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Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 ... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 w... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 li... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 i... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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. |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 l... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 a... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 in... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 k... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 actua... |
Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 m... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 ... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 tim... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 for... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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, thi... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 wa... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 ... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 ... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 par... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 ... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 soun... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 equ... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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? ... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 i... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 li... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 ... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 integra... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 shoul... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 inte... |
Integrating factors 1 First order differential equations Khan Academy.mp3 | 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 ... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 u... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 t... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 coul... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 th... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 w... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 so... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 wr... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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.... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 nume... |
Solving the logistic differential equation part 2 Khan Academy.mp3 | 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 equ... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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,... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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, whic... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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. An... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 th... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 depend... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 sh... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 f... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 verif... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 eq... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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, t... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 ... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 l... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 ... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 ... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 eq... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 equa... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 somethi... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 ... |
Slope field to visualize solutions First order differential equations Khan Academy.mp3 | 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 negat... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 18... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 wit... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 th... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 un... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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. ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 varia... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 si... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 co... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 nat... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 goi... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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? ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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, ... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | 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 f... |
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