plus 20. Plus 20. And so now all we have to do is we have to figure out what T gets us to a temperature of 40 degrees Celsius. So let's solve for that. So if we want this to be 40, 40 is equal to, and actually now I'm just going to stick to one color as we march through this part. So 40 is going to be equal to 60 E to the one half natural log of two thirds T power plus 20. Now let's see, we can subtract 20 from both sides."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So let's solve for that. So if we want this to be 40, 40 is equal to, and actually now I'm just going to stick to one color as we march through this part. So 40 is going to be equal to 60 E to the one half natural log of two thirds T power plus 20. Now let's see, we can subtract 20 from both sides. We get 20 is equal to 60 E, all that crazy business. One half natural log of two thirds times T. Now we can divide both sides by 60 and we get one third, 20 divided by 60 is one third, is equal to E to the one half natural log of two thirds times T. Now let's see, we can take the natural log of both sides. So we get the natural log of one third is equal to one half natural log of two thirds times T. And then home stretch to solve for T, you just divide both sides by one half natural log of two thirds."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Now let's see, we can subtract 20 from both sides. We get 20 is equal to 60 E, all that crazy business. One half natural log of two thirds times T. Now we can divide both sides by 60 and we get one third, 20 divided by 60 is one third, is equal to E to the one half natural log of two thirds times T. Now let's see, we can take the natural log of both sides. So we get the natural log of one third is equal to one half natural log of two thirds times T. And then home stretch to solve for T, you just divide both sides by one half natural log of two thirds. So we get T is equal to this, which is the natural log of one third divided by one half natural log of two thirds. Well if you divide by one half, that's the same thing as multiplying by two and then you're going to divide by natural log, natural log of two thirds. Two thirds."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So we get the natural log of one third is equal to one half natural log of two thirds times T. And then home stretch to solve for T, you just divide both sides by one half natural log of two thirds. So we get T is equal to this, which is the natural log of one third divided by one half natural log of two thirds. Well if you divide by one half, that's the same thing as multiplying by two and then you're going to divide by natural log, natural log of two thirds. Two thirds. Now let's see if this actually makes a sensical answer. So let me get a calculator out. Actually I could just use Google here."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Two thirds. Now let's see if this actually makes a sensical answer. So let me get a calculator out. Actually I could just use Google here. So I had, actually I already forgot what it was. Natural one, so I have natural log of one third over natural log of two thirds and the whole thing times two. So I can type, I can type two, two times the natural log of, natural log of one third, natural log of one third divided by, divided by the natural log of two thirds."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "Actually I could just use Google here. So I had, actually I already forgot what it was. Natural one, so I have natural log of one third over natural log of two thirds and the whole thing times two. So I can type, I can type two, two times the natural log of, natural log of one third, natural log of one third divided by, divided by the natural log of two thirds. Let's see what Google gets us. Alright, it didn't, what did I, where, how did I mess up? So this is equal to two times the natural log, oh okay, I messed up the parentheses."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So I can type, I can type two, two times the natural log of, natural log of one third, natural log of one third divided by, divided by the natural log of two thirds. Let's see what Google gets us. Alright, it didn't, what did I, where, how did I mess up? So this is equal to two times the natural log, oh okay, I messed up the parentheses. Let me make this clear. The natural log of one third divided by the natural log of two thirds. And then we're, there you go."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "So this is equal to two times the natural log, oh okay, I messed up the parentheses. Let me make this clear. The natural log of one third divided by the natural log of two thirds. And then we're, there you go. If we were to round to the nearest hundredth, it would be 5.42. So 5.42 minutes. Remember, everything we were doing were in minutes."}, {"video_title": "Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3", "Sentence": "And then we're, there you go. If we were to round to the nearest hundredth, it would be 5.42. So 5.42 minutes. Remember, everything we were doing were in minutes. So this right over here, this is approximately equal to 5.42, 5.42 minutes. And we are done. That's how long it will take us to cool to 40 degrees."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "For example, if you know that y is equal to f of x, you might write this as y prime. You might write this as dy dx, which you'll often hear me say is the derivative of y with respect to x, and that you could use the derivative of f with respect to x, because y is equal to our function. But then later on, especially when you start getting into differential equations, you see people start to treat this notation as an actual algebraic expression. For example, you will learn, or you might have already seen, if you're trying to solve the differential equation, the derivative of y with respect to x is equal to y, so the rate of change of y with respect to x is equal to the value of y itself. This is one of the most basic differential equations you might see. You'll see this technique where people say, well, let's just multiply both sides by dx, just treating dx like as if it's some algebraic expression. So you multiply both sides by dx, and then you have, so that would cancel out algebraically."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "For example, you will learn, or you might have already seen, if you're trying to solve the differential equation, the derivative of y with respect to x is equal to y, so the rate of change of y with respect to x is equal to the value of y itself. This is one of the most basic differential equations you might see. You'll see this technique where people say, well, let's just multiply both sides by dx, just treating dx like as if it's some algebraic expression. So you multiply both sides by dx, and then you have, so that would cancel out algebraically. And so you see people treat it like that. So you have dy is equal to y times dx, and then they'll say, okay, let's divide both sides by y, which is a reasonable thing to do. Y is an algebraic expression."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "So you multiply both sides by dx, and then you have, so that would cancel out algebraically. And so you see people treat it like that. So you have dy is equal to y times dx, and then they'll say, okay, let's divide both sides by y, which is a reasonable thing to do. Y is an algebraic expression. So if you divide both sides by y, you get one over y dy is equal to dx. And then folks will integrate both sides to find a general solution to this differential equation. But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using what we call differentials, so a dx or a dy, and treating them algebraically like this, treating them as algebraic expressions where I can just multiply both sides by just dx or dy or divide both sides by dx or dy."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "Y is an algebraic expression. So if you divide both sides by y, you get one over y dy is equal to dx. And then folks will integrate both sides to find a general solution to this differential equation. But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using what we call differentials, so a dx or a dy, and treating them algebraically like this, treating them as algebraic expressions where I can just multiply both sides by just dx or dy or divide both sides by dx or dy. And I don't normally say this, but the rigor you need to show that this is okay in this situation is not an easy thing to say. And so to just feel reasonably okay about doing this, this is a little bit hand-wavy. It's not super mathematically rigorous, but it has proven to be a useful tool for us to find these solutions."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using what we call differentials, so a dx or a dy, and treating them algebraically like this, treating them as algebraic expressions where I can just multiply both sides by just dx or dy or divide both sides by dx or dy. And I don't normally say this, but the rigor you need to show that this is okay in this situation is not an easy thing to say. And so to just feel reasonably okay about doing this, this is a little bit hand-wavy. It's not super mathematically rigorous, but it has proven to be a useful tool for us to find these solutions. And conceptually, the way that I think about a dy or a dx is this is the super small change in y in response to a super small change in x. And that's essentially what this definition of the limit is telling us, especially as delta x approaches zero, we're going to have a super small change in x as delta x approaches zero, and then we're gonna have a resulting super small change in y. So that's one way that you can feel a little bit better of, and this is actually one of the justifications for this type of notation, is you could view this, what's the resulting super small, or what's the super small change in y for a given super small change in x, which is giving us the sense of what's the limiting value of the slope as we go from the slope of a secant line to a tangent line."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "It's not super mathematically rigorous, but it has proven to be a useful tool for us to find these solutions. And conceptually, the way that I think about a dy or a dx is this is the super small change in y in response to a super small change in x. And that's essentially what this definition of the limit is telling us, especially as delta x approaches zero, we're going to have a super small change in x as delta x approaches zero, and then we're gonna have a resulting super small change in y. So that's one way that you can feel a little bit better of, and this is actually one of the justifications for this type of notation, is you could view this, what's the resulting super small, or what's the super small change in y for a given super small change in x, which is giving us the sense of what's the limiting value of the slope as we go from the slope of a secant line to a tangent line. And if you view it that way, you might feel a little bit better about using the differentials or treating them algebraically, whereas, okay, let me just multiply both sides by that super small change in x. So the big picture is this is a technique that you will often see in introductory differential equations classes, introductory multivariable classes, and introductory calculus classes, but it's not very mathematically rigorous to just treat differentials like algebraic expressions, but even though it's not very mathematically rigorous to do it willy-nilly like that, it has proven to be very useful. Now, as you get more sophisticated in your mathematics, there are rigorous definitions of a differential where you can get a better sense of where it is mathematically rigorous to use it and where it isn't."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'll now introduce you to the concept of the Laplace transform, and this is truly one of the most useful concepts that you'll learn, not just in differential equations, but really in mathematics. And especially if you're going to go into engineering, you'll find that the Laplace transform, besides helping you solve differential equations, also helps you transform functions or waveforms from the time domain to the frequency domain and study and understand a whole set of phenomena. But I won't get into all of that yet. Now I'll just teach you what it is, Laplace transform. I'll teach you what it is, make you comfortable with the mathematics of it, and then in a couple of videos from now, I'll actually show you how it is useful to use it to solve differential equations. We'll actually solve some of the differential equations we did before using the previous methods, but we'll keep doing it and it'll solve more and more difficult problems. So what is the Laplace transform?"}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now I'll just teach you what it is, Laplace transform. I'll teach you what it is, make you comfortable with the mathematics of it, and then in a couple of videos from now, I'll actually show you how it is useful to use it to solve differential equations. We'll actually solve some of the differential equations we did before using the previous methods, but we'll keep doing it and it'll solve more and more difficult problems. So what is the Laplace transform? Well, the Laplace transform, the notation is the L, like Laverne from Laverne and Shirley. Might be before many of your times, but I grew up on that. So actually, I think it was even reruns when I was a kid."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So what is the Laplace transform? Well, the Laplace transform, the notation is the L, like Laverne from Laverne and Shirley. Might be before many of your times, but I grew up on that. So actually, I think it was even reruns when I was a kid. Maybe, well anyway. So Laplace transform of some function, and here the convention, instead of saying f of x, people say f of t. And the reason is, is because in a lot of the differential equations or a lot of engineering, you actually are converting from a function of time to a function of frequency, and don't worry about that right now if it confuses you. But Laplace transform of a function of t, it transforms that function into some other function of s. And how does it do that?"}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So actually, I think it was even reruns when I was a kid. Maybe, well anyway. So Laplace transform of some function, and here the convention, instead of saying f of x, people say f of t. And the reason is, is because in a lot of the differential equations or a lot of engineering, you actually are converting from a function of time to a function of frequency, and don't worry about that right now if it confuses you. But Laplace transform of a function of t, it transforms that function into some other function of s. And how does it do that? Well actually, let me just do some mathematical notation that probably won't mean much to you. So what is a transform? Well, the way I think of it is kind of a function of functions."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But Laplace transform of a function of t, it transforms that function into some other function of s. And how does it do that? Well actually, let me just do some mathematical notation that probably won't mean much to you. So what is a transform? Well, the way I think of it is kind of a function of functions. A function will take you from one set of, well in what we've been dealing with, one set of numbers to another set of numbers. A transform will take you from one set of functions to another set of functions. Let me just define this."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, the way I think of it is kind of a function of functions. A function will take you from one set of, well in what we've been dealing with, one set of numbers to another set of numbers. A transform will take you from one set of functions to another set of functions. Let me just define this. The Laplace transform, for our purposes, is defined as the improper integral, and I know I haven't actually done improper integrals just yet, but I'll explain them in a few seconds, the improper integral from 0 to infinity of e to the minus st times f of t. So whatever's between the Laplace transform brackets dt. Now that might seem very daunting to you and very confusing, but I'll now do a couple of examples. So what is a Laplace transform?"}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let me just define this. The Laplace transform, for our purposes, is defined as the improper integral, and I know I haven't actually done improper integrals just yet, but I'll explain them in a few seconds, the improper integral from 0 to infinity of e to the minus st times f of t. So whatever's between the Laplace transform brackets dt. Now that might seem very daunting to you and very confusing, but I'll now do a couple of examples. So what is a Laplace transform? Well, let's say that f of t is equal to 1. So what is the Laplace transform of 1? So if f of t is equal to 1, it's just a constant function of time, well that will be equal to, and let me rewrite it, well actually let me just substitute exactly the way I wrote it here."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So what is a Laplace transform? Well, let's say that f of t is equal to 1. So what is the Laplace transform of 1? So if f of t is equal to 1, it's just a constant function of time, well that will be equal to, and let me rewrite it, well actually let me just substitute exactly the way I wrote it here. So that's the improper integral from 0 to infinity of e to the minus st times 1 here. Well, I don't have to rewrite it here, but there's a times 1 dt. And I know that infinity is probably bugging you right now, but we'll deal with that shortly."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if f of t is equal to 1, it's just a constant function of time, well that will be equal to, and let me rewrite it, well actually let me just substitute exactly the way I wrote it here. So that's the improper integral from 0 to infinity of e to the minus st times 1 here. Well, I don't have to rewrite it here, but there's a times 1 dt. And I know that infinity is probably bugging you right now, but we'll deal with that shortly. Actually, let's deal with that right now. This is the same thing. This is the same thing as the limit."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And I know that infinity is probably bugging you right now, but we'll deal with that shortly. Actually, let's deal with that right now. This is the same thing. This is the same thing as the limit. And let's say as a approaches infinity of the integral from 0 to a, e to the minus st dt. So this is, just so you feel a little bit more comfortable with it, you might have guessed that this is the same thing, because obviously you can't evaluate infinity, but you can take the limit as something approaches infinity. So anyway, let's take the antiderivative and evaluate this improper definite integral or this improper integral."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is the same thing as the limit. And let's say as a approaches infinity of the integral from 0 to a, e to the minus st dt. So this is, just so you feel a little bit more comfortable with it, you might have guessed that this is the same thing, because obviously you can't evaluate infinity, but you can take the limit as something approaches infinity. So anyway, let's take the antiderivative and evaluate this improper definite integral or this improper integral. So what's the antiderivative of e to the minus st with respect to dt? Well, it's equal to minus 1 over s e to the minus st. If you don't believe me, take the derivative of this."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So anyway, let's take the antiderivative and evaluate this improper definite integral or this improper integral. So what's the antiderivative of e to the minus st with respect to dt? Well, it's equal to minus 1 over s e to the minus st. If you don't believe me, take the derivative of this. You take minus s times that, that would all cancel out and you'd just be left with e to the minus st. Fair enough. And we're going to take, actually let me delete this here, this equal sign, because I could actually use some of that real estate. We are going to take the limit as a approaches infinity."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If you don't believe me, take the derivative of this. You take minus s times that, that would all cancel out and you'd just be left with e to the minus st. Fair enough. And we're going to take, actually let me delete this here, this equal sign, because I could actually use some of that real estate. We are going to take the limit as a approaches infinity. You don't always have to do this, but this is the first time we're dealing with improper integrals, so I figured I might as well remind you that we're taking a limit. And we're going to evaluate, we took the antiderivative, now we have to evaluate it at a minus the antiderivative evaluated at 0, and then take the limit of whatever that ends up being as a approaches infinity. So this is equal to the limit as a approaches infinity."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We are going to take the limit as a approaches infinity. You don't always have to do this, but this is the first time we're dealing with improper integrals, so I figured I might as well remind you that we're taking a limit. And we're going to evaluate, we took the antiderivative, now we have to evaluate it at a minus the antiderivative evaluated at 0, and then take the limit of whatever that ends up being as a approaches infinity. So this is equal to the limit as a approaches infinity. OK, if we substitute a in here first, we get minus 1 over s. Remember, we're dealing with t. We took the integral with respect to t. e to the minus s a, right? That's what happens when I put a in here. Minus, now what happens if when I put t equals 0 in here?"}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this is equal to the limit as a approaches infinity. OK, if we substitute a in here first, we get minus 1 over s. Remember, we're dealing with t. We took the integral with respect to t. e to the minus s a, right? That's what happens when I put a in here. Minus, now what happens if when I put t equals 0 in here? So when t equals 0, it becomes e to the minus s times 0, this whole thing becomes 1, and I'm just left with minus 1 over s. So I have minus 1 over s. Fair enough. And then let me scroll down a little bit. I wrote a little bit bigger than I wanted to, but that's OK."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Minus, now what happens if when I put t equals 0 in here? So when t equals 0, it becomes e to the minus s times 0, this whole thing becomes 1, and I'm just left with minus 1 over s. So I have minus 1 over s. Fair enough. And then let me scroll down a little bit. I wrote a little bit bigger than I wanted to, but that's OK. So this is going to be the limit as a approaches infinity of minus 1 over s e to the minus s a minus minus 1 over s, so plus 1 over s. So what's the limit as a approaches infinity? Well, what's this term going to do? As a approaches infinity, the exponent is going to get, if we assume that s is greater than 0, and we'll make that assumption for now."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I wrote a little bit bigger than I wanted to, but that's OK. So this is going to be the limit as a approaches infinity of minus 1 over s e to the minus s a minus minus 1 over s, so plus 1 over s. So what's the limit as a approaches infinity? Well, what's this term going to do? As a approaches infinity, the exponent is going to get, if we assume that s is greater than 0, and we'll make that assumption for now. Let me write that down explicitly. Let's assume that s is greater than 0. So if we assume that s is greater than 0, then as a approaches infinity, what's going to happen?"}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "As a approaches infinity, the exponent is going to get, if we assume that s is greater than 0, and we'll make that assumption for now. Let me write that down explicitly. Let's assume that s is greater than 0. So if we assume that s is greater than 0, then as a approaches infinity, what's going to happen? Well, this term is going to go to 0, right? e to the minus a googol is a very, very small number, and e to the minus googolplex is a very even smaller number. So then this e to the minus infinity approaches 0, so this term approaches 0."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if we assume that s is greater than 0, then as a approaches infinity, what's going to happen? Well, this term is going to go to 0, right? e to the minus a googol is a very, very small number, and e to the minus googolplex is a very even smaller number. So then this e to the minus infinity approaches 0, so this term approaches 0. This term isn't affected because it has no a in it, so we're just left with 1 over s. So there you go. This is a significant moment in your life. You have just been exposed to your first Laplace transform."}, {"video_title": "Laplace transform 1 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So then this e to the minus infinity approaches 0, so this term approaches 0. This term isn't affected because it has no a in it, so we're just left with 1 over s. So there you go. This is a significant moment in your life. You have just been exposed to your first Laplace transform. Eventually, and I'll show you in a few videos, there are whole tables of Laplace transforms, and eventually we'll prove all of them. But for now, we'll just work through some of the more basic ones. But this can be our first entry in our Laplace transform table."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's keep doing some Laplace transforms. And one, it's good to see where a lot of those Laplace transform tables you'll see later on actually come from. And it just makes you comfortable with the mathematics, which is really just kind of your second semester calculus mathematics. But it makes you comfortable with the whole notion of what we're doing. So first of all, let me just rewrite the definition of the Laplace transform. So it's the L from Laverne and Shirley. So Laplace transform of some function of t is equal to the improper integral from 0 to infinity of e to the minus st times our function."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But it makes you comfortable with the whole notion of what we're doing. So first of all, let me just rewrite the definition of the Laplace transform. So it's the L from Laverne and Shirley. So Laplace transform of some function of t is equal to the improper integral from 0 to infinity of e to the minus st times our function. Times our function of t and with respect to dt. So let's do another Laplace transform. Let's say that we want to take the Laplace transform."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So Laplace transform of some function of t is equal to the improper integral from 0 to infinity of e to the minus st times our function. Times our function of t and with respect to dt. So let's do another Laplace transform. Let's say that we want to take the Laplace transform. And now our function f of t, let's say it is e to the at. Laplace transform of e to the at. Well, we just substituted it into this definition of the Laplace transform."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's say that we want to take the Laplace transform. And now our function f of t, let's say it is e to the at. Laplace transform of e to the at. Well, we just substituted it into this definition of the Laplace transform. So that is equal to. And this is all going to be really good integration practice for us, especially integration by parts. Almost every Laplace transform problem turns into an integration by parts problem."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, we just substituted it into this definition of the Laplace transform. So that is equal to. And this is all going to be really good integration practice for us, especially integration by parts. Almost every Laplace transform problem turns into an integration by parts problem. Which as we learned long ago, integration by parts is just the reverse product rule. So anyway, this is equal to the integral from 0 to infinity, e to the minus st times e to the at. That's our f of t. dt."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Almost every Laplace transform problem turns into an integration by parts problem. Which as we learned long ago, integration by parts is just the reverse product rule. So anyway, this is equal to the integral from 0 to infinity, e to the minus st times e to the at. That's our f of t. dt. Well, this is equal to, just adding the exponents, because we're going to use the exponents. Equal to, just adding the exponents, because we have the same base, the integral from 0 to infinity of e to the what? a minus s. e to the a minus st dt."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That's our f of t. dt. Well, this is equal to, just adding the exponents, because we're going to use the exponents. Equal to, just adding the exponents, because we have the same base, the integral from 0 to infinity of e to the what? a minus s. e to the a minus st dt. And what's the antiderivative of this? Well, that's equal to what with respect to t? So we're at, it's equal to 1 over a minus s. That's just going to be a constant, right?"}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "a minus s. e to the a minus st dt. And what's the antiderivative of this? Well, that's equal to what with respect to t? So we're at, it's equal to 1 over a minus s. That's just going to be a constant, right? So we can just leave it out on the outside. 1 over a minus s times e to the a minus st. And we're going to evaluate that from t is equal to infinity, or the limit as t approaches infinity, to t is equal to 0. And I could have put this inside the brackets, but it's just a constant term, right?"}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we're at, it's equal to 1 over a minus s. That's just going to be a constant, right? So we can just leave it out on the outside. 1 over a minus s times e to the a minus st. And we're going to evaluate that from t is equal to infinity, or the limit as t approaches infinity, to t is equal to 0. And I could have put this inside the brackets, but it's just a constant term, right? a, none of them have t's in them, so I can just pull them out. And so this is equal to 1 over a minus s times. Now, what is the limit?"}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And I could have put this inside the brackets, but it's just a constant term, right? a, none of them have t's in them, so I can just pull them out. And so this is equal to 1 over a minus s times. Now, what is the limit? We essentially have to evaluate t at infinity. So what is the limit at infinity? Well, we have two cases here, right?"}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, what is the limit? We essentially have to evaluate t at infinity. So what is the limit at infinity? Well, we have two cases here, right? If this exponent, if this a minus s is a positive number, if a minus s is greater than 0, what's going to happen? Well, as we approach infinity, e to the infinity just gets bigger and bigger and bigger, right? Because it's e to an infinitely positive exponent."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, we have two cases here, right? If this exponent, if this a minus s is a positive number, if a minus s is greater than 0, what's going to happen? Well, as we approach infinity, e to the infinity just gets bigger and bigger and bigger, right? Because it's e to an infinitely positive exponent. So it actually, we don't get an answer. And when you do improper integrals, when you take the limit to infinity and it doesn't come to a finite number, the limit doesn't approach anything, that means that the limit, or actually the improper integral, diverges. And so there is no limit."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Because it's e to an infinitely positive exponent. So it actually, we don't get an answer. And when you do improper integrals, when you take the limit to infinity and it doesn't come to a finite number, the limit doesn't approach anything, that means that the limit, or actually the improper integral, diverges. And so there is no limit. And to some degree, we can say that the Laplace transform is not defined when a minus s is greater than 0, or when a is greater than s. Now, what happens if a minus s is less than 0? So if a minus s is less than 0, well then, this is going to be some negative number here, right? And then if we take e to an infinitely negative number, well then, that does approach something."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And so there is no limit. And to some degree, we can say that the Laplace transform is not defined when a minus s is greater than 0, or when a is greater than s. Now, what happens if a minus s is less than 0? So if a minus s is less than 0, well then, this is going to be some negative number here, right? And then if we take e to an infinitely negative number, well then, that does approach something. That approaches 0. And we saw that in the previous video. And I hope you understand what I'm saying, right?"}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then if we take e to an infinitely negative number, well then, that does approach something. That approaches 0. And we saw that in the previous video. And I hope you understand what I'm saying, right? e to an infinitely negative number is 0, or it approaches 0, while e to an infinitely positive number is just infinity, so it doesn't really converge on anything. So anyway, so if I assume that a minus s is less than 0, or a is less than s, and this is the assumption I will make, just so that I get, so that this integral, this improper integral, actually converges to something. So if a minus s is less than 0, then this is a negative number, e to the a minus s times, well, t, where t approaches infinity, will be 0, minus this integral evaluated at 0."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And I hope you understand what I'm saying, right? e to an infinitely negative number is 0, or it approaches 0, while e to an infinitely positive number is just infinity, so it doesn't really converge on anything. So anyway, so if I assume that a minus s is less than 0, or a is less than s, and this is the assumption I will make, just so that I get, so that this integral, this improper integral, actually converges to something. So if a minus s is less than 0, then this is a negative number, e to the a minus s times, well, t, where t approaches infinity, will be 0, minus this integral evaluated at 0. So when you evaluate this at 0, what happens? t equals 0, this whole thing becomes 0, e to the 0 is 1. e to the 0 is 1, and we are left with what? Minus 1 over a minus s, and that's just the same thing as 1 over s minus a."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if a minus s is less than 0, then this is a negative number, e to the a minus s times, well, t, where t approaches infinity, will be 0, minus this integral evaluated at 0. So when you evaluate this at 0, what happens? t equals 0, this whole thing becomes 0, e to the 0 is 1. e to the 0 is 1, and we are left with what? Minus 1 over a minus s, and that's just the same thing as 1 over s minus a. So we have our next entry in our Laplace transform table. We have our next entry, and that is the Laplace transform of e to the a t is equal to 1 over s minus a, as long as we make the assumption that s is greater than a. This is true when s is greater than a, or a is less than s. Either way, you could view it either way."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Minus 1 over a minus s, and that's just the same thing as 1 over s minus a. So we have our next entry in our Laplace transform table. We have our next entry, and that is the Laplace transform of e to the a t is equal to 1 over s minus a, as long as we make the assumption that s is greater than a. This is true when s is greater than a, or a is less than s. Either way, you could view it either way. So that's our second entry in our Laplace transform table. Fascinating. And actually, let's relate this to our previous entry in our Laplace transform table."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This is true when s is greater than a, or a is less than s. Either way, you could view it either way. So that's our second entry in our Laplace transform table. Fascinating. And actually, let's relate this to our previous entry in our Laplace transform table. What was our first entry in our Laplace transform table? It was Laplace transform of 1 is equal to 1 over s, right? Well, isn't 1 just the same thing as e to the 0?"}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And actually, let's relate this to our previous entry in our Laplace transform table. What was our first entry in our Laplace transform table? It was Laplace transform of 1 is equal to 1 over s, right? Well, isn't 1 just the same thing as e to the 0? So we could have said that this is the Laplace, I know I'm running out of space, but I'll do it here in purple. We could have said Laplace transform of 1 is the same thing as e to the 0 times t, right? And that equals 1 over s. And luckily, it's good to see that that is consistent."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, isn't 1 just the same thing as e to the 0? So we could have said that this is the Laplace, I know I'm running out of space, but I'll do it here in purple. We could have said Laplace transform of 1 is the same thing as e to the 0 times t, right? And that equals 1 over s. And luckily, it's good to see that that is consistent. And actually, remember, we even made the condition when s is greater than 0, right? We assumed that s is greater than 0 in this example. And lo and behold, here again, you say s is greater than 0."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And that equals 1 over s. And luckily, it's good to see that that is consistent. And actually, remember, we even made the condition when s is greater than 0, right? We assumed that s is greater than 0 in this example. And lo and behold, here again, you say s is greater than 0. This is completely consistent with this one, right? Because if a is equal to 0, then the Laplace transform of e to the 0 is just 1 over s minus 0. That's just 1 over s. And we have to assume that s is greater than 0."}, {"video_title": "Laplace transform 2 Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And lo and behold, here again, you say s is greater than 0. This is completely consistent with this one, right? Because if a is equal to 0, then the Laplace transform of e to the 0 is just 1 over s minus 0. That's just 1 over s. And we have to assume that s is greater than 0. So really, these are kind of the same entry in our Laplace transform table, but it's always nice in mathematics when we see that two results we got trying to do slightly different problems actually are, in some ways, connected or the same result. Anyway, I'll see you in the next video, and we'll keep trying to build our table of Laplace transforms in maybe three or four videos from now. I'll actually show you how these transforms are extremely useful in solving all sorts of differential equations."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "And I defined it to be, and I'll do the shifted version of it, you're already hopefully reasonably familiar with it. If I shift it, so Dirac delta of t minus c, we can say that it equals 0 when t does not equal c, so it equals 0 everywhere, but it essentially pops up to infinity. And we have to be careful with this infinity. I'm going to write it in quotes. It pops up to infinity, and we even saw in the previous video, it's kind of different degrees of infinity, because you can still multiply this by other numbers to get larger Dirac delta functions when t is equal to c. But more important than this, and this is kind of a pseudo-definition here, is the idea that when we take the integral, when we take the area under the curve over the entire x, or the entire t-axis, I guess we could say, when we take the area under this curve, and obviously it equals 0 everywhere except at t is equal to c, when we take this area, this is the important point, that the area is equal to 1. And so this is what I meant by kind of pseudo-infinity, because if I have 2 times the Dirac delta function, then if I'm taking the area under the curve of that, of 2 times the Dirac delta function, t minus c dt, this should be equal to 2 times the area of just under the Dirac delta function, 2 times from minus infinity to infinity of the delta function shifted by c dt, which is just 2 times, we already showed you, I just said by definition, this is 1. So this will be equal to 2."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "I'm going to write it in quotes. It pops up to infinity, and we even saw in the previous video, it's kind of different degrees of infinity, because you can still multiply this by other numbers to get larger Dirac delta functions when t is equal to c. But more important than this, and this is kind of a pseudo-definition here, is the idea that when we take the integral, when we take the area under the curve over the entire x, or the entire t-axis, I guess we could say, when we take the area under this curve, and obviously it equals 0 everywhere except at t is equal to c, when we take this area, this is the important point, that the area is equal to 1. And so this is what I meant by kind of pseudo-infinity, because if I have 2 times the Dirac delta function, then if I'm taking the area under the curve of that, of 2 times the Dirac delta function, t minus c dt, this should be equal to 2 times the area of just under the Dirac delta function, 2 times from minus infinity to infinity of the delta function shifted by c dt, which is just 2 times, we already showed you, I just said by definition, this is 1. So this will be equal to 2. So if I put a 2 out here, this infinity will have to be twice as high so that the area is now 2. That's why I put that infinity in parentheses. But it's an interesting function."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "So this will be equal to 2. So if I put a 2 out here, this infinity will have to be twice as high so that the area is now 2. That's why I put that infinity in parentheses. But it's an interesting function. I talked about it at the end of the last video, that it can help model things that kind of jar things all of a sudden, but they impart a fixed amount of impulse on something, a fixed amount of change in momentum. And we'll understand that a little bit more in the future. But let's kind of get the mathematical tools completely understood, and let's try to figure out what the Dirac delta function does when we multiply it, what it does to the Laplace transform when we multiply it times some function."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "But it's an interesting function. I talked about it at the end of the last video, that it can help model things that kind of jar things all of a sudden, but they impart a fixed amount of impulse on something, a fixed amount of change in momentum. And we'll understand that a little bit more in the future. But let's kind of get the mathematical tools completely understood, and let's try to figure out what the Dirac delta function does when we multiply it, what it does to the Laplace transform when we multiply it times some function. So let's say I have my Dirac delta function, and I'm going to shift it. That's a little bit more interesting. And if you want to unshift it, you just say, okay, well, c equals 0."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "But let's kind of get the mathematical tools completely understood, and let's try to figure out what the Dirac delta function does when we multiply it, what it does to the Laplace transform when we multiply it times some function. So let's say I have my Dirac delta function, and I'm going to shift it. That's a little bit more interesting. And if you want to unshift it, you just say, okay, well, c equals 0. What happens when c equals 0? And I'm going to shift it and multiply it times some arbitrary function f of t. If I wanted to figure out the Laplace transform of just the delta function by itself, I could say f of t is equal to 1. So let's take our Laplace transform of this, and we can just use the definition of the Laplace transform."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "And if you want to unshift it, you just say, okay, well, c equals 0. What happens when c equals 0? And I'm going to shift it and multiply it times some arbitrary function f of t. If I wanted to figure out the Laplace transform of just the delta function by itself, I could say f of t is equal to 1. So let's take our Laplace transform of this, and we can just use the definition of the Laplace transform. So this is equal to the area from 0 to infinity underneath, or we could call it the integral from 0 to infinity, of e to the minus st. That's just part of the Laplace transform definition, times this thing. And I'll just write it in this order, times f of t times our Dirac delta function, delta t minus c, and times dt. Now, here I'm going to make a little bit of an intuitive argument."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "So let's take our Laplace transform of this, and we can just use the definition of the Laplace transform. So this is equal to the area from 0 to infinity underneath, or we could call it the integral from 0 to infinity, of e to the minus st. That's just part of the Laplace transform definition, times this thing. And I'll just write it in this order, times f of t times our Dirac delta function, delta t minus c, and times dt. Now, here I'm going to make a little bit of an intuitive argument. A lot of the math we do is kind of, especially if you want to be very rigorous and formal, the Dirac delta function starts to break down a lot of tools that you might have not realized it would break down. But I think intuitively we can still work with it. So I'm going to solve this integral for you intuitively, and I think it'll make some sense."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Now, here I'm going to make a little bit of an intuitive argument. A lot of the math we do is kind of, especially if you want to be very rigorous and formal, the Dirac delta function starts to break down a lot of tools that you might have not realized it would break down. But I think intuitively we can still work with it. So I'm going to solve this integral for you intuitively, and I think it'll make some sense. So let's draw this. Let me draw this, what we're trying to do. So let me draw what we're trying to take the integral of."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "So I'm going to solve this integral for you intuitively, and I think it'll make some sense. So let's draw this. Let me draw this, what we're trying to do. So let me draw what we're trying to take the integral of. And we only care from 0 to infinity, so I'll only do it from 0 to infinity. And I'll assume that c is greater than 0, that the delta function pops up someplace in the positive t-axis. So what is this first part going to look like?"}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "So let me draw what we're trying to take the integral of. And we only care from 0 to infinity, so I'll only do it from 0 to infinity. And I'll assume that c is greater than 0, that the delta function pops up someplace in the positive t-axis. So what is this first part going to look like? What is that going to look like? e to the minus st times f of t. I don't know. It's going to be some function."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "So what is this first part going to look like? What is that going to look like? e to the minus st times f of t. I don't know. It's going to be some function. e to the minus st starts at 1 and drops down, but we're multiplying it times some arbitrary function. So I'll just draw it like this. Maybe it looks something like this."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "It's going to be some function. e to the minus st starts at 1 and drops down, but we're multiplying it times some arbitrary function. So I'll just draw it like this. Maybe it looks something like this. This right here is e to the minus st times f of t. And the f of t is what kind of gives it its arbitrary shape. Fair enough. Now let's graph our Dirac delta function."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Maybe it looks something like this. This right here is e to the minus st times f of t. And the f of t is what kind of gives it its arbitrary shape. Fair enough. Now let's graph our Dirac delta function. It's 0 everywhere except right at c, it pops up infinitely high. But we only draw an arrow that is of height 1 to show that its area is 1. Normally when you graph things you don't draw arrows, but this arrow shows that the area under this infinitely high thing is 1."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Now let's graph our Dirac delta function. It's 0 everywhere except right at c, it pops up infinitely high. But we only draw an arrow that is of height 1 to show that its area is 1. Normally when you graph things you don't draw arrows, but this arrow shows that the area under this infinitely high thing is 1. So we do a 1 there. So if we multiply this, we care about the area under this whole thing. When we multiply these two functions, when we multiply this times this times the delta function, let me write this, this is the delta function shifted to c. If I multiply that times that, what do I get?"}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Normally when you graph things you don't draw arrows, but this arrow shows that the area under this infinitely high thing is 1. So we do a 1 there. So if we multiply this, we care about the area under this whole thing. When we multiply these two functions, when we multiply this times this times the delta function, let me write this, this is the delta function shifted to c. If I multiply that times that, what do I get? This is kind of the key intuition here. Let me redraw my axes. Let me see if I can do it a little bit straighter."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "When we multiply these two functions, when we multiply this times this times the delta function, let me write this, this is the delta function shifted to c. If I multiply that times that, what do I get? This is kind of the key intuition here. Let me redraw my axes. Let me see if I can do it a little bit straighter. Don't judge me by the straightness of my axes. So that's t. So what happens when I multiply these two? Everywhere, when t equals anything other than c, the Dirac delta function is 0."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Let me see if I can do it a little bit straighter. Don't judge me by the straightness of my axes. So that's t. So what happens when I multiply these two? Everywhere, when t equals anything other than c, the Dirac delta function is 0. So it's 0 times anything. I don't care what this function is, it's going to be 0. So it's going to be 0 everywhere except something interesting happens at t is equal to c. At t equals c, what's the value of the function?"}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Everywhere, when t equals anything other than c, the Dirac delta function is 0. So it's 0 times anything. I don't care what this function is, it's going to be 0. So it's going to be 0 everywhere except something interesting happens at t is equal to c. At t equals c, what's the value of the function? Well, it's going to be the value of the Dirac delta function times whatever height this is. This is going to be this point right here, or this right there, that point, this is going to be this function evaluated at c. So it's going to be, I'll mark it right here on the y-axis or on the f of t, whatever you want to call it. This is going to be e to the minus sc times f of c. All I'm doing is I'm just evaluating this function at c. So that's the point right there."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "So it's going to be 0 everywhere except something interesting happens at t is equal to c. At t equals c, what's the value of the function? Well, it's going to be the value of the Dirac delta function times whatever height this is. This is going to be this point right here, or this right there, that point, this is going to be this function evaluated at c. So it's going to be, I'll mark it right here on the y-axis or on the f of t, whatever you want to call it. This is going to be e to the minus sc times f of c. All I'm doing is I'm just evaluating this function at c. So that's the point right there. So if you take this point, which is just some number, it could be 5, 5 times this, you're just getting 5 times the Dirac delta function. Or, in this case, it's not 5, it's this little more abstract thing. I could just draw it like this."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "This is going to be e to the minus sc times f of c. All I'm doing is I'm just evaluating this function at c. So that's the point right there. So if you take this point, which is just some number, it could be 5, 5 times this, you're just getting 5 times the Dirac delta function. Or, in this case, it's not 5, it's this little more abstract thing. I could just draw it like this. When I multiply this thing times my little delta function there, I get this. The height, it's a delta function, but it's scaled now. It's scaled, so now my new thing is going to look like this."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "I could just draw it like this. When I multiply this thing times my little delta function there, I get this. The height, it's a delta function, but it's scaled now. It's scaled, so now my new thing is going to look like this. If I just multiply that times that, I essentially get e to the minus sc times f of c. This might look like some fancy function, but it's just a number when we consider it in terms of t. S becomes something when we go into the plus world, but from t's point of view, it's just a constant. All of these are just constants, so this might as well just be 5. So it's this constant times my Dirac delta function, times delta of t minus c. When I multiply that thing times that thing, all I'm left with is this thing."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "It's scaled, so now my new thing is going to look like this. If I just multiply that times that, I essentially get e to the minus sc times f of c. This might look like some fancy function, but it's just a number when we consider it in terms of t. S becomes something when we go into the plus world, but from t's point of view, it's just a constant. All of these are just constants, so this might as well just be 5. So it's this constant times my Dirac delta function, times delta of t minus c. When I multiply that thing times that thing, all I'm left with is this thing. This height is still going to be infinitely high, but it's infinitely high scaled in such a way that its area is going to be not 1, and I'll show it to you. So what's the integral of this thing? Taking the integral of this thing from minus infinity to infinity, since this thing is this thing, it should be the same thing as taking the integral of this thing from minus infinity to infinity."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "So it's this constant times my Dirac delta function, times delta of t minus c. When I multiply that thing times that thing, all I'm left with is this thing. This height is still going to be infinitely high, but it's infinitely high scaled in such a way that its area is going to be not 1, and I'll show it to you. So what's the integral of this thing? Taking the integral of this thing from minus infinity to infinity, since this thing is this thing, it should be the same thing as taking the integral of this thing from minus infinity to infinity. So let's do that. Actually, we only have to do it from minus infinity. I said from 0 to infinity."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Taking the integral of this thing from minus infinity to infinity, since this thing is this thing, it should be the same thing as taking the integral of this thing from minus infinity to infinity. So let's do that. Actually, we only have to do it from minus infinity. I said from 0 to infinity. So if we take from 0 to infinity, what I'm saying is taking this integral is equivalent to taking this integral. So e to the minus sc f of c times my delta function, t minus c dt. Now, this thing right here, let me make this very clear."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "I said from 0 to infinity. So if we take from 0 to infinity, what I'm saying is taking this integral is equivalent to taking this integral. So e to the minus sc f of c times my delta function, t minus c dt. Now, this thing right here, let me make this very clear. I'm claiming that this is equivalent to this, because everywhere else the delta function zeros out this function, so we only care about this function, or e to the minus st f of t, when t is equal to c. And so that's why we were able to turn it into a constant. But since this is a constant, we can bring it out of the integral. And so this is equal to, I'm going to go backwards here just to kind of save space and still give you these things to look at."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Now, this thing right here, let me make this very clear. I'm claiming that this is equivalent to this, because everywhere else the delta function zeros out this function, so we only care about this function, or e to the minus st f of t, when t is equal to c. And so that's why we were able to turn it into a constant. But since this is a constant, we can bring it out of the integral. And so this is equal to, I'm going to go backwards here just to kind of save space and still give you these things to look at. If we take out the constants from inside of the integral, we get e to the minus sc times f of c times the integral from 0 to infinity of f of t minus c dt. Oh, sorry, not f of t minus c. Let me see how far back I can times. This is not an f. I have to be very careful."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "And so this is equal to, I'm going to go backwards here just to kind of save space and still give you these things to look at. If we take out the constants from inside of the integral, we get e to the minus sc times f of c times the integral from 0 to infinity of f of t minus c dt. Oh, sorry, not f of t minus c. Let me see how far back I can times. This is not an f. I have to be very careful. This is a delta. Let me do that in a different color. I took out the constant terms there, and it's going to be of delta of t minus c dt."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "This is not an f. I have to be very careful. This is a delta. Let me do that in a different color. I took out the constant terms there, and it's going to be of delta of t minus c dt. Let me get the right color, dt. Now, what is this thing by definition? This thing is 1."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "I took out the constant terms there, and it's going to be of delta of t minus c dt. Let me get the right color, dt. Now, what is this thing by definition? This thing is 1. I mean, we could put it from minus infinity to infinity. It doesn't matter. The only time where it has any error is right under c. So this thing is equal to 1."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "This thing is 1. I mean, we could put it from minus infinity to infinity. It doesn't matter. The only time where it has any error is right under c. So this thing is equal to 1. So this whole integral right there has been reduced to this right there because this is just equal to 1. So the Laplace transform of our shifted delta function times some other function is equal to e to the minus sc times f of c. Let me write that again down here. Let me write it all at once."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "The only time where it has any error is right under c. So this thing is equal to 1. So this whole integral right there has been reduced to this right there because this is just equal to 1. So the Laplace transform of our shifted delta function times some other function is equal to e to the minus sc times f of c. Let me write that again down here. Let me write it all at once. So the Laplace transform of our shifted delta function, t minus c, times some function f of t, it equals e to the minus c. Essentially, we're just evaluating e to the minus sc evaluated at c. So e to the minus cs times f of c. We're essentially just evaluating these things at c. This is what it equals. So from this, we can get a lot of interesting things. What is the Laplace transform?"}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "Let me write it all at once. So the Laplace transform of our shifted delta function, t minus c, times some function f of t, it equals e to the minus c. Essentially, we're just evaluating e to the minus sc evaluated at c. So e to the minus cs times f of c. We're essentially just evaluating these things at c. This is what it equals. So from this, we can get a lot of interesting things. What is the Laplace transform? Actually, what is the Laplace transform of just the plain vanilla delta function? Well, in this case, we have c is equal to 0 and f of t is equal to 1. It's just a constant term."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "What is the Laplace transform? Actually, what is the Laplace transform of just the plain vanilla delta function? Well, in this case, we have c is equal to 0 and f of t is equal to 1. It's just a constant term. So if we do that, then the Laplace transform of this thing is just going to be e to the minus 0 times s times 1, which is just equal to 1. So the Laplace transform of our delta function is 1, which is a nice, clean thing to find out. Then if we wanted to just figure out the Laplace transform of our shifted function, the Laplace transform of our shifted delta function, this is just a special case where f of t is equal to 1."}, {"video_title": "Laplace transform of the dirac delta function Laplace transform Khan Academy.mp3", "Sentence": "It's just a constant term. So if we do that, then the Laplace transform of this thing is just going to be e to the minus 0 times s times 1, which is just equal to 1. So the Laplace transform of our delta function is 1, which is a nice, clean thing to find out. Then if we wanted to just figure out the Laplace transform of our shifted function, the Laplace transform of our shifted delta function, this is just a special case where f of t is equal to 1. We could write it times 1, where f of t is equal to 1. So this is going to be equal to e to the minus cs times f of c, but f is just a constant. f is just 1 here, so times 1, or it's just e to the minus cs."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's keep building our table of Laplace transforms. And now we'll do a fairly hairy problem, so I'm going to have to focus so that I don't make a careless mistake. But let's say we want to take the Laplace transform, and this is a useful one. Actually, all of them we've done so far are useful. I'll tell you when we start doing not so useful ones. Let's say we want to take the Laplace transform of the sine of some constant times t. Well, our definition of the Laplace transform that says that it's the improper integral. Remember, the Laplace transform is just a definition, it's just a tool that has turned out to be extremely useful."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Actually, all of them we've done so far are useful. I'll tell you when we start doing not so useful ones. Let's say we want to take the Laplace transform of the sine of some constant times t. Well, our definition of the Laplace transform that says that it's the improper integral. Remember, the Laplace transform is just a definition, it's just a tool that has turned out to be extremely useful. And we'll do more on that intuition later on. But anyways, the integral from 0 to infinity of e to the minus st times whatever we're taking the Laplace transform of times sine of at dt. And now we have to go back and find our integration by parts neuron, and mine always disappears."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Remember, the Laplace transform is just a definition, it's just a tool that has turned out to be extremely useful. And we'll do more on that intuition later on. But anyways, the integral from 0 to infinity of e to the minus st times whatever we're taking the Laplace transform of times sine of at dt. And now we have to go back and find our integration by parts neuron, and mine always disappears. So we have to reprove integration by parts. I don't recommend you do this all the time. If you have to do this on an exam, you might want to memorize it before the exam."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And now we have to go back and find our integration by parts neuron, and mine always disappears. So we have to reprove integration by parts. I don't recommend you do this all the time. If you have to do this on an exam, you might want to memorize it before the exam. But always remember, integration by parts is just the product rule in reverse. So I'll just do that in this corner. So the product rule tells us we have two functions, u times v, and if I were to take the derivative of u times v, let's say that they're functions of t. These are both functions of t. I could have written u of x times v of x."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If you have to do this on an exam, you might want to memorize it before the exam. But always remember, integration by parts is just the product rule in reverse. So I'll just do that in this corner. So the product rule tells us we have two functions, u times v, and if I were to take the derivative of u times v, let's say that they're functions of t. These are both functions of t. I could have written u of x times v of x. That equals the derivative of the first times the second function plus the first function times the derivative of the second. Now if I were to integrate both sides, I get uv, this should be your view, is equal to the integral. Of u prime v with respect to dt, but I'm just doing a little bit of shorthand now."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So the product rule tells us we have two functions, u times v, and if I were to take the derivative of u times v, let's say that they're functions of t. These are both functions of t. I could have written u of x times v of x. That equals the derivative of the first times the second function plus the first function times the derivative of the second. Now if I were to integrate both sides, I get uv, this should be your view, is equal to the integral. Of u prime v with respect to dt, but I'm just doing a little bit of shorthand now. Plus the integral of uv prime. I'm just trying to help myself remember this thing. And let's take this and subtract it from both sides."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Of u prime v with respect to dt, but I'm just doing a little bit of shorthand now. Plus the integral of uv prime. I'm just trying to help myself remember this thing. And let's take this and subtract it from both sides. So we have this integral, the integral of u prime v is going to be equal to this, uv minus the integral of uv prime. And of course, this is a function of t. There's a dt here and all of that. But I just have to do this in the corner of my page a lot, because I always forget this."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And let's take this and subtract it from both sides. So we have this integral, the integral of u prime v is going to be equal to this, uv minus the integral of uv prime. And of course, this is a function of t. There's a dt here and all of that. But I just have to do this in the corner of my page a lot, because I always forget this. And with the primes and the integrals and all that, I always forget it. One way, if you did want to memorize it, you said OK, integration by parts says if I take the integral of the derivative of one thing and then just a regular function of another, it equals the two functions times each other minus the integral of the reverse. Here, when you take the subtraction, you're taking the one that had a derivative now doesn't, and the one that didn't have a derivative now does."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But I just have to do this in the corner of my page a lot, because I always forget this. And with the primes and the integrals and all that, I always forget it. One way, if you did want to memorize it, you said OK, integration by parts says if I take the integral of the derivative of one thing and then just a regular function of another, it equals the two functions times each other minus the integral of the reverse. Here, when you take the subtraction, you're taking the one that had a derivative now doesn't, and the one that didn't have a derivative now does. But anyway, let's apply that to our problem at hand, to this one. Well, let's make, we could go either way about it. Let's make u prime is equal to, so let's say u prime, we'll do our definition."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Here, when you take the subtraction, you're taking the one that had a derivative now doesn't, and the one that didn't have a derivative now does. But anyway, let's apply that to our problem at hand, to this one. Well, let's make, we could go either way about it. Let's make u prime is equal to, so let's say u prime, we'll do our definition. u prime is equal to e to the minus st, in which case u would be the antiderivative of that, which is equal to minus 1 over s e to the minus st. All right? And actually, just so, this is going to be an integration by parts twice problem. So I'm just actually going to define the Laplace transform as y."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's make u prime is equal to, so let's say u prime, we'll do our definition. u prime is equal to e to the minus st, in which case u would be the antiderivative of that, which is equal to minus 1 over s e to the minus st. All right? And actually, just so, this is going to be an integration by parts twice problem. So I'm just actually going to define the Laplace transform as y. That'll come in useful later on. And I think I actually did a very similar example to this when we did integration by parts. But anyway, back to the integration by parts."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So I'm just actually going to define the Laplace transform as y. That'll come in useful later on. And I think I actually did a very similar example to this when we did integration by parts. But anyway, back to the integration by parts. So that's u, and let me do v in a different color. So in v, if this is u prime, then this is v. So v is equal to sine of at. And then what is v prime?"}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But anyway, back to the integration by parts. So that's u, and let me do v in a different color. So in v, if this is u prime, then this is v. So v is equal to sine of at. And then what is v prime? Well, that's just a cosine of at, the chain rule. And now we're ready to do our integration. So the Laplace transform, and I'll just say that's y. y is equal to, right?"}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then what is v prime? Well, that's just a cosine of at, the chain rule. And now we're ready to do our integration. So the Laplace transform, and I'll just say that's y. y is equal to, right? y is what we're trying to solve for, the Laplace transform of sine of at. That is equal to u prime v. We had to find u prime in v, right? That's equal to that."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So the Laplace transform, and I'll just say that's y. y is equal to, right? y is what we're trying to solve for, the Laplace transform of sine of at. That is equal to u prime v. We had to find u prime in v, right? That's equal to that. The integral of u prime times v. That equals uv. So uv, so that's minus 1 over se to the minus st times v sine of at minus the integral. And when you do the integration by parts, this could be an indefinite integral, an improper integral, a definite integral, whatever."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That's equal to that. The integral of u prime times v. That equals uv. So uv, so that's minus 1 over se to the minus st times v sine of at minus the integral. And when you do the integration by parts, this could be an indefinite integral, an improper integral, a definite integral, whatever. But the boundaries stay. So we could still say from 0 to infinity. Of uv prime."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And when you do the integration by parts, this could be an indefinite integral, an improper integral, a definite integral, whatever. But the boundaries stay. So we could still say from 0 to infinity. Of uv prime. So u is minus 1 over se to the minus st times v prime. Times a cosine of at. Fair enough."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Of uv prime. So u is minus 1 over se to the minus st times v prime. Times a cosine of at. Fair enough. dt. Well now we have another hairy integral we need to solve. So this might involve another integration by parts."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Fair enough. dt. Well now we have another hairy integral we need to solve. So this might involve another integration by parts. And it does. So here, let's see if we can simplify it. Let's take the constants out first."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this might involve another integration by parts. And it does. So here, let's see if we can simplify it. Let's take the constants out first. Let me just rewrite this. So we get y is equal to minus e to the minus st over s sine of at. So you have a minus minus plus a over s. Plus a over s. Right?"}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's take the constants out first. Let me just rewrite this. So we get y is equal to minus e to the minus st over s sine of at. So you have a minus minus plus a over s. Plus a over s. Right? a divided by s. And these two negative signs cancel out. Times the integral from 0 to infinity e to the minus st cosine of at dt. Let's do another integration by parts."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So you have a minus minus plus a over s. Plus a over s. Right? a divided by s. And these two negative signs cancel out. Times the integral from 0 to infinity e to the minus st cosine of at dt. Let's do another integration by parts. And I'll do this in a purple color. Just so you know, this is our second integration by parts. Let's define, once again, u prime is equal to e to the minus st."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's do another integration by parts. And I'll do this in a purple color. Just so you know, this is our second integration by parts. Let's define, once again, u prime is equal to e to the minus st. So this is u prime. Then u is equal to minus 1 over se to the minus st. We'll make v equal to cosine of at. The hardest part about this is just not making careless mistakes."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let's define, once again, u prime is equal to e to the minus st. So this is u prime. Then u is equal to minus 1 over se to the minus st. We'll make v equal to cosine of at. The hardest part about this is just not making careless mistakes. And then v prime, I just want it to be in the same row, is equal to minus a sine of at. Right? Chain rule, derivative of cosine is minus sine."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "The hardest part about this is just not making careless mistakes. And then v prime, I just want it to be in the same row, is equal to minus a sine of at. Right? Chain rule, derivative of cosine is minus sine. So let's substitute that back in. And we get, this is going to get hairy. Actually, it already is hairy."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Chain rule, derivative of cosine is minus sine. So let's substitute that back in. And we get, this is going to get hairy. Actually, it already is hairy. y is equal to minus e to the minus st over s sine of at plus a over s times integration by parts. uv, so that's minus 1 over se to the minus st times v times cosine at minus the integral from 0 to infinity. This problem's making me hungry."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Actually, it already is hairy. y is equal to minus e to the minus st over s sine of at plus a over s times integration by parts. uv, so that's minus 1 over se to the minus st times v times cosine at minus the integral from 0 to infinity. This problem's making me hungry. It's taking so much glucose from my bloodstream. I'm focusing so much not to make careless mistakes. Anyway, integral from 0 to infinity."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This problem's making me hungry. It's taking so much glucose from my bloodstream. I'm focusing so much not to make careless mistakes. Anyway, integral from 0 to infinity. And now we have uv prime. So u is minus 1 over s e to the minus st. That's u. And then v prime times minus a."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Anyway, integral from 0 to infinity. And now we have uv prime. So u is minus 1 over s e to the minus st. That's u. And then v prime times minus a. So let's make that minus cancel out with this one so that becomes a plus. a sine of at dt. I'm starting to see the light at the end of the tunnel."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And then v prime times minus a. So let's make that minus cancel out with this one so that becomes a plus. a sine of at dt. I'm starting to see the light at the end of the tunnel. So then, let's simplify this thing. And of course, we're going to have to evaluate this whole thing, right? From infinity."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'm starting to see the light at the end of the tunnel. So then, let's simplify this thing. And of course, we're going to have to evaluate this whole thing, right? From infinity. Actually, we're going to have to evaluate everything. Let's just focus on the indefinite integral for now. We're going to have to take this whole thing and evaluate."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "From infinity. Actually, we're going to have to evaluate everything. Let's just focus on the indefinite integral for now. We're going to have to take this whole thing and evaluate. Let's just say that y is the antiderivative and then evaluate it from infinity to 0, from 0 to infinity. So this is equal to y is equal to minus e to the minus st over s sine of at. Now let's distribute this."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We're going to have to take this whole thing and evaluate. Let's just say that y is the antiderivative and then evaluate it from infinity to 0, from 0 to infinity. So this is equal to y is equal to minus e to the minus st over s sine of at. Now let's distribute this. Minus a over s squared e to the minus st cosine of at. Right? OK, now I want to make sure I don't make a careless mistake."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now let's distribute this. Minus a over s squared e to the minus st cosine of at. Right? OK, now I want to make sure I don't make a careless mistake. Now let's multiply this times this and take all the constants out. So we have an a and an s. a over s. There's a minus sign. We have a plus a to the s. So we'll have a minus a squared over s squared times the integral from 0."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "OK, now I want to make sure I don't make a careless mistake. Now let's multiply this times this and take all the constants out. So we have an a and an s. a over s. There's a minus sign. We have a plus a to the s. So we'll have a minus a squared over s squared times the integral from 0. Well, I said I'm just worrying about the indefinite integral right now and we'll evaluate the boundaries later. e to the minus st sine of at dt. Now this is the part, and we've done this before, it's a little bit of a trick with integration by parts."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We have a plus a to the s. So we'll have a minus a squared over s squared times the integral from 0. Well, I said I'm just worrying about the indefinite integral right now and we'll evaluate the boundaries later. e to the minus st sine of at dt. Now this is the part, and we've done this before, it's a little bit of a trick with integration by parts. But this expression, notice, is the same thing as our original y. Right? This is our original y and we're assuming we're doing the indefinite integral and we'll evaluate the boundaries later."}, {"video_title": "L{sin(at)}) - transform of sin(at) Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now this is the part, and we've done this before, it's a little bit of a trick with integration by parts. But this expression, notice, is the same thing as our original y. Right? This is our original y and we're assuming we're doing the indefinite integral and we'll evaluate the boundaries later. Although we could have kept the boundaries the whole time, but it would have made it even hairier. So we can rewrite this integral as y. In fact, that was our definition."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "So the first thing I want to introduce is just kind of a quick way of doing something. And that is, if I had the Laplace transform of the second derivative of y. Well, we've proved several videos ago that if I wanted to take the Laplace transform of the first derivative of y, that is equal to s times the Laplace transform of y minus y of 0. And we used this property in the last couple of videos to actually figure out the Laplace transform of the second derivative. Because if you say this is y prime, this is the antiderivative of it, then you can just pattern match. You could say, well, the Laplace transform of y prime that's just equal to s times the Laplace transform of y prime minus y prime of 0. This is the derivative of this, just like this is the derivative of this."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And we used this property in the last couple of videos to actually figure out the Laplace transform of the second derivative. Because if you say this is y prime, this is the antiderivative of it, then you can just pattern match. You could say, well, the Laplace transform of y prime that's just equal to s times the Laplace transform of y prime minus y prime of 0. This is the derivative of this, just like this is the derivative of this. I'll draw a line here just so you don't get confused. So Laplace transform of y prime prime is this thing. And now we can use this, which we proved several videos ago, to resubstitute it and get in terms of the Laplace transform of y."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "This is the derivative of this, just like this is the derivative of this. I'll draw a line here just so you don't get confused. So Laplace transform of y prime prime is this thing. And now we can use this, which we proved several videos ago, to resubstitute it and get in terms of the Laplace transform of y. So we can expand this part. The Laplace transform of the derivative of y, that's just equal to s times the Laplace transform of y minus y of 0. And then we have the outside."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And now we can use this, which we proved several videos ago, to resubstitute it and get in terms of the Laplace transform of y. So we can expand this part. The Laplace transform of the derivative of y, that's just equal to s times the Laplace transform of y minus y of 0. And then we have the outside. We have s minus y prime of 0. And then when you expand it all out, and we've done this before, you get s squared times the Laplace transform of y minus s times y of 0 minus y prime of 0. Now there's something interesting to note here."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And then we have the outside. We have s minus y prime of 0. And then when you expand it all out, and we've done this before, you get s squared times the Laplace transform of y minus s times y of 0 minus y prime of 0. Now there's something interesting to note here. And if you learn this, it'll make it a lot faster. You won't have to go through all this and risk making careless mistakes when you have scarce time and paper on your tests. Just notice that when you take the Laplace transform of the second derivative, what do we end up?"}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Now there's something interesting to note here. And if you learn this, it'll make it a lot faster. You won't have to go through all this and risk making careless mistakes when you have scarce time and paper on your tests. Just notice that when you take the Laplace transform of the second derivative, what do we end up? We end up with s squared. This was the second derivative, so we end up with s squared times the Laplace transform of y minus s times y of 0 minus 1 times y prime of 0. So every term, we started with s squared."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Just notice that when you take the Laplace transform of the second derivative, what do we end up? We end up with s squared. This was the second derivative, so we end up with s squared times the Laplace transform of y minus s times y of 0 minus 1 times y prime of 0. So every term, we started with s squared. And then every term, we lowered the degree of s1. And then everything except the first term is a negative sign. And then we started with the Laplace transform of y."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "So every term, we started with s squared. And then every term, we lowered the degree of s1. And then everything except the first term is a negative sign. And then we started with the Laplace transform of y. And then you can almost view the Laplace transform as a kind of integral. So we kind of take the derivative. So then you get y."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And then we started with the Laplace transform of y. And then you can almost view the Laplace transform as a kind of integral. So we kind of take the derivative. So then you get y. And then you take the derivative again. You get y prime. And of course, every other term is negative."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "So then you get y. And then you take the derivative again. You get y prime. And of course, every other term is negative. And these aren't the actual functions. These are those functions evaluated at 0. But that's a good way to help you hopefully remember how to do these."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And of course, every other term is negative. And these aren't the actual functions. These are those functions evaluated at 0. But that's a good way to help you hopefully remember how to do these. And once you get the hang of it, you can take the Laplace transform of any arbitrary function very, very quickly. Or any arbitrary derivative. So let's say we wanted to take the Laplace transform of, I don't know, this should hit the point home, the fourth derivative of y."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "But that's a good way to help you hopefully remember how to do these. And once you get the hang of it, you can take the Laplace transform of any arbitrary function very, very quickly. Or any arbitrary derivative. So let's say we wanted to take the Laplace transform of, I don't know, this should hit the point home, the fourth derivative of y. That 4 in parentheses means the fourth derivative. I could have drawn 4 prime marks, but either way. So what is this equal to?"}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "So let's say we wanted to take the Laplace transform of, I don't know, this should hit the point home, the fourth derivative of y. That 4 in parentheses means the fourth derivative. I could have drawn 4 prime marks, but either way. So what is this equal to? If we use this technique and substitute it, we're bound to make some form of careless mistake or other. And it would take us forever. And it would waste a lot of paper."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "So what is this equal to? If we use this technique and substitute it, we're bound to make some form of careless mistake or other. And it would take us forever. And it would waste a lot of paper. But now we see the pattern. And so we can just say, well, the Laplace transform of this, in terms of the Laplace transform of y, that's what we want to get to, is going to be s to the fourth times the Laplace transform of y. Now every other term's going to have a minus in front of it."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And it would waste a lot of paper. But now we see the pattern. And so we can just say, well, the Laplace transform of this, in terms of the Laplace transform of y, that's what we want to get to, is going to be s to the fourth times the Laplace transform of y. Now every other term's going to have a minus in front of it. Minus, lower the degree on the s, minus s to the third. And then you could kind of say, let's take some form of derivative so that you get y of 0. It's not a real derivative."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Now every other term's going to have a minus in front of it. Minus, lower the degree on the s, minus s to the third. And then you could kind of say, let's take some form of derivative so that you get y of 0. It's not a real derivative. The Laplace transform really isn't the antiderivative of y of 0. But anyway, I think you get the idea. And then we lower the degree on s again."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "It's not a real derivative. The Laplace transform really isn't the antiderivative of y of 0. But anyway, I think you get the idea. And then we lower the degree on s again. Minus s squared. Take the derivative. And of course, these aren't functions."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And then we lower the degree on s again. Minus s squared. Take the derivative. And of course, these aren't functions. But we're evaluating the derivative of that function now at 0. So y prime of 0. Minus, now we lower the degree one more, minus s times, this is an s, times y prime prime of 0."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And of course, these aren't functions. But we're evaluating the derivative of that function now at 0. So y prime of 0. Minus, now we lower the degree one more, minus s times, this is an s, times y prime prime of 0. We have one more term. Lower the degree on the s one more time. Then you get s to the 0, which is just 1."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Minus, now we lower the degree one more, minus s times, this is an s, times y prime prime of 0. We have one more term. Lower the degree on the s one more time. Then you get s to the 0, which is just 1. So minus 1 is a coefficient. And then you have y, the third derivative of y. Let me scroll over a little bit."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Then you get s to the 0, which is just 1. So minus 1 is a coefficient. And then you have y, the third derivative of y. Let me scroll over a little bit. The third derivative of y evaluated at 0. So I think you see the pattern now. And this is a much faster way of evaluating the Laplace transform of an arbitrary derivative of y, as opposed to keep going through that pattern over and over again."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Let me scroll over a little bit. The third derivative of y evaluated at 0. So I think you see the pattern now. And this is a much faster way of evaluating the Laplace transform of an arbitrary derivative of y, as opposed to keep going through that pattern over and over again. Another thing I want to introduce you to is just a notational savings. And it's just something that you'll see. So you might as well get used to it."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And this is a much faster way of evaluating the Laplace transform of an arbitrary derivative of y, as opposed to keep going through that pattern over and over again. Another thing I want to introduce you to is just a notational savings. And it's just something that you'll see. So you might as well get used to it. And it actually saves time over keep writing this curly l in this bracket. If I, the Laplace transform of y of t, I can write as, and people tend to write it as, well, it's going to be a function of s. And what they use is they use a capital Y to denote the function of s. And that, well, it makes sense. Because normally, when we're doing antiderivatives, you just take, you know, when we take, we often, when you learn the fundamental theorem of calculus, learn that the integral of f with respect to dx, you know, from 0 to x, is equal to capital F of x."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "So you might as well get used to it. And it actually saves time over keep writing this curly l in this bracket. If I, the Laplace transform of y of t, I can write as, and people tend to write it as, well, it's going to be a function of s. And what they use is they use a capital Y to denote the function of s. And that, well, it makes sense. Because normally, when we're doing antiderivatives, you just take, you know, when we take, we often, when you learn the fundamental theorem of calculus, learn that the integral of f with respect to dx, you know, from 0 to x, is equal to capital F of x. So it's kind of borrowing that notation, because this function of s is kind of an integral of y of t. The Laplace transform, to some degree, is a kind of, it's like a special type of integral, where you have a little exponential function in there to mess around with things a little bit. Anyway, I just want you to get used to this notation. When you see capital Y of s, that's the same thing as the Laplace transform of y of t. And you might also see it this way."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Because normally, when we're doing antiderivatives, you just take, you know, when we take, we often, when you learn the fundamental theorem of calculus, learn that the integral of f with respect to dx, you know, from 0 to x, is equal to capital F of x. So it's kind of borrowing that notation, because this function of s is kind of an integral of y of t. The Laplace transform, to some degree, is a kind of, it's like a special type of integral, where you have a little exponential function in there to mess around with things a little bit. Anyway, I just want you to get used to this notation. When you see capital Y of s, that's the same thing as the Laplace transform of y of t. And you might also see it this way. The Laplace transform of f of t is equal to capital F of s. And the clue that tells you that this isn't just a normal antiderivative is the fact that they're using that s as the independent variable. Because in general, s represents the frequency domain, and if people were to use s with just a general antiderivative, people would get confused, et cetera, et cetera. Anyway, I'm trying to think whether I have time to teach you more fascinating concepts of Laplace transform."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "When you see capital Y of s, that's the same thing as the Laplace transform of y of t. And you might also see it this way. The Laplace transform of f of t is equal to capital F of s. And the clue that tells you that this isn't just a normal antiderivative is the fact that they're using that s as the independent variable. Because in general, s represents the frequency domain, and if people were to use s with just a general antiderivative, people would get confused, et cetera, et cetera. Anyway, I'm trying to think whether I have time to teach you more fascinating concepts of Laplace transform. Well, sure, I think we do. So my next question for you, and now we'll teach you a couple more properties, and this will be helpful in taking Laplace transforms. What is the Laplace transform of e to the at times f of t?"}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Anyway, I'm trying to think whether I have time to teach you more fascinating concepts of Laplace transform. Well, sure, I think we do. So my next question for you, and now we'll teach you a couple more properties, and this will be helpful in taking Laplace transforms. What is the Laplace transform of e to the at times f of t? Fascinating. Well, let's just go back to our definition of the Laplace transform. It is the integral from 0 to infinity of e to the minus st times whatever we have between the curly brackets."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "What is the Laplace transform of e to the at times f of t? Fascinating. Well, let's just go back to our definition of the Laplace transform. It is the integral from 0 to infinity of e to the minus st times whatever we have between the curly brackets. So with the curly brackets, we have e to the at f of t dt. And now we can add these exponents, right? We have a similar base."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "It is the integral from 0 to infinity of e to the minus st times whatever we have between the curly brackets. So with the curly brackets, we have e to the at f of t dt. And now we can add these exponents, right? We have a similar base. So this is equal to what? This is equal to the integral from 0 to infinity. And let's see, I want to write it as, I could write it minus s plus a, but I'm going to write it as minus s minus at."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "We have a similar base. So this is equal to what? This is equal to the integral from 0 to infinity. And let's see, I want to write it as, I could write it minus s plus a, but I'm going to write it as minus s minus at. And you could expand this out, right? It becomes minus s plus a, which is exactly what we have here, times f of t dt. Now let me show you something."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And let's see, I want to write it as, I could write it minus s plus a, but I'm going to write it as minus s minus at. And you could expand this out, right? It becomes minus s plus a, which is exactly what we have here, times f of t dt. Now let me show you something. If I were to just take the Laplace transform of f of t, that is equal to some function of s. Whatever we essentially have right here for s, it becomes some function of that. So this is interesting. This is some function of s. Here all we did, to go from, well actually, let me rewrite this, so Laplace, which is equal to 0 to infinity e to the minus s t f of t dt."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Now let me show you something. If I were to just take the Laplace transform of f of t, that is equal to some function of s. Whatever we essentially have right here for s, it becomes some function of that. So this is interesting. This is some function of s. Here all we did, to go from, well actually, let me rewrite this, so Laplace, which is equal to 0 to infinity e to the minus s t f of t dt. The Laplace transform of just f of t is equal to this, which is some function of s, right? Well the Laplace transform of e to the at times f of t, it equals this. And what's the difference between this and this?"}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "This is some function of s. Here all we did, to go from, well actually, let me rewrite this, so Laplace, which is equal to 0 to infinity e to the minus s t f of t dt. The Laplace transform of just f of t is equal to this, which is some function of s, right? Well the Laplace transform of e to the at times f of t, it equals this. And what's the difference between this and this? What's the difference between the two? Well it's not much. Here, wherever I have an s, I have an s minus a here, right?"}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "And what's the difference between this and this? What's the difference between the two? Well it's not much. Here, wherever I have an s, I have an s minus a here, right? So if this is a function of s, what's this going to be? It's going to be that same function. Whatever the Laplace transform of f was, it's going to be that same function, but instead of s, it's going to be a function of s minus a."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Here, wherever I have an s, I have an s minus a here, right? So if this is a function of s, what's this going to be? It's going to be that same function. Whatever the Laplace transform of f was, it's going to be that same function, but instead of s, it's going to be a function of s minus a. And once again, how did I get that? Well I said the Laplace transform of f is a function of s, and it's equal to this. Well if I just replaced an s with an s minus a, I get this, which is a function of s minus a, which was the Laplace transform of e to the at times f of t. Maybe that's a little confusing."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Whatever the Laplace transform of f was, it's going to be that same function, but instead of s, it's going to be a function of s minus a. And once again, how did I get that? Well I said the Laplace transform of f is a function of s, and it's equal to this. Well if I just replaced an s with an s minus a, I get this, which is a function of s minus a, which was the Laplace transform of e to the at times f of t. Maybe that's a little confusing. Let me show you an example. So if I, let's just take the Laplace transform of cosine of 2t, we've shown is equal to, well I'll write the notation, it's equal to some function of s, and that function of s is s over s squared plus 4. We've shown that already."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "Well if I just replaced an s with an s minus a, I get this, which is a function of s minus a, which was the Laplace transform of e to the at times f of t. Maybe that's a little confusing. Let me show you an example. So if I, let's just take the Laplace transform of cosine of 2t, we've shown is equal to, well I'll write the notation, it's equal to some function of s, and that function of s is s over s squared plus 4. We've shown that already. And so the Laplace transform of e to the, I don't know, 3t times cosine of 2t is going to be equal to the same function, but instead of s, it's going to be a function of s minus a, so s minus 3, which is equal to s minus 3 over s minus 3 squared plus 4. Notice, when you just multiply something by this e to the 3t and then, or e to the at, you take the Laplace transform of it, you just, it's the same thing as the Laplace transform of this function, but everywhere where you had an s, you replace it with an s minus this a. Anyway, I hope I didn't confuse you too much with that last part."}, {"video_title": "Shifting transform by multiplying function by exponential Differential Equations Khan Academy.mp3", "Sentence": "We've shown that already. And so the Laplace transform of e to the, I don't know, 3t times cosine of 2t is going to be equal to the same function, but instead of s, it's going to be a function of s minus a, so s minus 3, which is equal to s minus 3 over s minus 3 squared plus 4. Notice, when you just multiply something by this e to the 3t and then, or e to the at, you take the Laplace transform of it, you just, it's the same thing as the Laplace transform of this function, but everywhere where you had an s, you replace it with an s minus this a. Anyway, I hope I didn't confuse you too much with that last part. I think my power adapter actually just went on. I hope the video keeps recording. I'll see you in the next one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But in order to just kind of make sure we don't get confused, I think it might be useful to review a little bit of everything that we've learned so far. So in the last video, we saw that the Laplace transform, well, let me just write something a little, the Laplace transform of f of t, let me just get some notation down, we can write that as big capital F of s, and I've told you that before. And so given that, in the last video, I showed you that if we had to deal with the unit step function, so if I said the Laplace transform of the unit step function that becomes one at some value c, times some shifted function, f of t minus c, in the last video, we saw that this is just equal to e to the minus c s times the Laplace transform of just this function right there so times the f of s, times f of s. And it's really important not to get this confused with another Laplace transform property or rule or whatever you wanna call it that we figured out, I think it was in one of the videos that I made last year, but if you're just following these in order, I think it's three or four videos ago. And that one told us that the Laplace transform, the Laplace transform of e to the a t times f of t, that this is equal to, instead of, and I wanna make this distinction very clear, here we shifted the f of t, and we got just a kind of a regular f of s. In this situation, when we multiply it times the e to the positive a t, we end up shifting, we end up shifting the actual transform. So this becomes f of s minus a. And these two rules or properties or whatever you wanna call them, they're very easy to confuse with each other. So we're gonna do a couple of examples that we're gonna have to figure out which one of these two apply."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And that one told us that the Laplace transform, the Laplace transform of e to the a t times f of t, that this is equal to, instead of, and I wanna make this distinction very clear, here we shifted the f of t, and we got just a kind of a regular f of s. In this situation, when we multiply it times the e to the positive a t, we end up shifting, we end up shifting the actual transform. So this becomes f of s minus a. And these two rules or properties or whatever you wanna call them, they're very easy to confuse with each other. So we're gonna do a couple of examples that we're gonna have to figure out which one of these two apply. And let's write all the other stuff that we learned as well. The very first thing we learned was that the Laplace transform of one was equal to one over s. We know that's a pretty straightforward one, easy to prove to yourself. And more generally, we learned that the Laplace transform of t to the n, where n is a positive integer, n is a positive integer, it equaled n factorial over n factorial over s to the n plus one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So we're gonna do a couple of examples that we're gonna have to figure out which one of these two apply. And let's write all the other stuff that we learned as well. The very first thing we learned was that the Laplace transform of one was equal to one over s. We know that's a pretty straightforward one, easy to prove to yourself. And more generally, we learned that the Laplace transform of t to the n, where n is a positive integer, n is a positive integer, it equaled n factorial over n factorial over s to the n plus one. And then we had our trig functions that we've gone over. Let me do this in a different color. Let me do it, I'll do it right here."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And more generally, we learned that the Laplace transform of t to the n, where n is a positive integer, n is a positive integer, it equaled n factorial over n factorial over s to the n plus one. And then we had our trig functions that we've gone over. Let me do this in a different color. Let me do it, I'll do it right here. The Laplace transform of sine of a t is equal to a over s squared plus a squared. And the Laplace transform of the cosine of a t is equal to s over s squared plus a squared. And you'll be amazed by how far we can go with just what I've written here."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let me do it, I'll do it right here. The Laplace transform of sine of a t is equal to a over s squared plus a squared. And the Laplace transform of the cosine of a t is equal to s over s squared plus a squared. And you'll be amazed by how far we can go with just what I've written here. In future videos, we're gonna broaden our toolkit even further. But just these right here, you can already do a whole set of Laplace transforms and inverse Laplace transforms. So let's try to do a few."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And you'll be amazed by how far we can go with just what I've written here. In future videos, we're gonna broaden our toolkit even further. But just these right here, you can already do a whole set of Laplace transforms and inverse Laplace transforms. So let's try to do a few. So let's say I were to give you the Laplace transform. And this is just the hard part. I think you know how to solve a differential equation if you know how to take the Laplace transforms and go back and forth."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's try to do a few. So let's say I were to give you the Laplace transform. And this is just the hard part. I think you know how to solve a differential equation if you know how to take the Laplace transforms and go back and forth. The hard part is just recognizing or inverting your Laplace transform. So let's say we had the Laplace transform of some function, f of s. Let's say it's three factorial over s minus two to the fourth. Now, your pattern matching or your pattern recognition part of your brain should immediately say, look, I have a Laplace transform of something that has a factorial in it and it's over an exponent."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I think you know how to solve a differential equation if you know how to take the Laplace transforms and go back and forth. The hard part is just recognizing or inverting your Laplace transform. So let's say we had the Laplace transform of some function, f of s. Let's say it's three factorial over s minus two to the fourth. Now, your pattern matching or your pattern recognition part of your brain should immediately say, look, I have a Laplace transform of something that has a factorial in it and it's over an exponent. This must be something related to this thing right here. If I just had the Laplace transform, let me write that down, the Laplace transform of, the Laplace transform, you see a three factorial and a fourth power, so it looks like n is equal to three. So if you write the Laplace transform of t to the three, this rule that we showed right here, this means that it would be equal to three factorial over s to the fourth."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, your pattern matching or your pattern recognition part of your brain should immediately say, look, I have a Laplace transform of something that has a factorial in it and it's over an exponent. This must be something related to this thing right here. If I just had the Laplace transform, let me write that down, the Laplace transform of, the Laplace transform, you see a three factorial and a fourth power, so it looks like n is equal to three. So if you write the Laplace transform of t to the three, this rule that we showed right here, this means that it would be equal to three factorial over s to the fourth. Now, this thing isn't exactly this thing. They're not quite the same thing. And just so, you know, I'm doing this to instruct you, but I find these when I'm actually doing them on an exam."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if you write the Laplace transform of t to the three, this rule that we showed right here, this means that it would be equal to three factorial over s to the fourth. Now, this thing isn't exactly this thing. They're not quite the same thing. And just so, you know, I'm doing this to instruct you, but I find these when I'm actually doing them on an exam. I remember when I did them when I first learned this, I would actually go through this step because I was, you know, you definitely don't wanna make a careless mistake and you definitely wanna kinda make sure you have a good handle on what you're doing. So you're like, okay, it's something related to this, but what's the difference between this expression right here and the expression that we're trying to take the inverse Laplace transform of? And this one here."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And just so, you know, I'm doing this to instruct you, but I find these when I'm actually doing them on an exam. I remember when I did them when I first learned this, I would actually go through this step because I was, you know, you definitely don't wanna make a careless mistake and you definitely wanna kinda make sure you have a good handle on what you're doing. So you're like, okay, it's something related to this, but what's the difference between this expression right here and the expression that we're trying to take the inverse Laplace transform of? And this one here. Well, we've shifted our s, right? If we call this expression right here f of s, we call this expression right here f of s, then what's this expression? This expression right here is f of s minus two, right?"}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And this one here. Well, we've shifted our s, right? If we call this expression right here f of s, we call this expression right here f of s, then what's this expression? This expression right here is f of s minus two, right? That expression is f of s minus two. So what are we dealing with here? So you see here you have a shifted f of s, right?"}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This expression right here is f of s minus two, right? That expression is f of s minus two. So what are we dealing with here? So you see here you have a shifted f of s, right? So in this case, a would be equal to two. So this is the Laplace transform of e to the at times our f of t. So this is the Laplace transform. So let me write this down."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So you see here you have a shifted f of s, right? So in this case, a would be equal to two. So this is the Laplace transform of e to the at times our f of t. So this is the Laplace transform. So let me write this down. This is the Laplace transform of e to the, and what's a? a is what we shifted by, right? It's what we shifted by, minus a, so you have a positive there."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let me write this down. This is the Laplace transform of e to the, and what's a? a is what we shifted by, right? It's what we shifted by, minus a, so you have a positive there. So e to the two t times the actual function. If this was just an f of s, what would f of t be? Well, we figured that out."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It's what we shifted by, minus a, so you have a positive there. So e to the two t times the actual function. If this was just an f of s, what would f of t be? Well, we figured that out. It was t to the three, t to the third power. So this, the Laplace transform of this is equal to that, or we could write that the inverse Laplace transform of three factorial over s minus two to the fourth minus two to the fourth is equal to e to the two t times t to the third. Now, if that seemed confusing to you, you can kind of go forward."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well, we figured that out. It was t to the three, t to the third power. So this, the Laplace transform of this is equal to that, or we could write that the inverse Laplace transform of three factorial over s minus two to the fourth minus two to the fourth is equal to e to the two t times t to the third. Now, if that seemed confusing to you, you can kind of go forward. You can kind of, let's apply this. Let's go the other direction, and maybe this will make it a little bit clearer for you. So let's go from this direction."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, if that seemed confusing to you, you can kind of go forward. You can kind of, let's apply this. Let's go the other direction, and maybe this will make it a little bit clearer for you. So let's go from this direction. If I had to take the Laplace transform of this thing, I'd say, okay, well, the Laplace transform of t to the third is easy. The Laplace transform of t to the third, I'll do it here, I think, yeah, the tool isn't working right there properly. Let me scroll up a little bit."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's go from this direction. If I had to take the Laplace transform of this thing, I'd say, okay, well, the Laplace transform of t to the third is easy. The Laplace transform of t to the third, I'll do it here, I think, yeah, the tool isn't working right there properly. Let me scroll up a little bit. So I could write it right here. So if I wanted to figure out the Laplace transform of e to the two t times t to the third, I'll say, well, you know, this e to the two t, I remember that it shifts something. So if I know that the Laplace transform of t to the third, this is an easy one, it's equal to three factorial over s to the fourth, that's three plus one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let me scroll up a little bit. So I could write it right here. So if I wanted to figure out the Laplace transform of e to the two t times t to the third, I'll say, well, you know, this e to the two t, I remember that it shifts something. So if I know that the Laplace transform of t to the third, this is an easy one, it's equal to three factorial over s to the fourth, that's three plus one. Then the Laplace transform of e to the two t times t to the third is going to be this shifted. This is equal to f of s. Then this is going to be f of s minus two. F of s minus two."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if I know that the Laplace transform of t to the third, this is an easy one, it's equal to three factorial over s to the fourth, that's three plus one. Then the Laplace transform of e to the two t times t to the third is going to be this shifted. This is equal to f of s. Then this is going to be f of s minus two. F of s minus two. So what's f of s minus two? It's going to be equal to three factorial over s minus two to the fourth. I think you're already getting an appreciation that the hardest thing about these Laplace transform problems are really kind of all of these shifts and kind of recognizing the patterns and recognizing what's your a and what's your c and being very careful about it so you don't make a careless mistake."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "F of s minus two. So what's f of s minus two? It's going to be equal to three factorial over s minus two to the fourth. I think you're already getting an appreciation that the hardest thing about these Laplace transform problems are really kind of all of these shifts and kind of recognizing the patterns and recognizing what's your a and what's your c and being very careful about it so you don't make a careless mistake. I think doing a lot of examples probably helps a lot. So let's do a couple of more to kind of make sure things really get hammered home in your brain. So let's try this one right here."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I think you're already getting an appreciation that the hardest thing about these Laplace transform problems are really kind of all of these shifts and kind of recognizing the patterns and recognizing what's your a and what's your c and being very careful about it so you don't make a careless mistake. I think doing a lot of examples probably helps a lot. So let's do a couple of more to kind of make sure things really get hammered home in your brain. So let's try this one right here. This looks like a little bit more complicated. They give us that the Laplace transform of some function is equal to two times s minus one times e to the minus two s, all of that over s squared minus two s plus two. Now this looks very daunting."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So let's try this one right here. This looks like a little bit more complicated. They give us that the Laplace transform of some function is equal to two times s minus one times e to the minus two s, all of that over s squared minus two s plus two. Now this looks very daunting. How do you do this? I have an e here, I have something shifted here, I have this polynomial in the denominator here. What can I do with this?"}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now this looks very daunting. How do you do this? I have an e here, I have something shifted here, I have this polynomial in the denominator here. What can I do with this? So the first thing when I look at these polynomials in the denominator, I say can I factor it somehow? And can I factor it out fairly simply? And actually in the exams that you'll find in differential equation class, they'll never give you something that's factorable into these weird numbers."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "What can I do with this? So the first thing when I look at these polynomials in the denominator, I say can I factor it somehow? And can I factor it out fairly simply? And actually in the exams that you'll find in differential equation class, they'll never give you something that's factorable into these weird numbers. It tends to be integers. When you say okay, what two numbers, they have to be positive. When you give their product, you get two."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And actually in the exams that you'll find in differential equation class, they'll never give you something that's factorable into these weird numbers. It tends to be integers. When you say okay, what two numbers, they have to be positive. When you give their product, you get two. And then when you add them, you get negative two. Well you can't have, or they could both be negative. But there's no two easy numbers, not one and two, none of those work."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "When you give their product, you get two. And then when you add them, you get negative two. Well you can't have, or they could both be negative. But there's no two easy numbers, not one and two, none of those work. So if you can't factor this out right, the next idea is maybe you can complete the square and maybe this will match one of the cosine or the sine formulas. So how can we complete the square in this denominator? Well this can be rewritten as, let me rewrite this as s squared minus two s, and I'm gonna put a plus two out here."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "But there's no two easy numbers, not one and two, none of those work. So if you can't factor this out right, the next idea is maybe you can complete the square and maybe this will match one of the cosine or the sine formulas. So how can we complete the square in this denominator? Well this can be rewritten as, let me rewrite this as s squared minus two s, and I'm gonna put a plus two out here. And this will hopefully, and you can watch my, I have a bunch of videos on completing the square if all of this looks foreign to you. And to complete the square, we just wanna turn this into a perfect square. So to turn this into a perfect square, so something when I add to itself twice becomes minus two."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Well this can be rewritten as, let me rewrite this as s squared minus two s, and I'm gonna put a plus two out here. And this will hopefully, and you can watch my, I have a bunch of videos on completing the square if all of this looks foreign to you. And to complete the square, we just wanna turn this into a perfect square. So to turn this into a perfect square, so something when I add to itself twice becomes minus two. And so that when I square it, when I add it to itself twice, it becomes minus two, it's minus one. And when I square, it'll become plus one. Plus one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So to turn this into a perfect square, so something when I add to itself twice becomes minus two. And so that when I square it, when I add it to itself twice, it becomes minus two, it's minus one. And when I square, it'll become plus one. Plus one. I can't just add plus one arbitrarily to some expression, I have to make it neutral, so let me subtract one. I haven't changed this. I added one and I subtracted one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Plus one. I can't just add plus one arbitrarily to some expression, I have to make it neutral, so let me subtract one. I haven't changed this. I added one and I subtracted one. A little bit of a primer on completing the square. But by doing this, I now can call this expression right here I can now say that this thing is s minus one squared, and then this stuff out here, this out here is two minus one, this is just plus one. So I can rewrite my entire expression now as two times s minus one times e to the minus two s. Make sure I'm not clipping off at the top."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I added one and I subtracted one. A little bit of a primer on completing the square. But by doing this, I now can call this expression right here I can now say that this thing is s minus one squared, and then this stuff out here, this out here is two minus one, this is just plus one. So I can rewrite my entire expression now as two times s minus one times e to the minus two s. Make sure I'm not clipping off at the top. E to the minus two s, all of that over s minus one squared plus one. So a couple of interesting things are starting to seem to be happening here. So if I just had, let's just do a couple of test Laplace transforms."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So I can rewrite my entire expression now as two times s minus one times e to the minus two s. Make sure I'm not clipping off at the top. E to the minus two s, all of that over s minus one squared plus one. So a couple of interesting things are starting to seem to be happening here. So if I just had, let's just do a couple of test Laplace transforms. If I did the Laplace transform of cosine of t, we know that this is equal to s over s squared plus one. Which this kind of looks like. If this was an s and this was an s squared plus one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So if I just had, let's just do a couple of test Laplace transforms. If I did the Laplace transform of cosine of t, we know that this is equal to s over s squared plus one. Which this kind of looks like. If this was an s and this was an s squared plus one. If this was f of s, then what is this? Well let's ignore this guy right here for a little bit. So what is, we know, actually from the last video, we saw well what if we took the Laplace transform, the Laplace transform of e to the, I'll call it one t, but let's say e to the, yeah let's write it, e to the one t times cosine of t. Well then this will just shift this Laplace transform by one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "If this was an s and this was an s squared plus one. If this was f of s, then what is this? Well let's ignore this guy right here for a little bit. So what is, we know, actually from the last video, we saw well what if we took the Laplace transform, the Laplace transform of e to the, I'll call it one t, but let's say e to the, yeah let's write it, e to the one t times cosine of t. Well then this will just shift this Laplace transform by one. It'll shift it by one to the right. So this will, wherever you see an s, you would put an s minus a one. So this will be equal to s minus one over s minus one squared plus one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So what is, we know, actually from the last video, we saw well what if we took the Laplace transform, the Laplace transform of e to the, I'll call it one t, but let's say e to the, yeah let's write it, e to the one t times cosine of t. Well then this will just shift this Laplace transform by one. It'll shift it by one to the right. So this will, wherever you see an s, you would put an s minus a one. So this will be equal to s minus one over s minus one squared plus one. We're getting close, we're getting close. We now figured out this part right here. Now in the previous video, I think it was two videos ago, or maybe it was the last video, I forget."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "So this will be equal to s minus one over s minus one squared plus one. We're getting close, we're getting close. We now figured out this part right here. Now in the previous video, I think it was two videos ago, or maybe it was the last video, I forget. Memory fails me. I showed you that if you have the Laplace transform of the unit step function of t times some f of t shifted by some value of c, then that this is equal to, this is equal to e to the minus cs times f of s. Right, okay, so if we know, and this can get very confusing. This can get very confusing, so I wanna be very careful here."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now in the previous video, I think it was two videos ago, or maybe it was the last video, I forget. Memory fails me. I showed you that if you have the Laplace transform of the unit step function of t times some f of t shifted by some value of c, then that this is equal to, this is equal to e to the minus cs times f of s. Right, okay, so if we know, and this can get very confusing. This can get very confusing, so I wanna be very careful here. Let's ignore all of this. I called this f of s before, but now I'm gonna backtrack a little bit. And let's just ignore this because I'm gonna redefine our f of s. Let's just ignore that for a second."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This can get very confusing, so I wanna be very careful here. Let's ignore all of this. I called this f of s before, but now I'm gonna backtrack a little bit. And let's just ignore this because I'm gonna redefine our f of s. Let's just ignore that for a second. Let's define our new f of t to be this. Let's say that that is f of t. Let's say f of t is equal to e to the t cosine of t. Then if you take the Laplace transform of that, that means that f of s is equal to s minus one over s minus one squared plus one. Nothing fancy there."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And let's just ignore this because I'm gonna redefine our f of s. Let's just ignore that for a second. Let's define our new f of t to be this. Let's say that that is f of t. Let's say f of t is equal to e to the t cosine of t. Then if you take the Laplace transform of that, that means that f of s is equal to s minus one over s minus one squared plus one. Nothing fancy there. I just defined our f of t as this, and then our f of s is that, right? Now, we have a situation here. This expression, let's ignore the two here."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Nothing fancy there. I just defined our f of t as this, and then our f of s is that, right? Now, we have a situation here. This expression, let's ignore the two here. The two is just kind of a scaling factor. This expression right here, we can rewrite as that expression is equal to, if this is our f of s, this expression right here is equal to two times our f of s times e to the minus two s. Or let me just write it. Let me switch the order just so we make it look right."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "This expression, let's ignore the two here. The two is just kind of a scaling factor. This expression right here, we can rewrite as that expression is equal to, if this is our f of s, this expression right here is equal to two times our f of s times e to the minus two s. Or let me just write it. Let me switch the order just so we make it look right. Two times e to the minus two s times f of s. Well, that looks just like this if our two was equal to our c, right? If our two was equal to our c. So what does that tell us? That tells us that the inverse Laplace transform, if we take the inverse Laplace transform, and let's ignore the two."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let me switch the order just so we make it look right. Two times e to the minus two s times f of s. Well, that looks just like this if our two was equal to our c, right? If our two was equal to our c. So what does that tell us? That tells us that the inverse Laplace transform, if we take the inverse Laplace transform, and let's ignore the two. Well, we could just write, let's do the inverse Laplace transform of the whole thing. The inverse Laplace transform of this thing is going to be equal to, we can just write the two there as a scaling factor. Two there times this thing, times the unit step function."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "That tells us that the inverse Laplace transform, if we take the inverse Laplace transform, and let's ignore the two. Well, we could just write, let's do the inverse Laplace transform of the whole thing. The inverse Laplace transform of this thing is going to be equal to, we can just write the two there as a scaling factor. Two there times this thing, times the unit step function. The unit step function, what's our c? If you just, you can just pattern match. You have a two here, you have a c, a minus c, a minus two."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Two there times this thing, times the unit step function. The unit step function, what's our c? If you just, you can just pattern match. You have a two here, you have a c, a minus c, a minus two. So c is two. The unit step function is zero until it gets to two times t, or of t, so, you know, and then it becomes one after t is equal to two, times our function shifted by two. By two."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "You have a two here, you have a c, a minus c, a minus two. So c is two. The unit step function is zero until it gets to two times t, or of t, so, you know, and then it becomes one after t is equal to two, times our function shifted by two. By two. Shifted by two. So this is our inverse Laplace transform. Now, what was our function?"}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "By two. Shifted by two. So this is our inverse Laplace transform. Now, what was our function? Our function was this thing right here. So if our inverse Laplace transform of that thing that I had written is this thing, and f of t, f of t is equal to e to the t cosine of t, then our inverse, let me write all of this down. Let me write kind of our big result."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Now, what was our function? Our function was this thing right here. So if our inverse Laplace transform of that thing that I had written is this thing, and f of t, f of t is equal to e to the t cosine of t, then our inverse, let me write all of this down. Let me write kind of our big result. We find we were established that the inverse Laplace transform of that big thing that I had written before, two times s minus one, times e to the minus two, sorry, e to the minus two s, wouldn't have a t there if we, over s squared minus two s plus two is equal to this thing where f of t is this, or we could just rewrite this as two, two times the unit step function starting at two, or that's when it becomes non-zero, t, times f of t minus two. F of t minus two is this with t being replaced by t minus two. I'll do it in another color just to ease the monotony."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "Let me write kind of our big result. We find we were established that the inverse Laplace transform of that big thing that I had written before, two times s minus one, times e to the minus two, sorry, e to the minus two s, wouldn't have a t there if we, over s squared minus two s plus two is equal to this thing where f of t is this, or we could just rewrite this as two, two times the unit step function starting at two, or that's when it becomes non-zero, t, times f of t minus two. F of t minus two is this with t being replaced by t minus two. I'll do it in another color just to ease the monotony. So it'd be e to the t minus two cosine of t minus two. Now you might be thinking, Sal, he must have taken all these baby steps with this problem because he's trying to explain it to me, but I'm taking baby steps with this problem so that I myself don't get confused, and I think it's essential that you do take these baby steps. And let's just think about what baby steps we took, and I really want to review this."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "I'll do it in another color just to ease the monotony. So it'd be e to the t minus two cosine of t minus two. Now you might be thinking, Sal, he must have taken all these baby steps with this problem because he's trying to explain it to me, but I'm taking baby steps with this problem so that I myself don't get confused, and I think it's essential that you do take these baby steps. And let's just think about what baby steps we took, and I really want to review this. This is actually a surprisingly good problem. I didn't realize it when I first decided to do it. We saw this thing."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "And let's just think about what baby steps we took, and I really want to review this. This is actually a surprisingly good problem. I didn't realize it when I first decided to do it. We saw this thing. We wanted to get this denominator into some form that is vaguely, vaguely useful to us, so I completed the square there, and then we rewrote our Laplace transform, our f of s like this, and then we used a little pattern recognition. We said, look, if I take the Laplace transform of cosine of t, I'd get s over s squared plus one, but this isn't s over s squared plus one. It's s minus one over s minus one squared plus one."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "We saw this thing. We wanted to get this denominator into some form that is vaguely, vaguely useful to us, so I completed the square there, and then we rewrote our Laplace transform, our f of s like this, and then we used a little pattern recognition. We said, look, if I take the Laplace transform of cosine of t, I'd get s over s squared plus one, but this isn't s over s squared plus one. It's s minus one over s minus one squared plus one. So we said, oh, well, that means that we're multiplying our original time domain function. We're multiplying our f of t times e to the one t, e to the one t, and that's what we got there, right? So the Laplace transform of e to the t cosine of t became s minus one over s minus one squared plus one, and then we had this e to the minus two s this entire time, and that's where we said, hey, if we have e to the minus two s in our Laplace transform, then when you take the inverse Laplace transform, it must be the step function times the shifted version of that function, and that's why I was very careful, and you had this two hanging out the whole time, and I could have used that any time, but just simple constants just scale, so you can just, you know, two times a function is equal to two times Laplace transform of that function and vice versa, so the twos are very easy to deal with, so I kind of ignored that most of the time, but that's why I was very careful."}, {"video_title": "Inverse Laplace examples Laplace transform Differential Equations Khan Academy.mp3", "Sentence": "It's s minus one over s minus one squared plus one. So we said, oh, well, that means that we're multiplying our original time domain function. We're multiplying our f of t times e to the one t, e to the one t, and that's what we got there, right? So the Laplace transform of e to the t cosine of t became s minus one over s minus one squared plus one, and then we had this e to the minus two s this entire time, and that's where we said, hey, if we have e to the minus two s in our Laplace transform, then when you take the inverse Laplace transform, it must be the step function times the shifted version of that function, and that's why I was very careful, and you had this two hanging out the whole time, and I could have used that any time, but just simple constants just scale, so you can just, you know, two times a function is equal to two times Laplace transform of that function and vice versa, so the twos are very easy to deal with, so I kind of ignored that most of the time, but that's why I was very careful. I redefined f of t to be this, f of s to be this, and said, gee, if I have, if f of s is this, and if I'm multiplying it times e to the minus two s, then what I'm essentially doing, I'm fitting this pattern right here, and so the answer to my problem is going to be the unit step function. I just throw the two out there. The two times the unit step function times my f of t shifted by c, and now we establish this was our f of t, so we just shifted it by c, we shifted it by two, and we got our final answer, so this is about as hard up to this point as you'll see an inverse Laplace transform problem, so hopefully you found that pretty interesting."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And if the roots of this characteristic equation are real, let's say we have two real roots, let me write that down, so the real scenario where the two solutions are going to be r1 and r2 where these are real numbers, then the general solution of this differential equation, and watch the previous videos if you don't remember this or if you don't feel like it's suitably proven to you, the general solution is y is equal to some constant times e to the first root x plus some other constant times e to the second root times x. And we did that in the last several videos and we even did some examples. Now my question to you is, what if the characteristic equation does not have real roots? What if they are complex? And just a little bit of review, what do I mean by that? Well if I wanted to figure out the roots of this and I didn't, you know, if I was lazy and I just wanted to do it without having to think, can I factor it, I would just immediately use the quadratic equation because that always works. And I would say, well the roots of my characteristic equation are negative b plus or minus the square root of b squared minus 4ac, all of that over 2a."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "What if they are complex? And just a little bit of review, what do I mean by that? Well if I wanted to figure out the roots of this and I didn't, you know, if I was lazy and I just wanted to do it without having to think, can I factor it, I would just immediately use the quadratic equation because that always works. And I would say, well the roots of my characteristic equation are negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. So what do I mean by non-real roots? Well if this expression right here, if this b squared minus 4ac, if that's a negative number, then I'm going to have to take the square root of a negative number so it will actually be a imaginary number. And so this whole term will actually become complex."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And I would say, well the roots of my characteristic equation are negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. So what do I mean by non-real roots? Well if this expression right here, if this b squared minus 4ac, if that's a negative number, then I'm going to have to take the square root of a negative number so it will actually be a imaginary number. And so this whole term will actually become complex. We'll have a real part and an imaginary part. And actually the two roots are going to be conjugates of each other, right? We can rewrite this in the real and imaginary parts."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And so this whole term will actually become complex. We'll have a real part and an imaginary part. And actually the two roots are going to be conjugates of each other, right? We can rewrite this in the real and imaginary parts. We could rewrite this as the roots are going to be equal to minus b over 2a plus or minus the square root of b squared minus 4ac over 2a. And if b squared minus 4ac is less than 0, this is going to be an imaginary number. So in that case, let's just think about what the roots look like generally and then we'll actually do some problems."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "We can rewrite this in the real and imaginary parts. We could rewrite this as the roots are going to be equal to minus b over 2a plus or minus the square root of b squared minus 4ac over 2a. And if b squared minus 4ac is less than 0, this is going to be an imaginary number. So in that case, let's just think about what the roots look like generally and then we'll actually do some problems. So let me go back up here. So then the roots aren't going to be two real numbers like that. The roots, we can write them as two complex numbers that are conjugates of each other."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So in that case, let's just think about what the roots look like generally and then we'll actually do some problems. So let me go back up here. So then the roots aren't going to be two real numbers like that. The roots, we can write them as two complex numbers that are conjugates of each other. And I think light blue is a suitable color for that. So in that situation, let me write this as complex roots. This is a complex roots scenario."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "The roots, we can write them as two complex numbers that are conjugates of each other. And I think light blue is a suitable color for that. So in that situation, let me write this as complex roots. This is a complex roots scenario. Then the roots of the characteristic equation are going to be some number. Let's call it lambda. Let's call it mu."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "This is a complex roots scenario. Then the roots of the characteristic equation are going to be some number. Let's call it lambda. Let's call it mu. I think that's the convention that people use. Actually, let me see what they tend to use. It really doesn't matter."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "Let's call it mu. I think that's the convention that people use. Actually, let me see what they tend to use. It really doesn't matter. Let's say it's lambda. So this number, some constant called lambda. And then plus or minus some imaginary number."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "It really doesn't matter. Let's say it's lambda. So this number, some constant called lambda. And then plus or minus some imaginary number. And so it's going to be some constant mu. That's just some constant. I'm not trying to be fancy, but this is, I think, the convention used in most differential equations books."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And then plus or minus some imaginary number. And so it's going to be some constant mu. That's just some constant. I'm not trying to be fancy, but this is, I think, the convention used in most differential equations books. So it's mu times i. So these are the two roots. And these are two roots, right?"}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "I'm not trying to be fancy, but this is, I think, the convention used in most differential equations books. So it's mu times i. So these are the two roots. And these are two roots, right? Because we have lambda plus mu i and lambda minus mu i. So these would be the two roots if b squared minus 4ac is less than 0. So let's see how that translates."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And these are two roots, right? Because we have lambda plus mu i and lambda minus mu i. So these would be the two roots if b squared minus 4ac is less than 0. So let's see how that translates. Let's see what happens when we take these two roots and we put them into our general solution. So just like we'd learned before, the general solution is going to be, I'll stay in the light blue, the general solution is going to be y is equal to c1 times e to the first root. Let's make that the plus version."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So let's see how that translates. Let's see what happens when we take these two roots and we put them into our general solution. So just like we'd learned before, the general solution is going to be, I'll stay in the light blue, the general solution is going to be y is equal to c1 times e to the first root. Let's make that the plus version. So lambda plus mu i. All of that times x plus c2 times e to the second root. So that's going to be lambda minus mu i times x."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "Let's make that the plus version. So lambda plus mu i. All of that times x plus c2 times e to the second root. So that's going to be lambda minus mu i times x. Let's see if we can do some simplification here, because that i there really kind of makes things kind of crazy. So let's see if we can do anything to either get rid of it or simplify it, et cetera. So let's multiply the x out."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So that's going to be lambda minus mu i times x. Let's see if we can do some simplification here, because that i there really kind of makes things kind of crazy. So let's see if we can do anything to either get rid of it or simplify it, et cetera. So let's multiply the x out. Just do some algebraic manipulation. I'm trying to use as much space as possible. So we get y is equal to c1 e to the what?"}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So let's multiply the x out. Just do some algebraic manipulation. I'm trying to use as much space as possible. So we get y is equal to c1 e to the what? Lambda x, just distributing that x, plus mu xi plus c2 times e to the lambda x minus mu xi. Just distributed the x's in both of the terms. And let's see what we can do."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So we get y is equal to c1 e to the what? Lambda x, just distributing that x, plus mu xi plus c2 times e to the lambda x minus mu xi. Just distributed the x's in both of the terms. And let's see what we can do. Well, when you add exponents, this is the exact same thing as y is equal to c1 e to the lambda x times e to the mu xi. If you have the same base and you're multiplying, you could just add exponents. So this is the same thing as that."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And let's see what we can do. Well, when you add exponents, this is the exact same thing as y is equal to c1 e to the lambda x times e to the mu xi. If you have the same base and you're multiplying, you could just add exponents. So this is the same thing as that. Plus c2 times e to the lambda x times e to the minus mu xi. And let's see, we have an e to the lambda x in both of these terms, so we can factor it out. So we get y is equal to, let me draw a line here."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So this is the same thing as that. Plus c2 times e to the lambda x times e to the minus mu xi. And let's see, we have an e to the lambda x in both of these terms, so we can factor it out. So we get y is equal to, let me draw a line here. I don't want you to get confused with all this quadratic equation stuff. y is equal to e to the lambda x times c1 e to the mu xi, that's an i, plus c2 times e to the minus mu xi. Now what can we do?"}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So we get y is equal to, let me draw a line here. I don't want you to get confused with all this quadratic equation stuff. y is equal to e to the lambda x times c1 e to the mu xi, that's an i, plus c2 times e to the minus mu xi. Now what can we do? And this is where it gets fun. If you watched the calculus playlist, especially when I talk about approximating functions with series, we came up with what I thought was the most amazing result in calculus, or in mathematics, just from a metaphysical point of view. And now we will actually use it for something that you'll hopefully see is vaguely useful."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "Now what can we do? And this is where it gets fun. If you watched the calculus playlist, especially when I talk about approximating functions with series, we came up with what I thought was the most amazing result in calculus, or in mathematics, just from a metaphysical point of view. And now we will actually use it for something that you'll hopefully see is vaguely useful. So here we have two terms that have something times e to the something times i. And we learned before Euler's formula. And what was Euler's formula?"}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And now we will actually use it for something that you'll hopefully see is vaguely useful. So here we have two terms that have something times e to the something times i. And we learned before Euler's formula. And what was Euler's formula? I'll write that in purple. That e to the i theta, or we could write e to the ix, is equal to cosine of x plus i sine of x. And what's amazing about that is if you put negative 1 in here, then you get e to the i pi is equal to negative 1."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And what was Euler's formula? I'll write that in purple. That e to the i theta, or we could write e to the ix, is equal to cosine of x plus i sine of x. And what's amazing about that is if you put negative 1 in here, then you get e to the i pi is equal to negative 1. If you substitute this, cosine of pi is 0. So I thought that was amazing. Or you could write e to the i 2 pi is equal to 1."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "And what's amazing about that is if you put negative 1 in here, then you get e to the i pi is equal to negative 1. If you substitute this, cosine of pi is 0. So I thought that was amazing. Or you could write e to the i 2 pi is equal to 1. That's pretty amazing as well. So in one equation you have all of the fundamental numbers of mathematics. So that's amazing."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "Or you could write e to the i 2 pi is equal to 1. That's pretty amazing as well. So in one equation you have all of the fundamental numbers of mathematics. So that's amazing. But let's get back down to earth and get practical. So let's see if we can use this to simplify Euler's. This is actually a definition."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So that's amazing. But let's get back down to earth and get practical. So let's see if we can use this to simplify Euler's. This is actually a definition. And the definition makes a lot of sense. Because when you do the power series approximation, or the Maclaurin series approximation of e to the x, it really looks like, or e to the i, right, e to the x, it really does look like cosine of x plus i times the power series approximation of x. But anyway, we won't go into that now."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "This is actually a definition. And the definition makes a lot of sense. Because when you do the power series approximation, or the Maclaurin series approximation of e to the x, it really looks like, or e to the i, right, e to the x, it really does look like cosine of x plus i times the power series approximation of x. But anyway, we won't go into that now. I have like six or seven videos on it. But let's use this to simplify this up here. So we can rewrite that as y is equal to e to the lambda x times, let's do the first one, c1."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "But anyway, we won't go into that now. I have like six or seven videos on it. But let's use this to simplify this up here. So we can rewrite that as y is equal to e to the lambda x times, let's do the first one, c1. It's e to the mu x i. So that can be rewritten, e to the mu x i, so instead of an x, we have a mu x, that will be equal to cosine of whatever is in front of the i. So cosine of mu x plus i sine of mu x."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So we can rewrite that as y is equal to e to the lambda x times, let's do the first one, c1. It's e to the mu x i. So that can be rewritten, e to the mu x i, so instead of an x, we have a mu x, that will be equal to cosine of whatever is in front of the i. So cosine of mu x plus i sine of mu x. And then plus c2. Times what? Times cosine of minus mu x plus i sine of minus mu x."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "So cosine of mu x plus i sine of mu x. And then plus c2. Times what? Times cosine of minus mu x plus i sine of minus mu x. And let's see if we can simplify this further. So one thing that you might want, let's distribute the c's. So now we get, I'll do it a different color."}, {"video_title": "Complex roots of the characteristic equations 1 Second order differential equations Khan Academy.mp3", "Sentence": "Times cosine of minus mu x plus i sine of minus mu x. And let's see if we can simplify this further. So one thing that you might want, let's distribute the c's. So now we get, I'll do it a different color. Actually, I'm running out of time. So I'll continue this in the next video. See you soon."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Modeling population. We're actually gonna go into some depth on this eventually, but here we're gonna start with simpler models. And we'll see, we will stumble on, using the logic of differential equations, things that you might have seen in your algebra or your pre-calculus class. So on some level, what we're going to do here is going to be review, but we're gonna get there using the power of modeling with differential equations. So let's just define some variables. Let's say that P is equal to our population. And let's say that T is, let's say that T is equal to the time that has passed in days."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "So on some level, what we're going to do here is going to be review, but we're gonna get there using the power of modeling with differential equations. So let's just define some variables. Let's say that P is equal to our population. And let's say that T is, let's say that T is equal to the time that has passed in days. In days. It could have been years or months. But let's say we're doing the population of insects that reproduce quite quickly."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And let's say that T is, let's say that T is equal to the time that has passed in days. In days. It could have been years or months. But let's say we're doing the population of insects that reproduce quite quickly. So days seem like a nice time span to care about. Now, what would be a reasonable model? Well, we could say that the rate of change, the rate of change of our population with respect to time, with respect to time, is, well, a reasonable thing to say is that it's going to be proportional to the actual population."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "But let's say we're doing the population of insects that reproduce quite quickly. So days seem like a nice time span to care about. Now, what would be a reasonable model? Well, we could say that the rate of change, the rate of change of our population with respect to time, with respect to time, is, well, a reasonable thing to say is that it's going to be proportional to the actual population. The actual population. Why is that reasonable? Well, the larger the population, the larger the rate at any given time."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Well, we could say that the rate of change, the rate of change of our population with respect to time, with respect to time, is, well, a reasonable thing to say is that it's going to be proportional to the actual population. The actual population. Why is that reasonable? Well, the larger the population, the larger the rate at any given time. If you have a thousand people, the rate at which they're reproducing is going to be more, or a thousand insects, is going to be more insects per second or per day or per year than if you only have 10 insects. So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population. And so, you know, sometimes you think of differential equations as these daunting, complex things, but notice, we've just been able to express a reasonably not-so-complicated idea."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Well, the larger the population, the larger the rate at any given time. If you have a thousand people, the rate at which they're reproducing is going to be more, or a thousand insects, is going to be more insects per second or per day or per year than if you only have 10 insects. So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population. And so, you know, sometimes you think of differential equations as these daunting, complex things, but notice, we've just been able to express a reasonably not-so-complicated idea. The rate of change of population is going to be proportional to the population. And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution. So how do we do that?"}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And so, you know, sometimes you think of differential equations as these daunting, complex things, but notice, we've just been able to express a reasonably not-so-complicated idea. The rate of change of population is going to be proportional to the population. And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution. So how do we do that? And I encourage you to pause the video at any time and see if you can solve this differential equation. So assuming you at least maybe have had an attempt at it. And you might immediately recognize that this is a separable differential equation."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "So how do we do that? And I encourage you to pause the video at any time and see if you can solve this differential equation. So assuming you at least maybe have had an attempt at it. And you might immediately recognize that this is a separable differential equation. And in separable differential equations, we want one variable and all the differentials involving that variable on one side, and the other variable and all the differentials involving the other variable on the other side, and then we can integrate both sides. And once again, dP, the derivative of P with respect to t, this isn't quite a fraction. This is the limit as our change in P over change in time."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And you might immediately recognize that this is a separable differential equation. And in separable differential equations, we want one variable and all the differentials involving that variable on one side, and the other variable and all the differentials involving the other variable on the other side, and then we can integrate both sides. And once again, dP, the derivative of P with respect to t, this isn't quite a fraction. This is the limit as our change in P over change in time. This is our instantaneous change. But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "This is the limit as our change in P over change in time. This is our instantaneous change. But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that. So we want to put all the P's and dP's on one side, and all the things that involve t or that I guess don't involve P on the other side. So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel. And then you can multiply both sides times dt."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "So let's do that. So we want to put all the P's and dP's on one side, and all the things that involve t or that I guess don't involve P on the other side. So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel. And then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity, which it really isn't a quantity. You really have to view this as a limit of that change in P over change in time."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity, which it really isn't a quantity. You really have to view this as a limit of that change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. But once again, for the sake of this, we can do this. And when we do that, we would be left with one over P dP is equal to K dt."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "You really have to view this as a limit of that change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. But once again, for the sake of this, we can do this. And when we do that, we would be left with one over P dP is equal to K dt. Now we can integrate both sides. Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's. And then if we integrate this side, we would get the natural log of the absolute value of our population."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And when we do that, we would be left with one over P dP is equal to K dt. Now we can integrate both sides. Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's. And then if we integrate this side, we would get the natural log of the absolute value of our population. And we could say plus some constant if we want, but we're gonna get a constant on this side as well. So we could just say that's going to be equal to K times T plus some constant. I'll just call that C1."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And then if we integrate this side, we would get the natural log of the absolute value of our population. And we could say plus some constant if we want, but we're gonna get a constant on this side as well. So we could just say that's going to be equal to K times T plus some constant. I'll just call that C1. And once again, I could have put a plus C2 here, but I could have then subtracted the constant from both sides, and I would just get the constant on the right-hand side. Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "I'll just call that C1. And once again, I could have put a plus C2 here, but I could have then subtracted the constant from both sides, and I would just get the constant on the right-hand side. Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here. That means, that's the same thing, that means that the absolute value of P is equal to E to all of this business. E to the, let me do the same colors, KT plus C1. Now, this right over here is the same thing, just using our exponent properties, this is the same thing as E to the KT, AE to the K times T times E to the C1."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Well, the natural log of the absolute value of P is equal to this thing right over here. That means, that's the same thing, that means that the absolute value of P is equal to E to all of this business. E to the, let me do the same colors, KT plus C1. Now, this right over here is the same thing, just using our exponent properties, this is the same thing as E to the KT, AE to the K times T times E to the C1. Now, this is just E to some constant, so we could just call this, let's just call that the constant C. So, this is all simplified to CE, CE to the KT, to the KT. And if we assume our population at any given time is positive, then we could get rid of this absolute value sign. Now, we have a general solution to this, frankly, fairly general differential equation, we just said proportional, we haven't given what the proportionality constant is, but we could say if we assume positive population, that the population is going to be equal to some constant C times E to the KT power."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Now, this right over here is the same thing, just using our exponent properties, this is the same thing as E to the KT, AE to the K times T times E to the C1. Now, this is just E to some constant, so we could just call this, let's just call that the constant C. So, this is all simplified to CE, CE to the KT, to the KT. And if we assume our population at any given time is positive, then we could get rid of this absolute value sign. Now, we have a general solution to this, frankly, fairly general differential equation, we just said proportional, we haven't given what the proportionality constant is, but we could say if we assume positive population, that the population is going to be equal to some constant C times E to the KT power. And the reason why I said that you've seen this before is this is just an exponential function, and it's very likely that in algebra or in pre-calculus class, you have modeled things with exponential functions, and my guess is that you've modeled things with populated, modeled things like population. The reason why this is interesting is you now see where this is coming from, the underlying logic that's just driven by the actual differential equation, the rate of change with respect to time of the population, well, maybe it's just proportional to population. So, I'll leave you there, and the next video, we'll do what you probably did in the 10th or 11th grade, or maybe later in your life, it doesn't matter when you did it, where we actually look at some initial conditions to find a particular solution."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "This is a separable differential equation, and we can integrate things quite easily. But as you will see as you go further in the world of differential equations, most differential equations are not so easy to solve. In fact, many of them are impossible to solve using analytic methods. And so given that, what do you do? You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up? And the answer to that question is no. You do not just give up, because we now have computers, and computers are really good at numerical methods, numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so given that, what do you do? You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up? And the answer to that question is no. You do not just give up, because we now have computers, and computers are really good at numerical methods, numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like. And so how do we do that? Well, in this video, we can explore one of the most straightforward numerical methods for approximating a particular solution. So what we do is, so let me draw a little table here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "You do not just give up, because we now have computers, and computers are really good at numerical methods, numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like. And so how do we do that? Well, in this video, we can explore one of the most straightforward numerical methods for approximating a particular solution. So what we do is, so let me draw a little table here. So a little table here. And so, actually let me give myself some, I'm gonna do it over here on the left-hand side. So a little table."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So what we do is, so let me draw a little table here. So a little table here. And so, actually let me give myself some, I'm gonna do it over here on the left-hand side. So a little table. So x and then y, with x, y, and then dy, dx. And you could set up a table like this to create a slope field. You could just pick all the, you could sample x's and y's in the xy plane, and then figure out, for a first-order differential equation like this, what is the slope going to be at that point, and you could construct a slope field."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So a little table. So x and then y, with x, y, and then dy, dx. And you could set up a table like this to create a slope field. You could just pick all the, you could sample x's and y's in the xy plane, and then figure out, for a first-order differential equation like this, what is the slope going to be at that point, and you could construct a slope field. And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition. We know that y of zero is equal to one. We know that the particular solution of this differential equation contains this point."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "You could just pick all the, you could sample x's and y's in the xy plane, and then figure out, for a first-order differential equation like this, what is the slope going to be at that point, and you could construct a slope field. And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition. We know that y of zero is equal to one. We know that the particular solution of this differential equation contains this point. So we're gonna start with that point. So we're gonna start with x is equal to zero, and let me do this in a different color. We're gonna start with x is equal to zero, y is equal to one, which is that point right over there."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "We know that the particular solution of this differential equation contains this point. So we're gonna start with that point. So we're gonna start with x is equal to zero, and let me do this in a different color. We're gonna start with x is equal to zero, y is equal to one, which is that point right over there. And we're gonna say, well, okay, what is the derivative at that point? Well, we know the derivative at any point or any solution to this differential equation, the derivative is going to be equal to the y value. So in this case, the derivative is going to be equal to y."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "We're gonna start with x is equal to zero, y is equal to one, which is that point right over there. And we're gonna say, well, okay, what is the derivative at that point? Well, we know the derivative at any point or any solution to this differential equation, the derivative is going to be equal to the y value. So in this case, the derivative is going to be equal to y. It's going to be equal to one. And in general, if the derivative, just like what we saw in the case of slope fields, as long as the derivative is expressed as a function of x's and y of x's, then you can figure out what the slope of the tangent line will be at that point. And so you say, okay, there's a slope of one at that point."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So in this case, the derivative is going to be equal to y. It's going to be equal to one. And in general, if the derivative, just like what we saw in the case of slope fields, as long as the derivative is expressed as a function of x's and y of x's, then you can figure out what the slope of the tangent line will be at that point. And so you say, okay, there's a slope of one at that point. So I can depict it like that. And instead of just keep doing that at a bunch of points, you say, okay, well, let's just, we know that the slope is changing, or it's probably changing for most cases, but let's just assume it's fixed until our next x, and then use that assumption to estimate what the next y would be. So what am I talking about here?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so you say, okay, there's a slope of one at that point. So I can depict it like that. And instead of just keep doing that at a bunch of points, you say, okay, well, let's just, we know that the slope is changing, or it's probably changing for most cases, but let's just assume it's fixed until our next x, and then use that assumption to estimate what the next y would be. So what am I talking about here? So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one. So we're gonna step from x equals zero now. We're gonna now step from that to x is equal to one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So what am I talking about here? So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one. So we're gonna step from x equals zero now. We're gonna now step from that to x is equal to one. So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x. So now let's say x is equal to one. So we've, our delta x is one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "We're gonna now step from that to x is equal to one. So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x. So now let's say x is equal to one. So we've, our delta x is one. So we've just added one here. And what we can do in our little approximation scheme here is let's just assume that that slope was constant over that interval. So where does that get us to?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we've, our delta x is one. So we've just added one here. And what we can do in our little approximation scheme here is let's just assume that that slope was constant over that interval. So where does that get us to? Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one. So then y is going to increase by one, and is going to get to two. And we see that point right over there."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So where does that get us to? Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one. So then y is going to increase by one, and is going to get to two. And we see that point right over there. And you already might see where this is going. Now, if this were actually a point on the curve, on the solution, and if it was satisfying this, what would then the derivative be? Well, the derivative is equal to y."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And we see that point right over there. And you already might see where this is going. Now, if this were actually a point on the curve, on the solution, and if it was satisfying this, what would then the derivative be? Well, the derivative is equal to y. The slope of the tangent line is going to be equal to y. So in this case, the slope of the tangent line is now going to be equal to two. And we could depict that, let me depict that in magenta here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, the derivative is equal to y. The slope of the tangent line is going to be equal to y. So in this case, the slope of the tangent line is now going to be equal to two. And we could depict that, let me depict that in magenta here. So it is going to be, it is going to be two. So it's gonna look, the slope of the tangent line there is going to be two. And so what does that tell us?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And we could depict that, let me depict that in magenta here. So it is going to be, it is going to be two. So it's gonna look, the slope of the tangent line there is going to be two. And so what does that tell us? Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be? Well, let's see. Now, for every one that we increase in the x direction, we should increase two in the y direction, because the slope is two."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so what does that tell us? Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be? Well, let's see. Now, for every one that we increase in the x direction, we should increase two in the y direction, because the slope is two. So the very next one should be four. Y is equal to four. So we could imagine, we have now kind of had a constant slope, and we get to that point right over there."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Now, for every one that we increase in the x direction, we should increase two in the y direction, because the slope is two. So the very next one should be four. Y is equal to four. So we could imagine, we have now kind of had a constant slope, and we get to that point right over there. And now we can do the same thing. Well, if we assume dy dx, based on the differential equation, has to be equal to y, we say, okay, the slope of the tangent line there is going to be the same thing as y. It's going to be four."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we could imagine, we have now kind of had a constant slope, and we get to that point right over there. And now we can do the same thing. Well, if we assume dy dx, based on the differential equation, has to be equal to y, we say, okay, the slope of the tangent line there is going to be the same thing as y. It's going to be four. And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one. We could have incremented by 10. We could have incremented by.01."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be four. And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one. We could have incremented by 10. We could have incremented by.01. And you could guess which one's going to give you a more accurate result. But if we step up by one now, and our slope is four, well, we're going to increase by, if we increased x by one, we're going to increase y by four. So we are going to get, we are going to get to eight."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "We could have incremented by.01. And you could guess which one's going to give you a more accurate result. But if we step up by one now, and our slope is four, well, we're going to increase by, if we increased x by one, we're going to increase y by four. So we are going to get, we are going to get to eight. And so we are at the.3,8, which is right over here. And so for this next stretch, the next stretch is going to look like that. And as you can see, just by doing this, we have been able to approximate what the particular solution looks like."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we are going to get, we are going to get to eight. And so we are at the.3,8, which is right over here. And so for this next stretch, the next stretch is going to look like that. And as you can see, just by doing this, we have been able to approximate what the particular solution looks like. And you might say, hey, Sal, well, you know, that's not so good of an approximation. And my reply to you is, well, yeah, I mean, depends on what your goals are. But I did this by hand."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And as you can see, just by doing this, we have been able to approximate what the particular solution looks like. And you might say, hey, Sal, well, you know, that's not so good of an approximation. And my reply to you is, well, yeah, I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand, I took fairly large delta x steps. If I wanted a better approximation, I could have lowered the delta x."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand, I took fairly large delta x steps. If I wanted a better approximation, I could have lowered the delta x. And let's do that. So let's take another scenario. So let's do another scenario where instead of delta x equals one, let's say delta x equals 1 1.5."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "If I wanted a better approximation, I could have lowered the delta x. And let's do that. So let's take another scenario. So let's do another scenario where instead of delta x equals one, let's say delta x equals 1 1.5. So once again, x, y, and the derivative of y with respect to x. So now let's say I want to take, so we know this first point, we're given this initial condition. When x is zero, y is one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do another scenario where instead of delta x equals one, let's say delta x equals 1 1.5. So once again, x, y, and the derivative of y with respect to x. So now let's say I want to take, so we know this first point, we're given this initial condition. When x is zero, y is one. And so the slope of the tangent line is going to be one. But then if we're incrementing by 1 1.5, so then when x is, I'll just write it as 0.5, 0.5, what is our new y going to be? Well, we're gonna assume that our slope from this to this is this slope right over here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "When x is zero, y is one. And so the slope of the tangent line is going to be one. But then if we're incrementing by 1 1.5, so then when x is, I'll just write it as 0.5, 0.5, what is our new y going to be? Well, we're gonna assume that our slope from this to this is this slope right over here. So our slope is one. So if we increase x by 0.5, we're gonna increase y by 0.5. And we're going to get to 1.5."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, we're gonna assume that our slope from this to this is this slope right over here. So our slope is one. So if we increase x by 0.5, we're gonna increase y by 0.5. And we're going to get to 1.5. So we're gonna, we get 0.5, 1.5. We get to that point right over there. Actually, you're having trouble seeing that."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And we're going to get to 1.5. So we're gonna, we get 0.5, 1.5. We get to that point right over there. Actually, you're having trouble seeing that. This stuff right over here is this point right over here. And now our new slope is going to be 1.5, which is going to look, which is going to look like, which is going to look like, actually not quite that steep. I don't want to overstate how good of an approximation it is and it's starting to get a little bit messy."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Actually, you're having trouble seeing that. This stuff right over here is this point right over here. And now our new slope is going to be 1.5, which is going to look, which is going to look like, which is going to look like, actually not quite that steep. I don't want to overstate how good of an approximation it is and it's starting to get a little bit messy. But it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment x by another 0.5, we get to one. So now if you increment by, if you increment by 0.5, and your slope is 1.5, your y is gonna increment by half of that, by 0.75."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "I don't want to overstate how good of an approximation it is and it's starting to get a little bit messy. But it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment x by another 0.5, we get to one. So now if you increment by, if you increment by 0.5, and your slope is 1.5, your y is gonna increment by half of that, by 0.75. And so you're gonna get to 2.25. So now you get to one, 2.25, which is this point right over here. And once again, this is a better approximation."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So now if you increment by, if you increment by 0.5, and your slope is 1.5, your y is gonna increment by half of that, by 0.75. And so you're gonna get to 2.25. So now you get to one, 2.25, which is this point right over here. And once again, this is a better approximation. Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on. Now in this one, y of, y of, y of one got us to two. In this one, y of one got us to 2.25."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And once again, this is a better approximation. Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on. Now in this one, y of, y of, y of one got us to two. In this one, y of one got us to 2.25. Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer. If we stepped by, if we stepped by 0.0001, we would get even closer and closer and closer. So there's a bunch of interesting things here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "In this one, y of one got us to 2.25. Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer. If we stepped by, if we stepped by 0.0001, we would get even closer and closer and closer. So there's a bunch of interesting things here. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved. And even if it's not the exact same solution or the same method, the idea that most differential equations are actually solved or, I guess you could say simulated with a numerical method because most of them actually cannot be solved in analytical form. Now you might be saying, hey, well what method is this one right over here called?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So there's a bunch of interesting things here. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved. And even if it's not the exact same solution or the same method, the idea that most differential equations are actually solved or, I guess you could say simulated with a numerical method because most of them actually cannot be solved in analytical form. Now you might be saying, hey, well what method is this one right over here called? Well this right over here is called Euler's, Euler's method after the famous Leonard Euler. Euler's, Euler's method. And not only is actually this one a good way of approximating what the solution to this or any differential equation is, but actually for this differential equation in particular, you can actually even use this to find, to find E with more and more and more precision."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And in this video, I'll show you one of those tricks. And it's useful beyond this, just because it's always good when, if maybe one day you become a mathematician or a physicist and you have an unsolved problem, some of these tricks that solve simpler problems back in your education might be a useful trick that solves some unsolved problem. So it's good to see it. And if you're taking differential equations, it might be on an exam. So it's good to learn. So we'll learn about integrating factors. So let's say we have an equation that has this form."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And if you're taking differential equations, it might be on an exam. So it's good to learn. So we'll learn about integrating factors. So let's say we have an equation that has this form. Let's say this is my differential equation. 3xy, I'm trying to write as neatly as possible, plus y squared plus x squared plus xy times y prime is equal to 0. So the first, especially since we've covered this in recent videos, whenever you see an equation of this form where you have some function of xy and then you have another function of x and y times y prime equals 0, you said, oh, this looks like this could be an exact differential equation."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So let's say we have an equation that has this form. Let's say this is my differential equation. 3xy, I'm trying to write as neatly as possible, plus y squared plus x squared plus xy times y prime is equal to 0. So the first, especially since we've covered this in recent videos, whenever you see an equation of this form where you have some function of xy and then you have another function of x and y times y prime equals 0, you said, oh, this looks like this could be an exact differential equation. And how do we test that? Well, we can take the partial derivative of this with respect to y. And we could call this function of x and y m. So the partial of that with respect to y, so m, the partial with respect to y, so it would be 3x plus 2y."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So the first, especially since we've covered this in recent videos, whenever you see an equation of this form where you have some function of xy and then you have another function of x and y times y prime equals 0, you said, oh, this looks like this could be an exact differential equation. And how do we test that? Well, we can take the partial derivative of this with respect to y. And we could call this function of x and y m. So the partial of that with respect to y, so m, the partial with respect to y, so it would be 3x plus 2y. And if this function right here, that expression right here, there, that's our function n, which is a function of x and y. We take the partial with respect to x, and we get that is equal to 2x plus y. And in order for this to have been an exact differential equation, the partial of this with respect to y would have to equal the partial of this with respect to x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And we could call this function of x and y m. So the partial of that with respect to y, so m, the partial with respect to y, so it would be 3x plus 2y. And if this function right here, that expression right here, there, that's our function n, which is a function of x and y. We take the partial with respect to x, and we get that is equal to 2x plus y. And in order for this to have been an exact differential equation, the partial of this with respect to y would have to equal the partial of this with respect to x. But we see here, just by looking at these two, they don't equal each other. They're not equal. So at least superficially, the way we looked at just now, this is not an exact differential equation."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And in order for this to have been an exact differential equation, the partial of this with respect to y would have to equal the partial of this with respect to x. But we see here, just by looking at these two, they don't equal each other. They're not equal. So at least superficially, the way we looked at just now, this is not an exact differential equation. But what if there were some factor, or I guess some function, that we could multiply both sides of this equation by that would make it an exact differential equation? So let's call that mu. So what I want to do is I want to multiply both sides of this equation by some function mu, and then see if I can solve for that function mu that would make it exact."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So at least superficially, the way we looked at just now, this is not an exact differential equation. But what if there were some factor, or I guess some function, that we could multiply both sides of this equation by that would make it an exact differential equation? So let's call that mu. So what I want to do is I want to multiply both sides of this equation by some function mu, and then see if I can solve for that function mu that would make it exact. So let's try to do that. So let's multiply both sides by mu. And just as a simplification, mu could be a function of x and y, it could be a function of x, it could be a function of just x, it could be a function of just y. I'll assume it's just a function of x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So what I want to do is I want to multiply both sides of this equation by some function mu, and then see if I can solve for that function mu that would make it exact. So let's try to do that. So let's multiply both sides by mu. And just as a simplification, mu could be a function of x and y, it could be a function of x, it could be a function of just x, it could be a function of just y. I'll assume it's just a function of x. You could assume it's just a function of y and try to solve it, or you could assume it's a function of x and y. If you assume it's a function of x and y, it becomes a lot harder to solve for. But that doesn't mean that there isn't one."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And just as a simplification, mu could be a function of x and y, it could be a function of x, it could be a function of just x, it could be a function of just y. I'll assume it's just a function of x. You could assume it's just a function of y and try to solve it, or you could assume it's a function of x and y. If you assume it's a function of x and y, it becomes a lot harder to solve for. But that doesn't mean that there isn't one. So let's say that mu is a function of x. And I'm going to multiply it by both sides of this equation. So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "But that doesn't mean that there isn't one. So let's say that mu is a function of x. And I'm going to multiply it by both sides of this equation. So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime. And then what's 0 times any function? Well, it's just going to be 0. 0 times mu of x is just going to be 0."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime. And then what's 0 times any function? Well, it's just going to be 0. 0 times mu of x is just going to be 0. But I did multiply the right-hand side times mu of x. And remember what we're doing. This mu of x, when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "0 times mu of x is just going to be 0. But I did multiply the right-hand side times mu of x. And remember what we're doing. This mu of x, when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation. So now if we consider this whole thing, our new m, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. So what's the partial derivative of this with respect to y? Well, if we're taking the partial with respect to y here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right?"}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "This mu of x, when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation. So now if we consider this whole thing, our new m, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. So what's the partial derivative of this with respect to y? Well, if we're taking the partial with respect to y here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right? When you take a partial with respect to y, x is just a constant, or a function of x can be viewed just as a constant. So the partial of this with respect to y is going to be equal to mu of x times 3x plus 2y. That's the partial of this with respect to y."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "Well, if we're taking the partial with respect to y here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right? When you take a partial with respect to y, x is just a constant, or a function of x can be viewed just as a constant. So the partial of this with respect to y is going to be equal to mu of x times 3x plus 2y. That's the partial of this with respect to y. And then what's the partial of this with respect to x? Well, here we'll use the product rule. So we'll take the derivative of the first expression with respect to x. Mu of x is no longer a constant anymore, since we're taking the partial with respect to x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "That's the partial of this with respect to y. And then what's the partial of this with respect to x? Well, here we'll use the product rule. So we'll take the derivative of the first expression with respect to x. Mu of x is no longer a constant anymore, since we're taking the partial with respect to x. So the derivative of mu of x with respect to x, well, that's just u prime of x, or mu prime, not u. Mu prime of x. Mu is a Greek letter, it's for the mu sound, but it looks a lot like a u. So mu prime of x times the second expression, x squared plus xy, plus just the first expression, this is just the product rule, mu of x, times the derivative of the second expression with respect to x. So times, ran out of space on that line, 2x plus y."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So we'll take the derivative of the first expression with respect to x. Mu of x is no longer a constant anymore, since we're taking the partial with respect to x. So the derivative of mu of x with respect to x, well, that's just u prime of x, or mu prime, not u. Mu prime of x. Mu is a Greek letter, it's for the mu sound, but it looks a lot like a u. So mu prime of x times the second expression, x squared plus xy, plus just the first expression, this is just the product rule, mu of x, times the derivative of the second expression with respect to x. So times, ran out of space on that line, 2x plus y. And now for this new equation, where I multiply both sides by mu, in order for this to be exact, these two things have to be equal to each other. So let's just remember the big picture. We're saying this is going to be exact."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So times, ran out of space on that line, 2x plus y. And now for this new equation, where I multiply both sides by mu, in order for this to be exact, these two things have to be equal to each other. So let's just remember the big picture. We're saying this is going to be exact. And now we're going to try to solve for mu. So let's see if we can do that. So let's see, we have, on this side, we have mu of x times 3x plus 2y."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "We're saying this is going to be exact. And now we're going to try to solve for mu. So let's see if we can do that. So let's see, we have, on this side, we have mu of x times 3x plus 2y. And let's subtract this expression from both sides. So it's minus mu of x times 2x plus y. You'll see a lot of these differential equation problems that get kind of hairy."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So let's see, we have, on this side, we have mu of x times 3x plus 2y. And let's subtract this expression from both sides. So it's minus mu of x times 2x plus y. You'll see a lot of these differential equation problems that get kind of hairy. They're really just a lot of algebra. And that equals, what do we have left? I'll write it in yellow."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "You'll see a lot of these differential equation problems that get kind of hairy. They're really just a lot of algebra. And that equals, what do we have left? I'll write it in yellow. That equals, I think I'm going to run out of space, so I'm going to do it a little bit lower. That equals just this term right here. That equals mu prime of x times x squared plus xy."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "I'll write it in yellow. That equals, I think I'm going to run out of space, so I'm going to do it a little bit lower. That equals just this term right here. That equals mu prime of x times x squared plus xy. And let's see, if we factor out a mu of x here, we get mu of x times 3x plus 2y minus 2x minus y is equal to mu prime of x, the derivative of mu with respect to x, times x squared plus xy. Now we could simplify this. So we get mu of x times 3x minus 2x is x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "That equals mu prime of x times x squared plus xy. And let's see, if we factor out a mu of x here, we get mu of x times 3x plus 2y minus 2x minus y is equal to mu prime of x, the derivative of mu with respect to x, times x squared plus xy. Now we could simplify this. So we get mu of x times 3x minus 2x is x. 2y minus y, so x plus y, is equal to, and I'm just going to simplify this side a little bit, is equal to mu prime of x. Let's factor out an x here. And the reason why I'm doing that is because it seems like if I factor out an x here, I'll get an x plus y."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So we get mu of x times 3x minus 2x is x. 2y minus y, so x plus y, is equal to, and I'm just going to simplify this side a little bit, is equal to mu prime of x. Let's factor out an x here. And the reason why I'm doing that is because it seems like if I factor out an x here, I'll get an x plus y. So this is mu prime of x times x times x plus y. x times x plus y is x squared plus xy. So that's why I did it. And I have this x plus y on both sides of this equation, which I will now divide both sides by."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And the reason why I'm doing that is because it seems like if I factor out an x here, I'll get an x plus y. So this is mu prime of x times x times x plus y. x times x plus y is x squared plus xy. So that's why I did it. And I have this x plus y on both sides of this equation, which I will now divide both sides by. So if you divide both sides by x plus y, we could maybe assume that it's not 0. We get, that simplifies things pretty dramatically, we get mu of x is equal to mu prime of x times x. And now, just the way my brain works, I like to rewrite this expression just kind of in our operator form, where we write instead of writing it u prime of x, we could write that as d mu dx."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And I have this x plus y on both sides of this equation, which I will now divide both sides by. So if you divide both sides by x plus y, we could maybe assume that it's not 0. We get, that simplifies things pretty dramatically, we get mu of x is equal to mu prime of x times x. And now, just the way my brain works, I like to rewrite this expression just kind of in our operator form, where we write instead of writing it u prime of x, we could write that as d mu dx. So let's do that. So we could write mu of x is equal to d, the derivative of mu with respect to x, times x. And this is actually a separable differential equation in of itself."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And now, just the way my brain works, I like to rewrite this expression just kind of in our operator form, where we write instead of writing it u prime of x, we could write that as d mu dx. So let's do that. So we could write mu of x is equal to d, the derivative of mu with respect to x, times x. And this is actually a separable differential equation in of itself. It's kind of a sub-differential equation to solve our broader one. We're just trying to figure out the integrating factor right here. So let's divide both sides by x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And this is actually a separable differential equation in of itself. It's kind of a sub-differential equation to solve our broader one. We're just trying to figure out the integrating factor right here. So let's divide both sides by x. So we get mu over x. I'm just doing, this is just a separable equation now. Is equal to d mu dx. And then let's divide both sides by mu of x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So let's divide both sides by x. So we get mu over x. I'm just doing, this is just a separable equation now. Is equal to d mu dx. And then let's divide both sides by mu of x. And we get 1 over x is equal to 1 over mu. That's mu of x. I'll just write 1 over mu right now for simplicity, times d mu dx. I'm actually going to go horizontal right here."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "And then let's divide both sides by mu of x. And we get 1 over x is equal to 1 over mu. That's mu of x. I'll just write 1 over mu right now for simplicity, times d mu dx. I'm actually going to go horizontal right here. Multiply both sides by dx, you get 1 over x dx is equal to 1 over mu of x d mu. Now you could integrate both sides of this and you'll get the natural log of the absolute value of x is equal to the natural log of the absolute value of mu, et cetera, et cetera. But it should be pretty clear from this that x is equal to mu, or mu is equal to x, right?"}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "I'm actually going to go horizontal right here. Multiply both sides by dx, you get 1 over x dx is equal to 1 over mu of x d mu. Now you could integrate both sides of this and you'll get the natural log of the absolute value of x is equal to the natural log of the absolute value of mu, et cetera, et cetera. But it should be pretty clear from this that x is equal to mu, or mu is equal to x, right? They're identical if you look at both sides of this equation. You can just change x for mu and it becomes the other side. So this is obviously telling us that mu of x is equal to x, right, or mu is equal to x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "But it should be pretty clear from this that x is equal to mu, or mu is equal to x, right? They're identical if you look at both sides of this equation. You can just change x for mu and it becomes the other side. So this is obviously telling us that mu of x is equal to x, right, or mu is equal to x. So we have our integrating factor. And if you want, you can take the antiderivative of both sides with the natural logs and all of that and you'll get the same answer. But this is just by looking at it, by inspection, you know that mu is equal to x."}, {"video_title": "Integrating factors 1 First order differential equations Khan Academy.mp3", "Sentence": "So this is obviously telling us that mu of x is equal to x, right, or mu is equal to x. So we have our integrating factor. And if you want, you can take the antiderivative of both sides with the natural logs and all of that and you'll get the same answer. But this is just by looking at it, by inspection, you know that mu is equal to x. Because both sides of this equation are completely the same. Anyway, we now have our integrating factor, but I am running out of time. So in the next video, we're not going to use this integrating factor."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So we left off in part one getting pretty close to finding our N of T that satisfies the logistic differential equation where its initial condition is between zero and K, and now we just have to really just do some algebra to finish things up. So we left with this, that for our N of T, this must be true. Now we could use a little bit of logarithm properties to rewrite this left-hand side as the logarithm of, have the logarithm of something minus the logarithm of something else, that's going to be the first something, the logarithm of the first something divided by the second something. One minus N over K, and this is of course going to be equal to all this business that we have, it's going to be equal to, actually let me just write it, it's going to be equal to, equal to R times T plus C, plus C, and now what we could do, this is the same thing as saying that E to the RT plus C is going to be equal to this thing, right over here. The natural log of this is equal to the exponent that I have to raise E to to get to this right over here, so I can just write that. We could just write, or another way of thinking about it is we could take E to, if this is equal to that, we could take E to this power on the left-hand side and E to this power on this right-hand side, and they should be equal. So E to this power is just going to be what's inside the parentheses, it's just going to be N, it's just going to be N over one minus, one minus N over K, and I'll do that in green color so you can just keep track of where things came from, is equal to E to this business, is equal to E to the RT, I'll do the T in white, RT plus C. Now, this I could rewrite if I want to."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "One minus N over K, and this is of course going to be equal to all this business that we have, it's going to be equal to, actually let me just write it, it's going to be equal to, equal to R times T plus C, plus C, and now what we could do, this is the same thing as saying that E to the RT plus C is going to be equal to this thing, right over here. The natural log of this is equal to the exponent that I have to raise E to to get to this right over here, so I can just write that. We could just write, or another way of thinking about it is we could take E to, if this is equal to that, we could take E to this power on the left-hand side and E to this power on this right-hand side, and they should be equal. So E to this power is just going to be what's inside the parentheses, it's just going to be N, it's just going to be N over one minus, one minus N over K, and I'll do that in green color so you can just keep track of where things came from, is equal to E to this business, is equal to E to the RT, I'll do the T in white, RT plus C. Now, this I could rewrite if I want to. If I have E to something plus something else, I could rewrite this as to the RT times E to the C, and just to simplify things, this is just going to be another constant here, and if this is C, I could call it C1, but I'm just going to call that a constant. So I could just say that that's going to be equal to some constant times E to the RT, and now we just need to solve for N, and once again, if at any point while we're working on this, you get inspired, feel free to solve for N. So let's see, one way to solve for N, let's see, if we could take the reciprocal of both sides of this, we're going to get one, let me draw a line here just so you know that we're, so let me draw a line. So if we take the reciprocal of both sides of this, we're going to get one over N, one minus N over K over light green N, is equal to, is equal to, and so let's, you know, we could say it's equal to, it's equal to one over C times E to the negative RT, but one over C, that's just going to be another constant."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So E to this power is just going to be what's inside the parentheses, it's just going to be N, it's just going to be N over one minus, one minus N over K, and I'll do that in green color so you can just keep track of where things came from, is equal to E to this business, is equal to E to the RT, I'll do the T in white, RT plus C. Now, this I could rewrite if I want to. If I have E to something plus something else, I could rewrite this as to the RT times E to the C, and just to simplify things, this is just going to be another constant here, and if this is C, I could call it C1, but I'm just going to call that a constant. So I could just say that that's going to be equal to some constant times E to the RT, and now we just need to solve for N, and once again, if at any point while we're working on this, you get inspired, feel free to solve for N. So let's see, one way to solve for N, let's see, if we could take the reciprocal of both sides of this, we're going to get one, let me draw a line here just so you know that we're, so let me draw a line. So if we take the reciprocal of both sides of this, we're going to get one over N, one minus N over K over light green N, is equal to, is equal to, and so let's, you know, we could say it's equal to, it's equal to one over C times E to the negative RT, but one over C, that's just going to be another constant. So I could write one over C here, when I take the reciprocal, but once again, this is just going to be another constant, so I'm going to be a little bit hand-wavy. Well, okay, we're going to get another constant here. I could have called, you know, this C1, this C2, I could call this C3 if I want to make it clear that these are not going to be the same number."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So if we take the reciprocal of both sides of this, we're going to get one over N, one minus N over K over light green N, is equal to, is equal to, and so let's, you know, we could say it's equal to, it's equal to one over C times E to the negative RT, but one over C, that's just going to be another constant. So I could write one over C here, when I take the reciprocal, but once again, this is just going to be another constant, so I'm going to be a little bit hand-wavy. Well, okay, we're going to get another constant here. I could have called, you know, this C1, this C2, I could call this C3 if I want to make it clear that these are not going to be the same number. This is E to this power, this is the reciprocal of that. Actually, maybe I'll do that just to make it a little bit, a little bit clearer. That would have been C1 right over there."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "I could have called, you know, this C1, this C2, I could call this C3 if I want to make it clear that these are not going to be the same number. This is E to this power, this is the reciprocal of that. Actually, maybe I'll do that just to make it a little bit, a little bit clearer. That would have been C1 right over there. And so this is going to be, sorry, the reciprocal of this is C3, and E to the negative, the reciprocal of E to the RT is E to the negative RT, E to the negative, E to the negative RE to the negative RT. And let's see, if we divide the numerator and the denominator by N, or if we divide, one way to think about it, we divide this term by N, we're going to get, so this term by N is going to be 1 over N, and then this term by N is just going to be minus 1 over K, so this is just going to be minus 1 over K is equal to this, is equal to this business. So copy and paste."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "That would have been C1 right over there. And so this is going to be, sorry, the reciprocal of this is C3, and E to the negative, the reciprocal of E to the RT is E to the negative RT, E to the negative, E to the negative RE to the negative RT. And let's see, if we divide the numerator and the denominator by N, or if we divide, one way to think about it, we divide this term by N, we're going to get, so this term by N is going to be 1 over N, and then this term by N is just going to be minus 1 over K, so this is just going to be minus 1 over K is equal to this, is equal to this business. So copy and paste. It's going to be equal to that. This is good algebra practice here. Now let's see, we could take this 1 over K, add it to both sides, so let's do that."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So copy and paste. It's going to be equal to that. This is good algebra practice here. Now let's see, we could take this 1 over K, add it to both sides, so let's do that. So let me just cut, so let me cut and paste it. I'm going to add it to both sides, so this should be a plus 1 over K. So plus 1 over K. And now to solve for N, I just take the reciprocal of both sides. So I'm going to get N, and I'll write it in kind of the function notation, N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to 1 over, is going to be equal to 1 over all of this business, is going to be equal to 1 over all of this business."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "Now let's see, we could take this 1 over K, add it to both sides, so let's do that. So let me just cut, so let me cut and paste it. I'm going to add it to both sides, so this should be a plus 1 over K. So plus 1 over K. And now to solve for N, I just take the reciprocal of both sides. So I'm going to get N, and I'll write it in kind of the function notation, N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to 1 over, is going to be equal to 1 over all of this business, is going to be equal to 1 over all of this business. So copy and paste is going to be equal to that. And that by itself, that by itself is already interesting. That by itself is already interesting."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So I'm going to get N, and I'll write it in kind of the function notation, N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to 1 over, is going to be equal to 1 over all of this business, is going to be equal to 1 over all of this business. So copy and paste is going to be equal to that. And that by itself, that by itself is already interesting. That by itself is already interesting. So I could write it like this, and if I want, if I don't like, let's see, well yeah, I could just, if I don't like having this K, you know, kind of a fraction in a fraction, I could rewrite it as, actually maybe I'll do it over here, N of T is equal to, I'll just multiply the numerator and the denominator by, I'll just multiply it, actually let me just leave it like that for now. And what I'm going to do, what I'm now going to do is I'm going to say, well look, you know we're assuming that N of 0 is N sub naught. So we're assuming that N of 0, N of 0 is equal to N sub naught."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "That by itself is already interesting. So I could write it like this, and if I want, if I don't like, let's see, well yeah, I could just, if I don't like having this K, you know, kind of a fraction in a fraction, I could rewrite it as, actually maybe I'll do it over here, N of T is equal to, I'll just multiply the numerator and the denominator by, I'll just multiply it, actually let me just leave it like that for now. And what I'm going to do, what I'm now going to do is I'm going to say, well look, you know we're assuming that N of 0 is N sub naught. So we're assuming that N of 0, N of 0 is equal to N sub naught. So let's write this thing, let's solve for the constant, let's figure out what this could be if we know what our initial condition is. So N of 0, N of 0 is going to be equal to, it's going to be equal to 1, 1 over, when T is 0, this is just going to be equal to 1, this is just going to be our constant, C3 plus 1 over K, plus 1 over the maximum population that our environment can handle, and that's going to be equal to N naught, and now we can solve for our constant. So we get, actually I'm probably going to need a lot of real estate for this, I can take the reciprocal of both sides again, so this is something that we're doing a lot of, C3 plus 1 over K is equal to 1 over N naught, just took the reciprocal of both sides, and so we get our constant, C3 is equal to 1 over N naught, minus 1 over K, and so we can rewrite our solution, which we'll call the logistic function, we get, this is fun now, N of T is equal to 1 over, our constant is this, so it's going to be, let me copy and paste this, so copy and paste, it's going to be that, that's our constant, so it's going to be that, times E to the negative RT, to the negative RT, plus 1 over K, and if we don't like having all of these denominators, all of these fractions in the denominator, why don't we multiply everything times, the numerator and the denominator by N naught K, so I'm going to multiply the numerator times N naught K, and I'm going to multiply the denominator by N naught K, and so what do we get?"}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So we're assuming that N of 0, N of 0 is equal to N sub naught. So let's write this thing, let's solve for the constant, let's figure out what this could be if we know what our initial condition is. So N of 0, N of 0 is going to be equal to, it's going to be equal to 1, 1 over, when T is 0, this is just going to be equal to 1, this is just going to be our constant, C3 plus 1 over K, plus 1 over the maximum population that our environment can handle, and that's going to be equal to N naught, and now we can solve for our constant. So we get, actually I'm probably going to need a lot of real estate for this, I can take the reciprocal of both sides again, so this is something that we're doing a lot of, C3 plus 1 over K is equal to 1 over N naught, just took the reciprocal of both sides, and so we get our constant, C3 is equal to 1 over N naught, minus 1 over K, and so we can rewrite our solution, which we'll call the logistic function, we get, this is fun now, N of T is equal to 1 over, our constant is this, so it's going to be, let me copy and paste this, so copy and paste, it's going to be that, that's our constant, so it's going to be that, times E to the negative RT, to the negative RT, plus 1 over K, and if we don't like having all of these denominators, all of these fractions in the denominator, why don't we multiply everything times, the numerator and the denominator by N naught K, so I'm going to multiply the numerator times N naught K, and I'm going to multiply the denominator by N naught K, and so what do we get? This is all going to be equal to, in the numerator, I have N naught times K, and in the denominator, I am going to have, let's see, if I multiply this term right over here, times N naught K, I'm going to have K, if I multiply this term times N naught K, I'm going to have N naught, so it's going to be minus N naught, minus N naught, times E to the negative RT, times E to the negative RT, negative RT, and then if I multiply this times N naught K, I'm going to get N naught, so plus N naught, and there you have it, we have found a solution for the logistic differential equation. We will call this the logistic function, and in future videos, we will explore it more, and we will see that it actually does, if you were to plot this, I encourage you to do so, either on the internet, you could try Wolfram-Alpha, or if you're on your graphing calculator, you will see that it has the exact properties that we want it to have. It starts at N naught, it starts to increase at increasing rates, but then it starts to slow down as we reach our maximum population, and so that is actually a very neat function."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And what we have plotted right over here is the slope field, or a slope field, for this differential equation. And we can verify that this indeed is a slope field for this differential equation. Let's draw a little table here. So let's just verify a few points. So let's say x, y, and dy, dx. So let's say we start with, I don't know, let's start with this point right over here, one comma one. When x is one and y is one, well, when I look at the differential equation, 1 6th times four minus one, so it's 1 6th times three, which is 3 6ths, which is 1 1.5."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So let's just verify a few points. So let's say x, y, and dy, dx. So let's say we start with, I don't know, let's start with this point right over here, one comma one. When x is one and y is one, well, when I look at the differential equation, 1 6th times four minus one, so it's 1 6th times three, which is 3 6ths, which is 1 1.5. And we see indeed on this slope field, they depicted the slope there. So if a solution goes to that point, right at that point, its slope would be 1 1.5. And as you see, it's actually only dependent on the y value."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "When x is one and y is one, well, when I look at the differential equation, 1 6th times four minus one, so it's 1 6th times three, which is 3 6ths, which is 1 1.5. And we see indeed on this slope field, they depicted the slope there. So if a solution goes to that point, right at that point, its slope would be 1 1.5. And as you see, it's actually only dependent on the y value. It doesn't matter what x is. As long as y is one, dy, dx is going to be 1 1.5. And you see that's why when x is 1 1.5 and y is one, you still have a slope of 1 1.5."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And as you see, it's actually only dependent on the y value. It doesn't matter what x is. As long as y is one, dy, dx is going to be 1 1.5. And you see that's why when x is 1 1.5 and y is one, you still have a slope of 1 1.5. And as long as y is one, all of these sampled points right over here all have a slope of 1 1.5. So just looking at that, that makes us feel that this slope field is consistent with this differential equation. But let's try a few other points just to feel a little bit better about it."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And you see that's why when x is 1 1.5 and y is one, you still have a slope of 1 1.5. And as long as y is one, all of these sampled points right over here all have a slope of 1 1.5. So just looking at that, that makes us feel that this slope field is consistent with this differential equation. But let's try a few other points just to feel a little bit better about it. And then we will use the slope field to actually visualize some solutions. So let's do an interesting point. Let's say we have this point."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "But let's try a few other points just to feel a little bit better about it. And then we will use the slope field to actually visualize some solutions. So let's do an interesting point. Let's say we have this point. Actually, no, that's at a half point. Let's say we have this, let's see, I want to do, let's say we do this point right over here. So that's x is equal to one and y is equal to six."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Let's say we have this point. Actually, no, that's at a half point. Let's say we have this, let's see, I want to do, let's say we do this point right over here. So that's x is equal to one and y is equal to six. And we see the way the differential equation is defined, it doesn't matter what our x is. It's really dependent on the y that's going to drive the slope. But we have six over six, which is one, times four minus six, which is negative two."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So that's x is equal to one and y is equal to six. And we see the way the differential equation is defined, it doesn't matter what our x is. It's really dependent on the y that's going to drive the slope. But we have six over six, which is one, times four minus six, which is negative two. So it's negative two. So we should have a slope of negative two. And it looks like that's what they depicted."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "But we have six over six, which is one, times four minus six, which is negative two. So it's negative two. So we should have a slope of negative two. And it looks like that's what they depicted. So as long as y is six, we should have a slope of negative two. Have a slope of negative two. And you see that in the slope field."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And it looks like that's what they depicted. So as long as y is six, we should have a slope of negative two. Have a slope of negative two. And you see that in the slope field. So hopefully you feel pretty good that this is the slope field for this differential equation. If you don't, I encourage you to keep, keep verifying these points here. But now let's actually use this slope field."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And you see that in the slope field. So hopefully you feel pretty good that this is the slope field for this differential equation. If you don't, I encourage you to keep, keep verifying these points here. But now let's actually use this slope field. Let's actually use this to visualize solutions to this differential equation based on points that the solution might go through. So let's say that we have a solution that goes through this point right over here. So what is that solution likely to look like?"}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "But now let's actually use this slope field. Let's actually use this to visualize solutions to this differential equation based on points that the solution might go through. So let's say that we have a solution that goes through this point right over here. So what is that solution likely to look like? And once again, this is going to be a rough approximation. Well, right at that point, it's going to have a slope, just as the slope field shows. And as our y increases, it looks like our slope, it looks like our slope, so at this point, I should be, actually let me, let me undo that."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So what is that solution likely to look like? And once again, this is going to be a rough approximation. Well, right at that point, it's going to have a slope, just as the slope field shows. And as our y increases, it looks like our slope, it looks like our slope, so at this point, I should be, actually let me, let me undo that. So this, if I keep going up at this point, when y is equal to two, I should be parallel to all of these, these segments on the slope field at y is equal to two. And then it looks like the slope starts to decrease as we approach y is equal to four. And so if I had a solution that went through this point, my guess is that it would look something, and then now the slope decreases again as we approach y is equal to zero."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And as our y increases, it looks like our slope, it looks like our slope, so at this point, I should be, actually let me, let me undo that. So this, if I keep going up at this point, when y is equal to two, I should be parallel to all of these, these segments on the slope field at y is equal to two. And then it looks like the slope starts to decrease as we approach y is equal to four. And so if I had a solution that went through this point, my guess is that it would look something, and then now the slope decreases again as we approach y is equal to zero. And of course, we see that because if when y equals zero, this whole thing is zero, so our derivative's going to be zero. So a reasonable solution might look something like this. So this gives us a clue."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And so if I had a solution that went through this point, my guess is that it would look something, and then now the slope decreases again as we approach y is equal to zero. And of course, we see that because if when y equals zero, this whole thing is zero, so our derivative's going to be zero. So a reasonable solution might look something like this. So this gives us a clue. Well, look, if a solution goes through this point, this right over here might, might be what it looks like. But what if it goes through, I don't know, what if it goes through this point right over here? Well then, it might look like, it might look like this by the same exact logic."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So this gives us a clue. Well, look, if a solution goes through this point, this right over here might, might be what it looks like. But what if it goes through, I don't know, what if it goes through this point right over here? Well then, it might look like, it might look like this by the same exact logic. So it might look like this. So just like that, we're starting to get a sense. We don't know the actual solution for this differential equation, but we're starting to get a sense of what, what type of functions, what type of functions or the class of functions that might satisfy the differential equation."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Well then, it might look like, it might look like this by the same exact logic. So it might look like this. So just like that, we're starting to get a sense. We don't know the actual solution for this differential equation, but we're starting to get a sense of what, what type of functions, what type of functions or the class of functions that might satisfy the differential equation. But what's interesting about this slope field is it looks like there's some, you know, there's some interesting stuff, you know, if we have, if our solution includes points between, where the y value's between zero and four, it looks like we're going to have solutions like this, but what if we had y values that were larger than that or that were less than that or exactly, exactly zero or four? So for example, what if we had a solution that went through this point right over here? Well, at that point right over here, the slope field tells us that our slope is zero, so our y value's not going to change."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "We don't know the actual solution for this differential equation, but we're starting to get a sense of what, what type of functions, what type of functions or the class of functions that might satisfy the differential equation. But what's interesting about this slope field is it looks like there's some, you know, there's some interesting stuff, you know, if we have, if our solution includes points between, where the y value's between zero and four, it looks like we're going to have solutions like this, but what if we had y values that were larger than that or that were less than that or exactly, exactly zero or four? So for example, what if we had a solution that went through this point right over here? Well, at that point right over here, the slope field tells us that our slope is zero, so our y value's not going to change. And as long as our y value doesn't change, our y value's going to stay at four, so our slope is going to stay zero. So we actually already found, this is actually a solution to the differential equation. It's y is equal to four is a solution to this differential equation."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Well, at that point right over here, the slope field tells us that our slope is zero, so our y value's not going to change. And as long as our y value doesn't change, our y value's going to stay at four, so our slope is going to stay zero. So we actually already found, this is actually a solution to the differential equation. It's y is equal to four is a solution to this differential equation. So y is equal to four. y is equal to four. And you can verify that that is a solution."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "It's y is equal to four is a solution to this differential equation. So y is equal to four. y is equal to four. And you can verify that that is a solution. When y is equal to four, this right-hand side is going to be zero, and the derivative is zero for y is equal to four. So that is a solution to the differential equation. And the same thing for y is equal to zero."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And you can verify that that is a solution. When y is equal to four, this right-hand side is going to be zero, and the derivative is zero for y is equal to four. So that is a solution to the differential equation. And the same thing for y is equal to zero. That is also a solution to the differential equation. Now, what if we included points, what if we included this point up here? And actually, let me do it in a different color so that you could see it."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And the same thing for y is equal to zero. That is also a solution to the differential equation. Now, what if we included points, what if we included this point up here? And actually, let me do it in a different color so that you could see it. Let's say our solution included that point. Well, then it might look something, it might look something like this. And once again, I'm just using the slope field as a guide to give me an idea of what the slope might be as my curve progresses, as my solution progresses."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And actually, let me do it in a different color so that you could see it. Let's say our solution included that point. Well, then it might look something, it might look something like this. And once again, I'm just using the slope field as a guide to give me an idea of what the slope might be as my curve progresses, as my solution progresses. So a solution that includes the point zero five might look something like this. And once again, it's just another clue. A solution that includes the point zero negative one and a half might look something like, might look something like this."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And once again, I'm just using the slope field as a guide to give me an idea of what the slope might be as my curve progresses, as my solution progresses. So a solution that includes the point zero five might look something like this. And once again, it's just another clue. A solution that includes the point zero negative one and a half might look something like, might look something like this. So anyway, hopefully this gives you a better appreciation for why slope fields are interesting. If you have a differential equation that just involves the first derivative and some x's and y's, this one only involves the first derivative and y's, we can plot a slope field like this, not too much trouble. We essentially just keep solving for the slopes."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "And what I have pictures here are some of the most known, or actually this gentleman right over here might be the most known person when people think about population and the limits to growth of population. This is Thomas Malthus. He was a British cleric and writer and scholar at the end of the 1700s, the end of the 18th century, early 19th century. And he really challenged the notion that population could grow indefinitely and that we would always, through technology, be able to feed ourselves. Really, that the environment would eventually put some caps on how much or where the population could grow to. And P.F. Verhulst, and I'm sure I'm mispronouncing his name here, he was a Belgian mathematician who read Malthus' work and tried to model the behavior that Malthus was talking about."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "And he really challenged the notion that population could grow indefinitely and that we would always, through technology, be able to feed ourselves. Really, that the environment would eventually put some caps on how much or where the population could grow to. And P.F. Verhulst, and I'm sure I'm mispronouncing his name here, he was a Belgian mathematician who read Malthus' work and tried to model the behavior that Malthus was talking about. That, okay, when there aren't environmental constraints, maybe population does grow somewhat exponentially, but then as it approaches the limits set by the environment, it's going to essentially asymptote towards some type of population. Malthus, in particular, he actually doesn't think it's just going to be a nice, clean asymptote. He actually thinks that the population would go above the limit and you would have these catastrophes, and then you would go crashing below the limit, and you would kind of oscillate right around the limit through these catastrophes."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Verhulst, and I'm sure I'm mispronouncing his name here, he was a Belgian mathematician who read Malthus' work and tried to model the behavior that Malthus was talking about. That, okay, when there aren't environmental constraints, maybe population does grow somewhat exponentially, but then as it approaches the limits set by the environment, it's going to essentially asymptote towards some type of population. Malthus, in particular, he actually doesn't think it's just going to be a nice, clean asymptote. He actually thinks that the population would go above the limit and you would have these catastrophes, and then you would go crashing below the limit, and you would kind of oscillate right around the limit through these catastrophes. As you can tell, Malthus was a fairly optimistic guy. Let's go through a little bit of the math and a little bit of the differential equations. Also, these aren't overly hairy differential equations to think about population."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "He actually thinks that the population would go above the limit and you would have these catastrophes, and then you would go crashing below the limit, and you would kind of oscillate right around the limit through these catastrophes. As you can tell, Malthus was a fairly optimistic guy. Let's go through a little bit of the math and a little bit of the differential equations. Also, these aren't overly hairy differential equations to think about population. The first way to think about population, and I'll express it as a differential equation. Actually, let me just set some variables here. Let's say that n is our population."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Also, these aren't overly hairy differential equations to think about population. The first way to think about population, and I'll express it as a differential equation. Actually, let me just set some variables here. Let's say that n is our population. That's our population. We are going to assume that n is a function of t. n as a function of t is what we're going to be thinking about in this and, frankly, the next series of videos. One way to think about how to model this is just, well, what is the rate of change of population with respect to time?"}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Let's say that n is our population. That's our population. We are going to assume that n is a function of t. n as a function of t is what we're going to be thinking about in this and, frankly, the next series of videos. One way to think about how to model this is just, well, what is the rate of change of population with respect to time? How does that relate to things? We could say, okay, well, what is the rate of change of population with respect to time? D capital N dt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "One way to think about how to model this is just, well, what is the rate of change of population with respect to time? How does that relate to things? We could say, okay, well, what is the rate of change of population with respect to time? D capital N dt. One way to think about it is it's going to be proportional to the population. You could say, well, maybe this is going to be some proportionality constant times the population times the population itself. This makes sense."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "D capital N dt. One way to think about it is it's going to be proportional to the population. You could say, well, maybe this is going to be some proportionality constant times the population times the population itself. This makes sense. If the population is smaller, then you're not going to have as much change per unit time as if the population is larger. The larger the population, the more it's going to grow in a particular unit of time. This is actually a fairly simple to solve differential equation."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This makes sense. If the population is smaller, then you're not going to have as much change per unit time as if the population is larger. The larger the population, the more it's going to grow in a particular unit of time. This is actually a fairly simple to solve differential equation. You might have done it before. I encourage you to pause this video if you feel inspired to do so. I'll solve it right here."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This is actually a fairly simple to solve differential equation. You might have done it before. I encourage you to pause this video if you feel inspired to do so. I'll solve it right here. You'll see that we get an exponential function here for n. Let's do that. Let's solve this and we'll essentially separate the variables, separate the n from the t's, although we only see a dt here, but I'll do that in a second. If I divide both sides by n, I get 1 over n. If I multiply both sides by dt, if you think about the dt as something that you can multiply, I'm going to divide both sides by n and multiply both sides by dt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "I'll solve it right here. You'll see that we get an exponential function here for n. Let's do that. Let's solve this and we'll essentially separate the variables, separate the n from the t's, although we only see a dt here, but I'll do that in a second. If I divide both sides by n, I get 1 over n. If I multiply both sides by dt, if you think about the dt as something that you can multiply, I'm going to divide both sides by n and multiply both sides by dt. I'm going to get 1 over n dn on the left-hand side. On the right-hand side, I'm going to get r times dt. Notice I got the dt onto the right-hand side by multiplying both sides of that."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "If I divide both sides by n, I get 1 over n. If I multiply both sides by dt, if you think about the dt as something that you can multiply, I'm going to divide both sides by n and multiply both sides by dt. I'm going to get 1 over n dn on the left-hand side. On the right-hand side, I'm going to get r times dt. Notice I got the dt onto the right-hand side by multiplying both sides of that. Then I divided both sides by n and I got the 1 over n right over here. Now what we can do is we can take the antiderivative of both sides. What do we get on the left-hand side?"}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Notice I got the dt onto the right-hand side by multiplying both sides of that. Then I divided both sides by n and I got the 1 over n right over here. Now what we can do is we can take the antiderivative of both sides. What do we get on the left-hand side? This is just going to be the natural log of the absolute value of our population. Actually, if we assume that the population is always going to be nonzero, then we can actually take these absolute value off, but I'll do that in a second. That's going to be equal to r times t. We could have added a constant here, but I'm just going to do it on one side."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "What do we get on the left-hand side? This is just going to be the natural log of the absolute value of our population. Actually, if we assume that the population is always going to be nonzero, then we can actually take these absolute value off, but I'll do that in a second. That's going to be equal to r times t. We could have added a constant here, but I'm just going to do it on one side. r times t plus c. Now if we actually want to solve for n, we could take, if this is equal to this, then e to this power should be the same as e to this power. Another way of thinking about it, e, the natural log of the absolute value of n is equal to this, is another way of saying that e to this is going to be equal to that. Actually, let me just do it this way."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "That's going to be equal to r times t. We could have added a constant here, but I'm just going to do it on one side. r times t plus c. Now if we actually want to solve for n, we could take, if this is equal to this, then e to this power should be the same as e to this power. Another way of thinking about it, e, the natural log of the absolute value of n is equal to this, is another way of saying that e to this is going to be equal to that. Actually, let me just do it this way. Let me just take e to this power and to that power. If that's equal to that, then e to that power should be the same as e to that power. We're going to be left with e to the natural log of the absolute value of n. That's going to give you the absolute value of n. Let's just assume n positive."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Actually, let me just do it this way. Let me just take e to this power and to that power. If that's equal to that, then e to that power should be the same as e to that power. We're going to be left with e to the natural log of the absolute value of n. That's going to give you the absolute value of n. Let's just assume n positive. Let's just assume population is greater than zero. Then we could, this left-hand side right over here, we'll just simplify to n. Then our right-hand side, it's going to be, well, it could be e to the rt plus c, or this is the same thing. This thing right over here is the same thing as e to the rt, e to the r times t times, actually, let me do that e in that same color, e to the rt times e to the c. I'm just taking e to the sum of these two exponents."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "We're going to be left with e to the natural log of the absolute value of n. That's going to give you the absolute value of n. Let's just assume n positive. Let's just assume population is greater than zero. Then we could, this left-hand side right over here, we'll just simplify to n. Then our right-hand side, it's going to be, well, it could be e to the rt plus c, or this is the same thing. This thing right over here is the same thing as e to the rt, e to the r times t times, actually, let me do that e in that same color, e to the rt times e to the c. I'm just taking e to the sum of these two exponents. That's going to be e to the rt times e to the c. If we want to, we could just say, hey, you know what? This is still just going to be some arbitrary constant here. Actually, let's just call this c. It's e to the rt times c, or we could say c times e to the rt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This thing right over here is the same thing as e to the rt, e to the r times t times, actually, let me do that e in that same color, e to the rt times e to the c. I'm just taking e to the sum of these two exponents. That's going to be e to the rt times e to the c. If we want to, we could just say, hey, you know what? This is still just going to be some arbitrary constant here. Actually, let's just call this c. It's e to the rt times c, or we could say c times e to the rt. C times e to the rt. Notice, we have solved the differential equation. We haven't gotten to this less than optimistic reality of Malthus where we're limiting it."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Actually, let's just call this c. It's e to the rt times c, or we could say c times e to the rt. C times e to the rt. Notice, we have solved the differential equation. We haven't gotten to this less than optimistic reality of Malthus where we're limiting it. This is just, hey, if we just assume population is going to, the rate of change of population with respect to time is going to be proportional to population, when we solve that differential equation, we get that population is a function of time. Actually, let me make it explicit that this is a function of time. Let me just move the n over a little bit."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "We haven't gotten to this less than optimistic reality of Malthus where we're limiting it. This is just, hey, if we just assume population is going to, the rate of change of population with respect to time is going to be proportional to population, when we solve that differential equation, we get that population is a function of time. Actually, let me make it explicit that this is a function of time. Let me just move the n over a little bit. Let me write it this way. N of t is going to be equal to this. This was our solution to this differential equation."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Let me just move the n over a little bit. Let me write it this way. N of t is going to be equal to this. This was our solution to this differential equation. Once again, this is just going to grow forever. If we know the initial conditions, let's say that we knew that n of zero, when time is equal to zero, let's just say that's n sub naught. What would c be?"}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This was our solution to this differential equation. Once again, this is just going to grow forever. If we know the initial conditions, let's say that we knew that n of zero, when time is equal to zero, let's just say that's n sub naught. What would c be? Well, n of zero is going to be equal to c, c times e to the zero power. E to the zero power is just one, so it's just going to be equal to c. C is equal to n sub naught. Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be careful, n naught, our initial population, times e to the rt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "What would c be? Well, n of zero is going to be equal to c, c times e to the zero power. E to the zero power is just one, so it's just going to be equal to c. C is equal to n sub naught. Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be careful, n naught, our initial population, times e to the rt. Now once again, this is an exponential. Essentially our population is going to look like this. If I were to graph it, it's going to look, if that's my time axis, if that's my n axis right over here, I could say it's y equals n axis, however I want to denote it."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be careful, n naught, our initial population, times e to the rt. Now once again, this is an exponential. Essentially our population is going to look like this. If I were to graph it, it's going to look, if that's my time axis, if that's my n axis right over here, I could say it's y equals n axis, however I want to denote it. That would be n naught, and it's going to grow exponentially from there. The rate of this exponential function is going to be dictated by this constant right over there, but it's going to look something like this, and it's just going to grow faster and faster and faster forever and ever and ever and ever. Now as I mentioned at the beginning of the video, Malthus does not believe that this is going to be true."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "If I were to graph it, it's going to look, if that's my time axis, if that's my n axis right over here, I could say it's y equals n axis, however I want to denote it. That would be n naught, and it's going to grow exponentially from there. The rate of this exponential function is going to be dictated by this constant right over there, but it's going to look something like this, and it's just going to grow faster and faster and faster forever and ever and ever and ever. Now as I mentioned at the beginning of the video, Malthus does not believe that this is going to be true. He thinks that we're going to hit some natural limits that are going to start to constrain the population. In his mind, a more natural or a more realistic function to model population would look something like this, or even potentially something that you keep crashing around that kind of limit. What we'll see in the next video is that P.F."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "And we just tried out the solution y is equal to e to the rx. And we figured out that if you try that out, that it works for particular r's. And those r's we figured out in the last one were minus 2 and minus 3. But it came out of factoring this characteristic equation. And watch the last video if you forgot how we got that characteristic equation. And we ended up with this general solution for this differential equation. And you could try it out if you don't believe me that it works."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "But it came out of factoring this characteristic equation. And watch the last video if you forgot how we got that characteristic equation. And we ended up with this general solution for this differential equation. And you could try it out if you don't believe me that it works. But what if we don't want the general solution, we want to find the particular solution? Well, then we need initial conditions. So let's do this differential equation with some initial conditions."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "And you could try it out if you don't believe me that it works. But what if we don't want the general solution, we want to find the particular solution? Well, then we need initial conditions. So let's do this differential equation with some initial conditions. So let's say the initial conditions are, let me scroll this down, we have the solution that we figured out in the last video. And let me rewrite the differential equation. So it was the second derivative plus 5 times the first derivative plus 6 times the function is equal to 0."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "So let's do this differential equation with some initial conditions. So let's say the initial conditions are, let me scroll this down, we have the solution that we figured out in the last video. And let me rewrite the differential equation. So it was the second derivative plus 5 times the first derivative plus 6 times the function is equal to 0. And the initial conditions we're given is that y of 0 is equal to 2. And the first derivative at 0, or y prime at 0, is equal to 3. So they're giving us, what does y equal at the point 0?"}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "So it was the second derivative plus 5 times the first derivative plus 6 times the function is equal to 0. And the initial conditions we're given is that y of 0 is equal to 2. And the first derivative at 0, or y prime at 0, is equal to 3. So they're giving us, what does y equal at the point 0? And what is the slope at 0, at x is equal to 0? And the slope is 3. So how do we use these to solve for c1 and c2?"}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "So they're giving us, what does y equal at the point 0? And what is the slope at 0, at x is equal to 0? And the slope is 3. So how do we use these to solve for c1 and c2? Well, let's just use the first initial condition. y of 0 is equal to 2. So y of 0 is equal to 2, which is equal to, essentially, just substitute 0 into this equation."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "So how do we use these to solve for c1 and c2? Well, let's just use the first initial condition. y of 0 is equal to 2. So y of 0 is equal to 2, which is equal to, essentially, just substitute 0 into this equation. So it's c1 times e to the minus 2 times 0. What's e to the, that's essentially e to the 0, right? So that's just 1."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "So y of 0 is equal to 2, which is equal to, essentially, just substitute 0 into this equation. So it's c1 times e to the minus 2 times 0. What's e to the, that's essentially e to the 0, right? So that's just 1. So it's c1 times 1, which is just c1. Plus c2 times e to the minus 3 times 0. This is e to the 0, so it's just 1."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "So that's just 1. So it's c1 times 1, which is just c1. Plus c2 times e to the minus 3 times 0. This is e to the 0, so it's just 1. So plus c2. So the first equation we get when we substitute our first initial condition is essentially c1 plus c2 is equal to 2. Now let's apply our second initial condition that tells us the slope at x is equal to 0."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "This is e to the 0, so it's just 1. So plus c2. So the first equation we get when we substitute our first initial condition is essentially c1 plus c2 is equal to 2. Now let's apply our second initial condition that tells us the slope at x is equal to 0. So y prime of 0. So this is our general solution. Let's take its derivative."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "Now let's apply our second initial condition that tells us the slope at x is equal to 0. So y prime of 0. So this is our general solution. Let's take its derivative. And then we can use this. So y prime of x is equal to what? The derivative of this is equal to minus 2 c1 times e to the minus 2x."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "Let's take its derivative. And then we can use this. So y prime of x is equal to what? The derivative of this is equal to minus 2 c1 times e to the minus 2x. And what's the derivative of this? It's minus 3 c2 times e to the minus 3x. And now we can use our initial condition."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "The derivative of this is equal to minus 2 c1 times e to the minus 2x. And what's the derivative of this? It's minus 3 c2 times e to the minus 3x. And now we can use our initial condition. y prime at 0. So when x is equal to 0, what's the right-hand side equal? It's minus 2 times c1."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "And now we can use our initial condition. y prime at 0. So when x is equal to 0, what's the right-hand side equal? It's minus 2 times c1. And then e to the minus 0, e to the 0, so that's just 1. Minus 3 c2. And then once again, x is 0."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "It's minus 2 times c1. And then e to the minus 0, e to the 0, so that's just 1. Minus 3 c2. And then once again, x is 0. So e to the minus 3 times 0, that's just 1. So it's just 1 times minus 3 c2. And it tells us that when x is equal to 0, what does this whole derivative equal?"}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "And then once again, x is 0. So e to the minus 3 times 0, that's just 1. So it's just 1 times minus 3 c2. And it tells us that when x is equal to 0, what does this whole derivative equal? Well, that equals 3. y prime of 0 is equal to 3. So now we go back into your first year of algebra. We have two equations, two linear equations with two unknowns."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "And it tells us that when x is equal to 0, what does this whole derivative equal? Well, that equals 3. y prime of 0 is equal to 3. So now we go back into your first year of algebra. We have two equations, two linear equations with two unknowns. And we could solve. Let me write them in a form that you're probably more used to. So the first one is c1 plus c2 is equal to 2."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "We have two equations, two linear equations with two unknowns. And we could solve. Let me write them in a form that you're probably more used to. So the first one is c1 plus c2 is equal to 2. And the second one is minus 2 c1 minus 3 c2 is equal to 3. So what can we do? Let's multiply this top equation by 2."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "So the first one is c1 plus c2 is equal to 2. And the second one is minus 2 c1 minus 3 c2 is equal to 3. So what can we do? Let's multiply this top equation by 2. There's a ton of ways to solve this. But if you multiply the top equation times 2, you'll get, and I'll do this in a different color, just so that it's changed. I'm just multiplying the top one by 2."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "Let's multiply this top equation by 2. There's a ton of ways to solve this. But if you multiply the top equation times 2, you'll get, and I'll do this in a different color, just so that it's changed. I'm just multiplying the top one by 2. You get 2 c1 plus 2 c2 is equal to 4. And now we can add these two equations. Let's see, minus 2 c1 plus 2, those cancel out."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "I'm just multiplying the top one by 2. You get 2 c1 plus 2 c2 is equal to 4. And now we can add these two equations. Let's see, minus 2 c1 plus 2, those cancel out. So minus 3 plus 2, you get minus c2 is equal to 7. Or we could say that c2 is equal to minus 7. And now we could substitute back in here."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "Let's see, minus 2 c1 plus 2, those cancel out. So minus 3 plus 2, you get minus c2 is equal to 7. Or we could say that c2 is equal to minus 7. And now we could substitute back in here. We have c1 plus c2, c2 is minus 7, so minus 7 is equal to 9. Or we know that c1 plus 2 c2, that's minus 7, is equal to 2. I'm just substituting back to this differential equation."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "And now we could substitute back in here. We have c1 plus c2, c2 is minus 7, so minus 7 is equal to 9. Or we know that c1 plus 2 c2, that's minus 7, is equal to 2. I'm just substituting back to this differential equation. Not a differential, it's just a simple linear equation. And then we get c1 is equal to 9. And now we have our particular solution to the differential equation."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "I'm just substituting back to this differential equation. Not a differential, it's just a simple linear equation. And then we get c1 is equal to 9. And now we have our particular solution to the differential equation. So this was our general solution. We can just substitute our c1's and our c2's back in. We have our particular solution for those initial conditions, and I think that warrants a different color."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "And now we have our particular solution to the differential equation. So this was our general solution. We can just substitute our c1's and our c2's back in. We have our particular solution for those initial conditions, and I think that warrants a different color. So our particular solution is y of x is equal to c1, which we figured out is 9e to the minus 2x plus c2. Well, c2 is minus 7. Minus 7e to the minus 3x."}, {"video_title": "2nd order linear homogeneous differential equations 3 Khan Academy.mp3", "Sentence": "We have our particular solution for those initial conditions, and I think that warrants a different color. So our particular solution is y of x is equal to c1, which we figured out is 9e to the minus 2x plus c2. Well, c2 is minus 7. Minus 7e to the minus 3x. That is the particular solution to our original differential equation. And it might be a good exercise for you to actually test it out. This particular solution to this differential equation."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "Now that we know a little bit about the convolution integral and how it applies to the Laplace transform, let's actually try to solve an actual differential equation using what we know. So I have this equation here, this initial value problem, where it says that the second derivative of y plus 2 times the first derivative of y plus 2 times y is equal to sine of alpha t, and they give us some initial conditions. They tell us that y of 0 is equal to 0, and that y prime of 0 is equal to 0. And that's nice and convenient that those initial conditions tend to make the problem pretty clean. But let's get to the problem. So the first thing we do is we take the Laplace transform of both sides of this equation. The Laplace transform of the second derivative of y is just s squared."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "And that's nice and convenient that those initial conditions tend to make the problem pretty clean. But let's get to the problem. So the first thing we do is we take the Laplace transform of both sides of this equation. The Laplace transform of the second derivative of y is just s squared. This should be a bit of second nature to you by now. It's s squared times the Laplace transform of y, which I'll just write as capital Y of s, minus s. So we start with the same degree as the number of derivatives we're taking, and then we decrement that every time. Minus s times y of 0."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "The Laplace transform of the second derivative of y is just s squared. This should be a bit of second nature to you by now. It's s squared times the Laplace transform of y, which I'll just write as capital Y of s, minus s. So we start with the same degree as the number of derivatives we're taking, and then we decrement that every time. Minus s times y of 0. You kind of think of this as the integral, and you take the derivative 1, so this isn't exactly the derivative of that, minus, you decrement that 1, you just have a 1 there, y prime of 0. And that's the Laplace transform of the second derivative. Now we have to do the Laplace transform of 2 times the first derivative."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "Minus s times y of 0. You kind of think of this as the integral, and you take the derivative 1, so this isn't exactly the derivative of that, minus, you decrement that 1, you just have a 1 there, y prime of 0. And that's the Laplace transform of the second derivative. Now we have to do the Laplace transform of 2 times the first derivative. That's just going to be equal to plus 2 times s y of s, s times the Laplace transform of y, that's that there, minus y of 0, and we just have 1 left. The Laplace transform of 2y, that's just equal to plus 2 times the Laplace transform of y, and then that's going to be equal to the Laplace transform of sine of alpha t. We've done that multiple times so far. That's just alpha over s squared plus alpha squared."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "Now we have to do the Laplace transform of 2 times the first derivative. That's just going to be equal to plus 2 times s y of s, s times the Laplace transform of y, that's that there, minus y of 0, and we just have 1 left. The Laplace transform of 2y, that's just equal to plus 2 times the Laplace transform of y, and then that's going to be equal to the Laplace transform of sine of alpha t. We've done that multiple times so far. That's just alpha over s squared plus alpha squared. Now the next thing we want to do is we want to separate out the Laplace transform of y terms, or the y of s terms. Actually, even better, let's get rid of these initial conditions. y of 0 and y prime of 0 is 0, so this term is 0, that term is 0, and that term is 0."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "That's just alpha over s squared plus alpha squared. Now the next thing we want to do is we want to separate out the Laplace transform of y terms, or the y of s terms. Actually, even better, let's get rid of these initial conditions. y of 0 and y prime of 0 is 0, so this term is 0, that term is 0, and that term is 0. So our whole expression, and I can get rid of the colors now, is just becomes, let me pick a nice color here, becomes s squared times y of s plus 2s y of s, that's that term right there, plus 2y of s is equal to the right-hand side, is equal to alpha over s squared plus alpha squared. Now let's factor out the y of s, or the Laplace transform of y, and so we get s squared plus 2s plus 2, all of that times y of s is equal to this right-hand side, is equal to alpha over s squared plus alpha squared. Now we can divide both sides of this equation by this thing right here, by that right there, and we get y of s, the Laplace transform of y, is equal to this thing, alpha over s squared plus alpha squared, times 1 over s squared plus 2s plus 2."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "y of 0 and y prime of 0 is 0, so this term is 0, that term is 0, and that term is 0. So our whole expression, and I can get rid of the colors now, is just becomes, let me pick a nice color here, becomes s squared times y of s plus 2s y of s, that's that term right there, plus 2y of s is equal to the right-hand side, is equal to alpha over s squared plus alpha squared. Now let's factor out the y of s, or the Laplace transform of y, and so we get s squared plus 2s plus 2, all of that times y of s is equal to this right-hand side, is equal to alpha over s squared plus alpha squared. Now we can divide both sides of this equation by this thing right here, by that right there, and we get y of s, the Laplace transform of y, is equal to this thing, alpha over s squared plus alpha squared, times 1 over s squared plus 2s plus 2. I could just say divided by this, but it works out the same either way. Now what can we do here? Remember, I was doing this in the context of convolution, so I want to look for a Laplace transform that looks like the product of two Laplace transforms."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "Now we can divide both sides of this equation by this thing right here, by that right there, and we get y of s, the Laplace transform of y, is equal to this thing, alpha over s squared plus alpha squared, times 1 over s squared plus 2s plus 2. I could just say divided by this, but it works out the same either way. Now what can we do here? Remember, I was doing this in the context of convolution, so I want to look for a Laplace transform that looks like the product of two Laplace transforms. I know what the inverse Laplace transform of this is. In fact, I just took it. It's sine of alpha t. So if I could figure out the inverse Laplace transform of this, I could at least express our function y of t at least as a convolution integral, even if I don't necessarily solve the integral."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "Remember, I was doing this in the context of convolution, so I want to look for a Laplace transform that looks like the product of two Laplace transforms. I know what the inverse Laplace transform of this is. In fact, I just took it. It's sine of alpha t. So if I could figure out the inverse Laplace transform of this, I could at least express our function y of t at least as a convolution integral, even if I don't necessarily solve the integral. From there, it's just calculus, or if it's an unsolvable integral, we can just use a computer or something. Although you could actually use a computer to solve this, so you might skip some steps even going through this. But anyway, let's just try to get this in terms of a convolution integral."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "It's sine of alpha t. So if I could figure out the inverse Laplace transform of this, I could at least express our function y of t at least as a convolution integral, even if I don't necessarily solve the integral. From there, it's just calculus, or if it's an unsolvable integral, we can just use a computer or something. Although you could actually use a computer to solve this, so you might skip some steps even going through this. But anyway, let's just try to get this in terms of a convolution integral. So what can I do with this? Let's see, this isn't a perfect square. So if this isn't a perfect square, the next best thing is to try to complete the square here."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "But anyway, let's just try to get this in terms of a convolution integral. So what can I do with this? Let's see, this isn't a perfect square. So if this isn't a perfect square, the next best thing is to try to complete the square here. So let's try to write this as a s squared plus 2s plus something plus 2. I just rewrote it like this. And if I wrote this as s squared plus 2s plus 1, that becomes s plus 1 squared."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "So if this isn't a perfect square, the next best thing is to try to complete the square here. So let's try to write this as a s squared plus 2s plus something plus 2. I just rewrote it like this. And if I wrote this as s squared plus 2s plus 1, that becomes s plus 1 squared. But if I add a 1, I have to also subtract a 1. I can't just add 1s arbitrarily to things. If I add 1, I have to subtract 1 to cancel out with that 1."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "And if I wrote this as s squared plus 2s plus 1, that becomes s plus 1 squared. But if I add a 1, I have to also subtract a 1. I can't just add 1s arbitrarily to things. If I add 1, I have to subtract 1 to cancel out with that 1. So I really haven't changed this at all. I just rewrote it like this. But this now, I can rewrite this term right here as s plus 1 squared, and then this becomes plus 1."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "If I add 1, I have to subtract 1 to cancel out with that 1. So I really haven't changed this at all. I just rewrote it like this. But this now, I can rewrite this term right here as s plus 1 squared, and then this becomes plus 1. That's this term right here, is the plus 1. So I could rewrite my whole y of s is now equal to alpha over s squared plus alpha squared times 1 over this thing, s plus 1 squared plus 1. Now, I already said I know what the inverse Laplace transform of this thing is."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "But this now, I can rewrite this term right here as s plus 1 squared, and then this becomes plus 1. That's this term right here, is the plus 1. So I could rewrite my whole y of s is now equal to alpha over s squared plus alpha squared times 1 over this thing, s plus 1 squared plus 1. Now, I already said I know what the inverse Laplace transform of this thing is. Now I just have to figure out what the inverse Laplace transform of this thing is, of this blue thing in the blue box, and then I can express it as a convolution integral. And how do I do that? I could just do it right now."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "Now, I already said I know what the inverse Laplace transform of this thing is. Now I just have to figure out what the inverse Laplace transform of this thing is, of this blue thing in the blue box, and then I can express it as a convolution integral. And how do I do that? I could just do it right now. I could just immediately say that y of t, let me write this down, y of t. So the inverse is equal to the inverse Laplace transform of, obviously, of y of s. Let me write that down. y of s, which is equal to the inverse Laplace transform of these two things, the inverse Laplace transform of alpha over s squared plus alpha squared times 1 over s plus 1 squared plus 1. And now the convolution theorem tells us that this is going to be equal to the inverse Laplace transform of this first term in the product, so the inverse Laplace transform of that first term, alpha over s squared plus alpha squared convoluted with, I'll do the little convolution sign there, I was about to say convulsion, they're not too different, convoluted with the inverse Laplace transform of this term, inverse Laplace transform of 1 over s plus 1 squared plus 1."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "I could just do it right now. I could just immediately say that y of t, let me write this down, y of t. So the inverse is equal to the inverse Laplace transform of, obviously, of y of s. Let me write that down. y of s, which is equal to the inverse Laplace transform of these two things, the inverse Laplace transform of alpha over s squared plus alpha squared times 1 over s plus 1 squared plus 1. And now the convolution theorem tells us that this is going to be equal to the inverse Laplace transform of this first term in the product, so the inverse Laplace transform of that first term, alpha over s squared plus alpha squared convoluted with, I'll do the little convolution sign there, I was about to say convulsion, they're not too different, convoluted with the inverse Laplace transform of this term, inverse Laplace transform of 1 over s plus 1 squared plus 1. That's all, if I have the product of two Laplace transforms and I can take each of them independently and I can invert them, the inverse Laplace transform of their product is going to be the convolution of the inverse Laplace transforms of each of the terms. And what I just said confused me a bit, so I don't want to confuse you, but I think you get the idea. I have these two things, I recognize these independently, I can independently take the inverse of each of these things, so the inverse Laplace transform of their product is going to be the convolution of each of their inverse transforms."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "And now the convolution theorem tells us that this is going to be equal to the inverse Laplace transform of this first term in the product, so the inverse Laplace transform of that first term, alpha over s squared plus alpha squared convoluted with, I'll do the little convolution sign there, I was about to say convulsion, they're not too different, convoluted with the inverse Laplace transform of this term, inverse Laplace transform of 1 over s plus 1 squared plus 1. That's all, if I have the product of two Laplace transforms and I can take each of them independently and I can invert them, the inverse Laplace transform of their product is going to be the convolution of the inverse Laplace transforms of each of the terms. And what I just said confused me a bit, so I don't want to confuse you, but I think you get the idea. I have these two things, I recognize these independently, I can independently take the inverse of each of these things, so the inverse Laplace transform of their product is going to be the convolution of each of their inverse transforms. Now what's this over here? Why does this at the beginning of the problem? The inverse Laplace transform of this right here is sine of alpha t. Now, and then we're going to convolute that with the inverse Laplace transform of this right here."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "I have these two things, I recognize these independently, I can independently take the inverse of each of these things, so the inverse Laplace transform of their product is going to be the convolution of each of their inverse transforms. Now what's this over here? Why does this at the beginning of the problem? The inverse Laplace transform of this right here is sine of alpha t. Now, and then we're going to convolute that with the inverse Laplace transform of this right here. Let's do a little bit of work on the side just to make sure we get this right. So the Laplace transform of sine of t is equal to 1 over s squared plus 1. That looks like this, but I was shifted."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "The inverse Laplace transform of this right here is sine of alpha t. Now, and then we're going to convolute that with the inverse Laplace transform of this right here. Let's do a little bit of work on the side just to make sure we get this right. So the Laplace transform of sine of t is equal to 1 over s squared plus 1. That looks like this, but I was shifted. I was shifted by minus 1. You might remember that the Laplace transform of e to the at sine of t, when you multiply e to the at times anything, you're shifting its Laplace transform. So that will be equal to 1 over s minus a squared plus 1."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "That looks like this, but I was shifted. I was shifted by minus 1. You might remember that the Laplace transform of e to the at sine of t, when you multiply e to the at times anything, you're shifting its Laplace transform. So that will be equal to 1 over s minus a squared plus 1. And now we have something that looks just, we essentially shifted it by a. So now we have something that looks very similar to this. If we just set our a to be equal to negative 1, then it fits this pattern."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "So that will be equal to 1 over s minus a squared plus 1. And now we have something that looks just, we essentially shifted it by a. So now we have something that looks very similar to this. If we just set our a to be equal to negative 1, then it fits this pattern. This is s minus negative 1. So the inverse Laplace transform of this thing right here is just e to the a, which is minus 1, so minus 1t times sine of t. So this is the solution to our differential equation, even though it's not in a pleasant form to look at. And we can, if we want to, express it as an integral."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "If we just set our a to be equal to negative 1, then it fits this pattern. This is s minus negative 1. So the inverse Laplace transform of this thing right here is just e to the a, which is minus 1, so minus 1t times sine of t. So this is the solution to our differential equation, even though it's not in a pleasant form to look at. And we can, if we want to, express it as an integral. I'm not going to actually solve the integral in this problem, because it gets hairy and it's not even clear that, well, I won't even attempt to do it. But I just want to get into a form. And from there, it's just integral calculus."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "And we can, if we want to, express it as an integral. I'm not going to actually solve the integral in this problem, because it gets hairy and it's not even clear that, well, I won't even attempt to do it. But I just want to get into a form. And from there, it's just integral calculus. Or maybe a computer. Let's see. So this is equal to, what's the convolution of these two things?"}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "And from there, it's just integral calculus. Or maybe a computer. Let's see. So this is equal to, what's the convolution of these two things? It's the integral from 0 to t of sine of the first function of t minus tau. So I could write this. Well, I can actually switch, and I haven't shown you this, but we can switch the order either way."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "So this is equal to, what's the convolution of these two things? It's the integral from 0 to t of sine of the first function of t minus tau. So I could write this. Well, I can actually switch, and I haven't shown you this, but we can switch the order either way. But actually, let me just do it this way. So it's sine, I could write this as sine of t minus tau times alpha. I'm taking the sine of all of those things."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "Well, I can actually switch, and I haven't shown you this, but we can switch the order either way. But actually, let me just do it this way. So it's sine, I could write this as sine of t minus tau times alpha. I'm taking the sine of all of those things. Times e to the minus tau sine of tau d tau. That's one way that if I wanted to express the solution of this differential equation as an integral, I could write it like that. And it actually should be kind of obvious to you that this could go either way."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "I'm taking the sine of all of those things. Times e to the minus tau sine of tau d tau. That's one way that if I wanted to express the solution of this differential equation as an integral, I could write it like that. And it actually should be kind of obvious to you that this could go either way. Because when I wrote it, when it was a product up here, obviously, a product is, the order does not matter. I could write this term first, or I could write that term first. So regardless of which term is written first, I could take the same principle would apply."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "And it actually should be kind of obvious to you that this could go either way. Because when I wrote it, when it was a product up here, obviously, a product is, the order does not matter. I could write this term first, or I could write that term first. So regardless of which term is written first, I could take the same principle would apply. And I'll formally prove it in a future video. So we could have also done it the other way. We could have written this expression as e to the minus t sine of t convoluted with sine of alpha t. And that would be equal to the integral from 0 to t of e to the minus t minus tau sine of t minus tau times sine of alpha tau d tau."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "So regardless of which term is written first, I could take the same principle would apply. And I'll formally prove it in a future video. So we could have also done it the other way. We could have written this expression as e to the minus t sine of t convoluted with sine of alpha t. And that would be equal to the integral from 0 to t of e to the minus t minus tau sine of t minus tau times sine of alpha tau d tau. So these are equivalent. Either of these would be an acceptable answer. If you have a test where, and normally on a test like this, the teacher won't expect you to actually evaluate these integrals."}, {"video_title": "Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3", "Sentence": "We could have written this expression as e to the minus t sine of t convoluted with sine of alpha t. And that would be equal to the integral from 0 to t of e to the minus t minus tau sine of t minus tau times sine of alpha tau d tau. So these are equivalent. Either of these would be an acceptable answer. If you have a test where, and normally on a test like this, the teacher won't expect you to actually evaluate these integrals. The teacher will just say, get it into an integral just to kind of see whether you know how to convolute things and get your solution to the differential equation at least into this form. Because from here it really is just, I won't say basic calculus, but it's non-differential equations. So hopefully this second example with the convolution to solve an inverse transform clarified things up a little bit."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "Because I think you know how to solve essentially the homogenous version. So same problem as we did in the last video. The second derivative of y minus 3 times the first derivative of y minus 4 times the function. And now in the last example, we had the non-homogenous part was 3e to the 2x, but we're tired of dealing with exponent functions, so let's make it a trigonometric function. So let's say it equals 2 sine of x. So the first step you do is what we've been doing, is you essentially solve the homogenous equation. So this left-hand side is equal to 0."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And now in the last example, we had the non-homogenous part was 3e to the 2x, but we're tired of dealing with exponent functions, so let's make it a trigonometric function. So let's say it equals 2 sine of x. So the first step you do is what we've been doing, is you essentially solve the homogenous equation. So this left-hand side is equal to 0. You do that by getting the characteristic equation, r squared minus 3r minus 4 is equal to 0. You get the solutions r is equal to 4, r is equal to minus 1. And then you get that general solution."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So this left-hand side is equal to 0. You do that by getting the characteristic equation, r squared minus 3r minus 4 is equal to 0. You get the solutions r is equal to 4, r is equal to minus 1. And then you get that general solution. And we did this in the last video. You get the general solution of the homogenous, maybe we'll call this the homogenous solution, y homogenous. We got c1e to the 4x plus c2e to the minus x."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And then you get that general solution. And we did this in the last video. You get the general solution of the homogenous, maybe we'll call this the homogenous solution, y homogenous. We got c1e to the 4x plus c2e to the minus x. And that's all in good. But in order to get the general solution of this non-homogenous equation, I have to take the solution of the homogenous equation, if this were equal to 0, and then add that to a particular solution that satisfies this equation, that satisfies that when you take the second derivative, minus 3 times the first, minus 4 times the function, I actually get 2 sine of x. And here, once again, we'll use undetermined coefficients."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "We got c1e to the 4x plus c2e to the minus x. And that's all in good. But in order to get the general solution of this non-homogenous equation, I have to take the solution of the homogenous equation, if this were equal to 0, and then add that to a particular solution that satisfies this equation, that satisfies that when you take the second derivative, minus 3 times the first, minus 4 times the function, I actually get 2 sine of x. And here, once again, we'll use undetermined coefficients. And undetermined coefficients, just think to yourself, what function, when I take its second and first derivatives and add and subtract multiples of them to each other, will I get sine of x? Well, two functions end up with sine of x when you take the first and second derivatives, and that's sine and cosine of x. So it's a good guess."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And here, once again, we'll use undetermined coefficients. And undetermined coefficients, just think to yourself, what function, when I take its second and first derivatives and add and subtract multiples of them to each other, will I get sine of x? Well, two functions end up with sine of x when you take the first and second derivatives, and that's sine and cosine of x. So it's a good guess. And that's really what you're doing in the method of undetermined coefficients. You take a guess of a particular solution, and then you solve for the undetermined coefficients. So let's say that our guess is y is equal to, I don't know, some coefficient times sine of x."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So it's a good guess. And that's really what you're doing in the method of undetermined coefficients. You take a guess of a particular solution, and then you solve for the undetermined coefficients. So let's say that our guess is y is equal to, I don't know, some coefficient times sine of x. And if this was sine of 2x, I'd put a times sine of 2x here, just because I want its derivatives to, I still want, no matter what happens here, I want the sine of 2x's or maybe cosine of 2x's to still exist. If this was a sine of 2x, there's nothing I could do to a sine of x, or nothing at least trivial that I could do to a sine of x that would end up with a sine of 2x. So whatever's here, I want here."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So let's say that our guess is y is equal to, I don't know, some coefficient times sine of x. And if this was sine of 2x, I'd put a times sine of 2x here, just because I want its derivatives to, I still want, no matter what happens here, I want the sine of 2x's or maybe cosine of 2x's to still exist. If this was a sine of 2x, there's nothing I could do to a sine of x, or nothing at least trivial that I could do to a sine of x that would end up with a sine of 2x. So whatever's here, I want here. Plus cosine of 2x. Plus b, some coefficient, undetermined coefficient, times cosine of x. And once again, if this was sine of 2x, I'd want a cosine of 2x here."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So whatever's here, I want here. Plus cosine of 2x. Plus b, some coefficient, undetermined coefficient, times cosine of x. And once again, if this was sine of 2x, I'd want a cosine of 2x here. So let's figure out its first and second derivatives. So the first derivative of this, y prime, is equal to a cosine of x. Cosine derivative is minus sine, so minus b sine of x. And then the second derivative, I'll write down here, the second derivative is equal to what?"}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And once again, if this was sine of 2x, I'd want a cosine of 2x here. So let's figure out its first and second derivatives. So the first derivative of this, y prime, is equal to a cosine of x. Cosine derivative is minus sine, so minus b sine of x. And then the second derivative, I'll write down here, the second derivative is equal to what? Derivative of cosine is minus sine, so minus a sine of x minus b cosine of x. I think you're starting to see the hardest thing in most differential equations problems is not making careless mistakes. It's a lot of algebra and a lot of fairly basic calculus, and the real trick is to not make careless mistakes. Every time I say that, I tend to make one."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And then the second derivative, I'll write down here, the second derivative is equal to what? Derivative of cosine is minus sine, so minus a sine of x minus b cosine of x. I think you're starting to see the hardest thing in most differential equations problems is not making careless mistakes. It's a lot of algebra and a lot of fairly basic calculus, and the real trick is to not make careless mistakes. Every time I say that, I tend to make one. So I'm going to focus extra right now. So anyway, let's take these and substitute them back into this non-homogenous equation. Let's see if I can solve for a and b."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "Every time I say that, I tend to make one. So I'm going to focus extra right now. So anyway, let's take these and substitute them back into this non-homogenous equation. Let's see if I can solve for a and b. So the second derivative is that. From that, I'm going to just rewrite it, just so that you see what I'm doing. So I'm going to take the second derivative, y prime prime, so that's minus a sine of x minus b cosine of x. I'm going to add minus 3 times the first derivative to that, and I'm going to write the sines under the sines and the cosines under the cosines."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "Let's see if I can solve for a and b. So the second derivative is that. From that, I'm going to just rewrite it, just so that you see what I'm doing. So I'm going to take the second derivative, y prime prime, so that's minus a sine of x minus b cosine of x. I'm going to add minus 3 times the first derivative to that, and I'm going to write the sines under the sines and the cosines under the cosines. So minus 3 times this. So the sine is plus 3b sine of x. So 3b sine of x minus 3 times this."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So I'm going to take the second derivative, y prime prime, so that's minus a sine of x minus b cosine of x. I'm going to add minus 3 times the first derivative to that, and I'm going to write the sines under the sines and the cosines under the cosines. So minus 3 times this. So the sine is plus 3b sine of x. So 3b sine of x minus 3 times this. So minus 3a cosine of x. And then minus 4 times our original function. So minus 4a sine of x minus 4 times this."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So 3b sine of x minus 3 times this. So minus 3a cosine of x. And then minus 4 times our original function. So minus 4a sine of x minus 4 times this. Minus 4b cosine of x. When I take the sum of all of those, that's essentially the left-hand side of this equation. When I take the sum of all of that, that is equal to 2 sine of x. I could have written them out in a line, but it would have just been more confusing."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So minus 4a sine of x minus 4 times this. Minus 4b cosine of x. When I take the sum of all of those, that's essentially the left-hand side of this equation. When I take the sum of all of that, that is equal to 2 sine of x. I could have written them out in a line, but it would have just been more confusing. And now this makes it easy to add up the sine of x's and the cosine of x's. So if I add up all the coefficients on the sine of x, I get minus a plus 3b minus 4a. So that looks like minus 5a plus 3b sine of x."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "When I take the sum of all of that, that is equal to 2 sine of x. I could have written them out in a line, but it would have just been more confusing. And now this makes it easy to add up the sine of x's and the cosine of x's. So if I add up all the coefficients on the sine of x, I get minus a plus 3b minus 4a. So that looks like minus 5a plus 3b sine of x. Plus, and now what are the coefficients here? What are all the coefficients here? I have minus b and then I have another minus 4b."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So that looks like minus 5a plus 3b sine of x. Plus, and now what are the coefficients here? What are all the coefficients here? I have minus b and then I have another minus 4b. So minus 5b and then minus 3a. So minus 3a minus 5b. Ran out of space."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "I have minus b and then I have another minus 4b. So minus 5b and then minus 3a. So minus 3a minus 5b. Ran out of space. Cosine of x. This is the cosine of x should go right here. So anyway, how do I solve for a and b?"}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "Ran out of space. Cosine of x. This is the cosine of x should go right here. So anyway, how do I solve for a and b? Well, I have minus 5a plus 3b is equal to whatever coefficients in front of sine of x here. So minus 5a plus 3b must be equal to 2. And then minus 3a minus 5b is a coefficient on cosine of x, although I kind of squeezed in the cosine of x here."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So anyway, how do I solve for a and b? Well, I have minus 5a plus 3b is equal to whatever coefficients in front of sine of x here. So minus 5a plus 3b must be equal to 2. And then minus 3a minus 5b is a coefficient on cosine of x, although I kind of squeezed in the cosine of x here. So this must be equal to whatever the coefficient on cosine of x is on the right-hand side. Well, the coefficient of cosine of x on the right-hand side is 0. So that sets up a system of two unknowns with two equations, a linear system."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And then minus 3a minus 5b is a coefficient on cosine of x, although I kind of squeezed in the cosine of x here. So this must be equal to whatever the coefficient on cosine of x is on the right-hand side. Well, the coefficient of cosine of x on the right-hand side is 0. So that sets up a system of two unknowns with two equations, a linear system. So we get minus 5a plus 3b is equal to 2. And we get minus 3a minus 5b is equal to 0. And let's see if I can simplify this a little bit."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So that sets up a system of two unknowns with two equations, a linear system. So we get minus 5a plus 3b is equal to 2. And we get minus 3a minus 5b is equal to 0. And let's see if I can simplify this a little bit. See, this is a system of two unknowns, two equations. If I multiply the top equation by 5 thirds, I get minus 25 over 3a plus 5b is equal to 5 thirds times this. 5 thirds times 2 is 10 thirds."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And let's see if I can simplify this a little bit. See, this is a system of two unknowns, two equations. If I multiply the top equation by 5 thirds, I get minus 25 over 3a plus 5b is equal to 5 thirds times this. 5 thirds times 2 is 10 thirds. And the bottom equation is I have minus 3a minus 5b is equal to 0. Let's add the two equations. I get 10 thirds is equal to these cancel out."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "5 thirds times 2 is 10 thirds. And the bottom equation is I have minus 3a minus 5b is equal to 0. Let's add the two equations. I get 10 thirds is equal to these cancel out. And let's see. That's minus 25 over 3 minus 9 over 3a is equal to 10 thirds. It's getting a little bit messier than I like, but we'll soldier on."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "I get 10 thirds is equal to these cancel out. And let's see. That's minus 25 over 3 minus 9 over 3a is equal to 10 thirds. It's getting a little bit messier than I like, but we'll soldier on. So let's see. Minus 25 minus 9. What's minus 25 minus 9?"}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "It's getting a little bit messier than I like, but we'll soldier on. So let's see. Minus 25 minus 9. What's minus 25 minus 9? So that is 34. So we get 34 over 3a is equal to 10 over 3. Or let's see."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "What's minus 25 minus 9? So that is 34. So we get 34 over 3a is equal to 10 over 3. Or let's see. We can multiply both sides by 3, divide both sides by 34. a is equal to 10 over 34, which is equal to 5 over 17. Nice, ugly number. 5 over 17."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "Or let's see. We can multiply both sides by 3, divide both sides by 34. a is equal to 10 over 34, which is equal to 5 over 17. Nice, ugly number. 5 over 17. And now we can solve for b. So let's see. minus 3 times a."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "5 over 17. And now we can solve for b. So let's see. minus 3 times a. This color is nauseating me. Minus 3 times a. 5 over 17 minus 5b is equal to 0."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "minus 3 times a. This color is nauseating me. Minus 3 times a. 5 over 17 minus 5b is equal to 0. So that's what? Minus 15 over 17 is equal to plus 5b. I just took this, put it on the right hand side."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "5 over 17 minus 5b is equal to 0. So that's what? Minus 15 over 17 is equal to plus 5b. I just took this, put it on the right hand side. And then divide both sides by 5. Actually, let me make sure. Oh, you know what?"}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "I just took this, put it on the right hand side. And then divide both sides by 5. Actually, let me make sure. Oh, you know what? I realize I made a close mistake here. Minus 25 minus 9. That's minus 34 over 3."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "Oh, you know what? I realize I made a close mistake here. Minus 25 minus 9. That's minus 34 over 3. So minus 34a is equal to 10. a is equal to minus 10 over 34, or minus 5 over 17. So minus 3 times minus 5 over 17. So that's equal to plus 15 over 17 is equal to plus 5b."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "That's minus 34 over 3. So minus 34a is equal to 10. a is equal to minus 10 over 34, or minus 5 over 17. So minus 3 times minus 5 over 17. So that's equal to plus 15 over 17 is equal to plus 5b. And then we get b is equal to 3 over 17. That was hairy. And notice, the hard part was not losing your negative signs."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "So that's equal to plus 15 over 17 is equal to plus 5b. And then we get b is equal to 3 over 17. That was hairy. And notice, the hard part was not losing your negative signs. But anyway, we now have our particular solution to this. Our particular solution is, let me try to write it in a non-nauseating color, although I think I picked a nauseating one. The particular solution is a minus 5 over 17 sine of x, right, that was a coefficient on sine of x, plus b plus 3 over 17 times cosine of x."}, {"video_title": "Undetermined coefficients 2 Second order differential equations Khan Academy.mp3", "Sentence": "And notice, the hard part was not losing your negative signs. But anyway, we now have our particular solution to this. Our particular solution is, let me try to write it in a non-nauseating color, although I think I picked a nauseating one. The particular solution is a minus 5 over 17 sine of x, right, that was a coefficient on sine of x, plus b plus 3 over 17 times cosine of x. And if we look at our original problem, the general solution now to this non-homogenous equation would be this, which is the general solution to the homogenous equation, which we've done many videos on, plus now our particular solution that we solved using the method of undetermined coefficients. So if you just take that and add it to that, you're done. And I am out of time."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "We saw that if we wanted to solve this, and we didn't want one of the constant solutions of n of t equals zero or n of t is equal to k, and we did this in the last few videos, we got the solution that n of t, n of t is equal to, is equal to our initial n, n naught, times our maximum population, all of that over, all of that over our initial population, our initial population, plus the difference between our maximum population and the initial population, so k minus n naught, k minus n naught, times e, e to the negative rt, e to the negative rt. That this right over here, this logistic function, this logistic function is a non-constant solution, and it's the interesting one we care about if we're gonna model population to the logistic differential equation. Now that we've done all of that work to come up with this, let's actually apply it. That was the whole goal, is to model population growth. Let's come up with some assumptions. Let's first think about, well, let's say that I have an island. Let's say this is my island, and I start settling it with 100 people."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "That was the whole goal, is to model population growth. Let's come up with some assumptions. Let's first think about, well, let's say that I have an island. Let's say this is my island, and I start settling it with 100 people. I'm essentially saying n naught, let me do that in the n naught color, so I'm saying n naught is equal to 100. Let's say this environment, given what current technology of farming and agriculture and the availability of water and whatever else, let's say it can only support, let's say it can only, I don't wanna do that in green, let's say it can only support 1,000 people, max. So you get the idea."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "Let's say this is my island, and I start settling it with 100 people. I'm essentially saying n naught, let me do that in the n naught color, so I'm saying n naught is equal to 100. Let's say this environment, given what current technology of farming and agriculture and the availability of water and whatever else, let's say it can only support, let's say it can only, I don't wanna do that in green, let's say it can only support 1,000 people, max. So you get the idea. So we get K, capital K, is equal to 1,000. That's the limit to the population. And so now we have to think about what is r going to be."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "So you get the idea. So we get K, capital K, is equal to 1,000. That's the limit to the population. And so now we have to think about what is r going to be. So we have to come up with some assumptions. So let's say in a generation, in a generation which is about 20 years, or I'll just assume, in 20 years, I think it's reasonable that the population grows by, let's say the population grows by 50%. In 20 years, you have 50% growth, 50% increase, increase in the actual population."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "And so now we have to think about what is r going to be. So we have to come up with some assumptions. So let's say in a generation, in a generation which is about 20 years, or I'll just assume, in 20 years, I think it's reasonable that the population grows by, let's say the population grows by 50%. In 20 years, you have 50% growth, 50% increase, increase in the actual population. So what would you have to have your annual increase in order for after 20 years to grow by 50%? Well, to think about that, I'll get out my calculator. And one way to think about it, growing by 50%, that means that you are at 1.5, your original population."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "In 20 years, you have 50% growth, 50% increase, increase in the actual population. So what would you have to have your annual increase in order for after 20 years to grow by 50%? Well, to think about that, I'll get out my calculator. And one way to think about it, growing by 50%, that means that you are at 1.5, your original population. And if I take that to the 1 20th power, so the 1 20th, well, I'll just do 1 divided by 20th, this essentially says, well, how much am I going to grow by, or what is going to, this is telling me that I'm gonna grow by a factor of 1.02 every year, 1.02048. And so one way to think about it is, every year, if I grow by, if every year, if every year I grow by, if every year I grow by 0.020, I'll just round, five, then over 20 years, as this compounds, I will have grown by 50%, so that would be our R. This is essentially how much we are going to grow each year. Let me write that."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "And one way to think about it, growing by 50%, that means that you are at 1.5, your original population. And if I take that to the 1 20th power, so the 1 20th, well, I'll just do 1 divided by 20th, this essentially says, well, how much am I going to grow by, or what is going to, this is telling me that I'm gonna grow by a factor of 1.02 every year, 1.02048. And so one way to think about it is, every year, if I grow by, if every year, if every year I grow by, if every year I grow by 0.020, I'll just round, five, then over 20 years, as this compounds, I will have grown by 50%, so that would be our R. This is essentially how much we are going to grow each year. Let me write that. Growth each year. And we're going to assume that our T here is in years. So we're going to assume that our T is in years."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "Let me write that. Growth each year. And we're going to assume that our T here is in years. So we're going to assume that our T is in years. So T is in years. So what would our logistic function look like, given all of these assumptions? We would have N of T, let me, N of T is equal to, is equal to N naught times K. That's going to be 100 times 1,000."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "So we're going to assume that our T is in years. So T is in years. So what would our logistic function look like, given all of these assumptions? We would have N of T, let me, N of T is equal to, is equal to N naught times K. That's going to be 100 times 1,000. That's going to be 100, 100 times 1,000, my initial population times my maximum population, divided by my initial population, my initial population, plus the difference between my final and initial, so it's 1,000 minus 100, so that's going to be, this right over here is going to be 900, 900 times E to the negative R. So the negative 0.0205 times T. So it will be equal to that. And to verify that this actually is, this actually does describe population in a way that we thought the logistic differential equation would, let's actually plot it. So let me actually pause this video and then plot it."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "We would have N of T, let me, N of T is equal to, is equal to N naught times K. That's going to be 100 times 1,000. That's going to be 100, 100 times 1,000, my initial population times my maximum population, divided by my initial population, my initial population, plus the difference between my final and initial, so it's 1,000 minus 100, so that's going to be, this right over here is going to be 900, 900 times E to the negative R. So the negative 0.0205 times T. So it will be equal to that. And to verify that this actually is, this actually does describe population in a way that we thought the logistic differential equation would, let's actually plot it. So let me actually pause this video and then plot it. So there you go, I made a plot and I copied and pasted it here, and we see the behavior that we wanted, the behavior that we wanted to see. We see the population right over here, it's at year zero is starting at 100. Let me do this in a color that you're more likely to see."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "So let me actually pause this video and then plot it. So there you go, I made a plot and I copied and pasted it here, and we see the behavior that we wanted, the behavior that we wanted to see. We see the population right over here, it's at year zero is starting at 100. Let me do this in a color that you're more likely to see. Population starts at 100. And we can see that, let's see, after 20 years, our population looks like it's almost grown to 150. So it looks like, at least in the beginning, this term, this right over here is dominating."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "Let me do this in a color that you're more likely to see. Population starts at 100. And we can see that, let's see, after 20 years, our population looks like it's almost grown to 150. So it looks like, at least in the beginning, this term, this right over here is dominating. We are growing by this 0.0205, that would be 2.05% per year, which gets us close to 50% growth. And we see that's what's happening initially. So we go from 100 to 150 in the first 20 years, in the first generation."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "So it looks like, at least in the beginning, this term, this right over here is dominating. We are growing by this 0.0205, that would be 2.05% per year, which gets us close to 50% growth. And we see that's what's happening initially. So we go from 100 to 150 in the first 20 years, in the first generation. And then in the next generation, we should add another 75 if we weren't kind of being constrained by the environment. So 150 plus 75 would be 225. And it looks like we got, after 20 years, to about 200, so we're a little bit slower."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "So we go from 100 to 150 in the first 20 years, in the first generation. And then in the next generation, we should add another 75 if we weren't kind of being constrained by the environment. So 150 plus 75 would be 225. And it looks like we got, after 20 years, to about 200, so we're a little bit slower. We're a little bit slower than kind of the pure exponential growth. But we're, you know, the pure exponential growth would probably have us tracking something closer to here. But still growing pretty well."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "And it looks like we got, after 20 years, to about 200, so we're a little bit slower. We're a little bit slower than kind of the pure exponential growth. But we're, you know, the pure exponential growth would probably have us tracking something closer to here. But still growing pretty well. But then as our population gets larger and larger and larger, as we're getting closer and closer to the maximum population, our rate of growth, our rate of growth is approaching zero. So we constantly approach our maximum population, but we never quite get there. It's really an asymptote."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "But still growing pretty well. But then as our population gets larger and larger and larger, as we're getting closer and closer to the maximum population, our rate of growth, our rate of growth is approaching zero. So we constantly approach our maximum population, but we never quite get there. It's really an asymptote. We're just approaching it as time goes on and on and on. But you can kind of set your own threshold if you want to say, okay, when do we get to kind of 90% of maximum population? Well, that looks like that happens, 90% of maximum population happened after 210 years on this island."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "It's really an asymptote. We're just approaching it as time goes on and on and on. But you can kind of set your own threshold if you want to say, okay, when do we get to kind of 90% of maximum population? Well, that looks like that happens, 90% of maximum population happened after 210 years on this island. So on a human scale, that seems like a long time, many generations, but I guess on a cosmic scale, it's not that long, not even a cosmic scale, even just slightly longer than a human scale. And so it'll happen, well, this describes what's happening. And so this is a pretty interesting model and I'd be interested to see how it actually compares with actual data out there for actual population growth."}, {"video_title": "Logistic function application First order differential equations Khan Academy.mp3", "Sentence": "Well, that looks like that happens, 90% of maximum population happened after 210 years on this island. So on a human scale, that seems like a long time, many generations, but I guess on a cosmic scale, it's not that long, not even a cosmic scale, even just slightly longer than a human scale. And so it'll happen, well, this describes what's happening. And so this is a pretty interesting model and I'd be interested to see how it actually compares with actual data out there for actual population growth. With that said, it's not this, you know, everything that we've done so far has always assumed, we're kind of assuming this idea of a Malthusian limit, but what we've learned from human history is that this Malthusian limit seems to get, keeps getting pushed higher and higher based on the improvement of technology, that we're able to grow more crops in a certain amount of area, we have better rule of laws, so people don't kill each other as often, we have better control of water and irrigation and all of these things, so that we're able to increase the limits then far beyond what we thought. I would guess if you told, if you told Thomas Malthus that in the year 2014, we have 7 billion people on the earth, he would have said that's far beyond the Malthusian limit. He probably would have guessed that the Malthusian limit was more like a billion or 2 billion, given the technology at the time, and we're already at 7 billion and as technology improves and agriculture improves and rule of law improves and everything improves, we might be able to get, who knows, there might be a time, we might think it's crazy for there to be 20 billion people on the planet, but given today's technology, but if technology improves and optimistic scenarios, that maybe we could keep going."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "We have the second derivative of y plus 4 times the first derivative plus 4y is equal to 0. And we're asked to find the general solution to this differential equation. So the first thing we do, like we've done in the last several videos, we'll get the characteristic equation. That's r squared plus 4r plus 4 is equal to 0. This one's fairly easy to factor. We don't need the quadratic equation here. This is r plus 2 times r plus 2."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "That's r squared plus 4r plus 4 is equal to 0. This one's fairly easy to factor. We don't need the quadratic equation here. This is r plus 2 times r plus 2. And now something interesting happens, something that we haven't seen before. The two roots of our characteristic equation are actually the same number. r is equal to minus 2."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "This is r plus 2 times r plus 2. And now something interesting happens, something that we haven't seen before. The two roots of our characteristic equation are actually the same number. r is equal to minus 2. So you could say we only have one solution, or one root, or a repeated root, however you want to say it. We only have one r that satisfies the characteristic equation. So you might say, well, that's fine."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "r is equal to minus 2. So you could say we only have one solution, or one root, or a repeated root, however you want to say it. We only have one r that satisfies the characteristic equation. So you might say, well, that's fine. Maybe my general solution is just y is equal to some constant times e to the minus 2x using my one solution. And my reply to you is, this is a solution. And if you don't believe me, you can test it out."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So you might say, well, that's fine. Maybe my general solution is just y is equal to some constant times e to the minus 2x using my one solution. And my reply to you is, this is a solution. And if you don't believe me, you can test it out. But it's not the general solution. And why do I say that? Because this is a second order differential equation."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And if you don't believe me, you can test it out. But it's not the general solution. And why do I say that? Because this is a second order differential equation. And if someone wanted a particular solution, they would have to give you two initial conditions. The two initial conditions we've been using so far are what y of 0 equals and what y prime of 0 equals. They could give you what y of 5 equals, who knows."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "Because this is a second order differential equation. And if someone wanted a particular solution, they would have to give you two initial conditions. The two initial conditions we've been using so far are what y of 0 equals and what y prime of 0 equals. They could give you what y of 5 equals, who knows. But in general, when you have a second order differential equation, they have to give you two initial conditions. Now the problem with this solution, and why it's not the general solution, is if you use one of these initial conditions, you can solve for a c. You'll get an answer. You'll solve for that c. But then there's nothing to do with the second initial condition."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "They could give you what y of 5 equals, who knows. But in general, when you have a second order differential equation, they have to give you two initial conditions. Now the problem with this solution, and why it's not the general solution, is if you use one of these initial conditions, you can solve for a c. You'll get an answer. You'll solve for that c. But then there's nothing to do with the second initial condition. And in fact, except for only in one particular case, whatever c you get for the first initial condition, this equation won't be true for the second initial condition. And you could try it out. I mean, if we said y of 0 is equal to a and y prime of 0 is equal to, I don't know, 5a, let's see if these work."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "You'll solve for that c. But then there's nothing to do with the second initial condition. And in fact, except for only in one particular case, whatever c you get for the first initial condition, this equation won't be true for the second initial condition. And you could try it out. I mean, if we said y of 0 is equal to a and y prime of 0 is equal to, I don't know, 5a, let's see if these work. If y of 0 is equal to a, so that tells us that a is equal to c times e to the minus 2 times 0. So e to the 0. Or c is equal to a."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "I mean, if we said y of 0 is equal to a and y prime of 0 is equal to, I don't know, 5a, let's see if these work. If y of 0 is equal to a, so that tells us that a is equal to c times e to the minus 2 times 0. So e to the 0. Or c is equal to a. So if you just had this first initial condition, say fine, my particular solution is y is equal to a times e to the minus 2x. But let's see if this particular solution satisfies the second initial condition. So what's the derivative of this?"}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "Or c is equal to a. So if you just had this first initial condition, say fine, my particular solution is y is equal to a times e to the minus 2x. But let's see if this particular solution satisfies the second initial condition. So what's the derivative of this? y prime is equal to minus 2a e to the minus 2x. And it says that 5a, this initial condition says that 5a is equal to minus 2a times e to the minus 2 times 0. So e to the 0."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So what's the derivative of this? y prime is equal to minus 2a e to the minus 2x. And it says that 5a, this initial condition says that 5a is equal to minus 2a times e to the minus 2 times 0. So e to the 0. Or another way of saying that, e to the 0 is just 1. It says that 5a is equal to minus 2a, which we know is not true. So note, when we only have this general or pseudo-general solution, it can only satisfy generally one of the initial conditions, and if we're really lucky, both initial conditions."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So e to the 0. Or another way of saying that, e to the 0 is just 1. It says that 5a is equal to minus 2a, which we know is not true. So note, when we only have this general or pseudo-general solution, it can only satisfy generally one of the initial conditions, and if we're really lucky, both initial conditions. So that at least gives you an intuitive feel of why this isn't the general solution. So let me clean that up a little bit so that I have a feeling I'll have to use this real estate. So what do we do?"}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So note, when we only have this general or pseudo-general solution, it can only satisfy generally one of the initial conditions, and if we're really lucky, both initial conditions. So that at least gives you an intuitive feel of why this isn't the general solution. So let me clean that up a little bit so that I have a feeling I'll have to use this real estate. So what do we do? So we can use a technique called reduction of order, and it really just says, well, let's just guess a second solution. In general, when we first thought about these linear constant coefficient differential equations, we said, well, e to rx might be a good guess. And why is that?"}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So what do we do? So we can use a technique called reduction of order, and it really just says, well, let's just guess a second solution. In general, when we first thought about these linear constant coefficient differential equations, we said, well, e to rx might be a good guess. And why is that? Because all of the derivatives of e are kind of multiples of the original function, and that's why we used it. So if we're looking for a second solution, it doesn't hurt to kind of make the same guess. In order to be a little bit more general, let's make our guess for our second solution."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And why is that? Because all of the derivatives of e are kind of multiples of the original function, and that's why we used it. So if we're looking for a second solution, it doesn't hurt to kind of make the same guess. In order to be a little bit more general, let's make our guess for our second solution. I'll call this g for guess. Let's say it's some function of x times our first solution, e to the minus 2x. I could say some function of x times c times e to the minus 2x, but this c is kind of encapsulated."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "In order to be a little bit more general, let's make our guess for our second solution. I'll call this g for guess. Let's say it's some function of x times our first solution, e to the minus 2x. I could say some function of x times c times e to the minus 2x, but this c is kind of encapsulated. It could be part of this some random function of x. So let's be as general as possible. So let's assume that this is a solution, and then substitute it back into our original differential equation and see if we can actually solve for v that makes it all work."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "I could say some function of x times c times e to the minus 2x, but this c is kind of encapsulated. It could be part of this some random function of x. So let's be as general as possible. So let's assume that this is a solution, and then substitute it back into our original differential equation and see if we can actually solve for v that makes it all work. So before we do that, let's get its first and second derivatives. So the first derivative of g is equal to, well, this is a product rule. And I'll drop the v of x."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So let's assume that this is a solution, and then substitute it back into our original differential equation and see if we can actually solve for v that makes it all work. So before we do that, let's get its first and second derivatives. So the first derivative of g is equal to, well, this is a product rule. And I'll drop the v of x. We know that v is a function and not a constant. So product rule, derivative of the first, v prime times the second expression, e to the minus 2x, plus the first function or expression times the derivative of the second. So minus 2 times e to the minus 2x."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And I'll drop the v of x. We know that v is a function and not a constant. So product rule, derivative of the first, v prime times the second expression, e to the minus 2x, plus the first function or expression times the derivative of the second. So minus 2 times e to the minus 2x. So times minus 2 e to the minus 2x, or just to write it a little bit neater, g prime is equal to v prime e to the minus 2x minus 2v e to the minus 2x. Now we have to get the second derivative. I'll do that in a different color just to fight the monotony of it."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So minus 2 times e to the minus 2x. So times minus 2 e to the minus 2x, or just to write it a little bit neater, g prime is equal to v prime e to the minus 2x minus 2v e to the minus 2x. Now we have to get the second derivative. I'll do that in a different color just to fight the monotony of it. So the second derivative, we're going to have to do the product rule twice. Derivative of this first expression is going to be v prime prime e to the minus 2x minus 2v prime e to the minus 2x, that was just the product rule again. And then the derivative of the second expression is going to be, let's see, the derivative of the first one is v prime, so it's going to be minus 2v prime e to the minus 2x times, or no, plus, which is the product rule, minus 2 times minus 2, plus 4v e to the minus 2x."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "I'll do that in a different color just to fight the monotony of it. So the second derivative, we're going to have to do the product rule twice. Derivative of this first expression is going to be v prime prime e to the minus 2x minus 2v prime e to the minus 2x, that was just the product rule again. And then the derivative of the second expression is going to be, let's see, the derivative of the first one is v prime, so it's going to be minus 2v prime e to the minus 2x times, or no, plus, which is the product rule, minus 2 times minus 2, plus 4v e to the minus 2x. I hope I haven't made a careless mistake. And we can simplify this a little bit. So we get the second derivative of g, which is our guess solution, is equal to the second derivative of v prime e to the minus 2x minus 4v prime e to the minus 2x plus 4v e to the minus 2x."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And then the derivative of the second expression is going to be, let's see, the derivative of the first one is v prime, so it's going to be minus 2v prime e to the minus 2x times, or no, plus, which is the product rule, minus 2 times minus 2, plus 4v e to the minus 2x. I hope I haven't made a careless mistake. And we can simplify this a little bit. So we get the second derivative of g, which is our guess solution, is equal to the second derivative of v prime e to the minus 2x minus 4v prime e to the minus 2x plus 4v e to the minus 2x. And now, before we substitute it into this, we can just make one observation that'll just make the algebra a little bit simpler. Notice that g is something times e to the minus 2x. g prime is, we could factor out an e to the minus 2x."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So we get the second derivative of g, which is our guess solution, is equal to the second derivative of v prime e to the minus 2x minus 4v prime e to the minus 2x plus 4v e to the minus 2x. And now, before we substitute it into this, we can just make one observation that'll just make the algebra a little bit simpler. Notice that g is something times e to the minus 2x. g prime is, we could factor out an e to the minus 2x. And g prime prime, we can factor out an e to the minus 2x. So let's factor them out, essentially. So when we write this, we can write, so the second derivative is g prime prime, which we can write as, and I'm going to try to do this, it's e to the minus 2x."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "g prime is, we could factor out an e to the minus 2x. And g prime prime, we can factor out an e to the minus 2x. So let's factor them out, essentially. So when we write this, we can write, so the second derivative is g prime prime, which we can write as, and I'm going to try to do this, it's e to the minus 2x. Times the second derivative. So now we can get rid of the e to the minus 2x terms. So that's v prime prime minus 4v prime plus 4v."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So when we write this, we can write, so the second derivative is g prime prime, which we can write as, and I'm going to try to do this, it's e to the minus 2x. Times the second derivative. So now we can get rid of the e to the minus 2x terms. So that's v prime prime minus 4v prime plus 4v. If I just distribute this out, I get the second derivative, which is this. Plus 4 times the first derivative, and I'm also going to factor out the e to the minus 2x. So plus 4 times this."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So that's v prime prime minus 4v prime plus 4v. If I just distribute this out, I get the second derivative, which is this. Plus 4 times the first derivative, and I'm also going to factor out the e to the minus 2x. So plus 4 times this. So it's going to be plus 4v prime minus 8v. Once again, I factored out the e to the minus 2x. Plus 4 times y."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So plus 4 times this. So it's going to be plus 4v prime minus 8v. Once again, I factored out the e to the minus 2x. Plus 4 times y. So plus 4 times, we factored out the e to the minus 2x, so plus 4 times v. I did that because if I didn't do that, I'd be writing e to the minus 2x, and I'd probably make a careless mistake, and I'd run out of space, et cetera. But anyway, I essentially, to get this, I just substituted the second derivative, the first derivative, and g back into the differential equation. And we know that that has to equal 0."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "Plus 4 times y. So plus 4 times, we factored out the e to the minus 2x, so plus 4 times v. I did that because if I didn't do that, I'd be writing e to the minus 2x, and I'd probably make a careless mistake, and I'd run out of space, et cetera. But anyway, I essentially, to get this, I just substituted the second derivative, the first derivative, and g back into the differential equation. And we know that that has to equal 0. Let's see if we can simplify this a little bit more. And then hopefully solve for v. So let's see. Some things are popping out at me."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And we know that that has to equal 0. Let's see if we can simplify this a little bit more. And then hopefully solve for v. So let's see. Some things are popping out at me. So I have plus 4v, plus 4v, plus 4v. That's plus 8v minus 8v. Right?"}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "Some things are popping out at me. So I have plus 4v, plus 4v, plus 4v. That's plus 8v minus 8v. Right? So plus 4, minus 8, plus 4, those cancel out. It's plus 8, minus 8, those cancel out. And then I also have minus 4v prime, plus 4v prime, so those cancel out."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "Right? So plus 4, minus 8, plus 4, those cancel out. It's plus 8, minus 8, those cancel out. And then I also have minus 4v prime, plus 4v prime, so those cancel out. And lo and behold, we've done some serious simplification. It ends up being e to the minus 2x times v prime prime. We could call it v prime prime of x, now that we've saved so much space, is equal to 0."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And then I also have minus 4v prime, plus 4v prime, so those cancel out. And lo and behold, we've done some serious simplification. It ends up being e to the minus 2x times v prime prime. We could call it v prime prime of x, now that we've saved so much space, is equal to 0. We know this can never equal to 0. So essentially, we have now established that this expression has to be equal to 0. And we get a separable second order differential equation."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "We could call it v prime prime of x, now that we've saved so much space, is equal to 0. We know this can never equal to 0. So essentially, we have now established that this expression has to be equal to 0. And we get a separable second order differential equation. We get that the second derivative of v with respect to x, or it's a function of x, is equal to 0. So now we just have to differentiate both sides of this equation twice. You differentiate once, you get what?"}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And we get a separable second order differential equation. We get that the second derivative of v with respect to x, or it's a function of x, is equal to 0. So now we just have to differentiate both sides of this equation twice. You differentiate once, you get what? v prime of x is equal to, let's call it c1. And if we were to take the antiderivative of both sides again, we get v of x is equal to c1x plus some other c2. Now, remember, what was our guess?"}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "You differentiate once, you get what? v prime of x is equal to, let's call it c1. And if we were to take the antiderivative of both sides again, we get v of x is equal to c1x plus some other c2. Now, remember, what was our guess? Our guess was that our solution was going to be, or our general solution was going to be, some arbitrary function v times that first solution we found, e to the minus 2x. And when we actually took that guess and we substituted it in, we actually were able to solve for that v. And we got that v is equal to this. So this is interesting."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "Now, remember, what was our guess? Our guess was that our solution was going to be, or our general solution was going to be, some arbitrary function v times that first solution we found, e to the minus 2x. And when we actually took that guess and we substituted it in, we actually were able to solve for that v. And we got that v is equal to this. So this is interesting. So what is g, or what is our guess function, equal? It's no longer a guess. We've kind of established that it works."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "So this is interesting. So what is g, or what is our guess function, equal? It's no longer a guess. We've kind of established that it works. g, which we can call our solution, is equal to v of x times e to the minus 2x. Well, that equals this, c1x plus c2 e to the minus 2x. That equals c1x e to the minus 2x plus c2 e to the minus 2x."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "We've kind of established that it works. g, which we can call our solution, is equal to v of x times e to the minus 2x. Well, that equals this, c1x plus c2 e to the minus 2x. That equals c1x e to the minus 2x plus c2 e to the minus 2x. This is an e. And now we have a truly general solution. We have two constants, so we can satisfy two initial conditions. And if we're looking for a pattern, this is the pattern."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "That equals c1x e to the minus 2x plus c2 e to the minus 2x. This is an e. And now we have a truly general solution. We have two constants, so we can satisfy two initial conditions. And if we're looking for a pattern, this is the pattern. When you have a repeated root of your characteristic equation, the general solution is going to be, you're going to use that e to the, whatever root is, twice. But one time, you're going to have an x in front of it. And this works every time for second order, homogenous, constant coefficient, linear equations."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And if we're looking for a pattern, this is the pattern. When you have a repeated root of your characteristic equation, the general solution is going to be, you're going to use that e to the, whatever root is, twice. But one time, you're going to have an x in front of it. And this works every time for second order, homogenous, constant coefficient, linear equations. I will see you in the next video."}, {"video_title": "Repeated roots of the characteristic equation Second order differential equations Khan Academy.mp3", "Sentence": "And this works every time for second order, homogenous, constant coefficient, linear equations. I will see you in the next video."}]