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Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 d... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 sol... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 ... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 tim... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 w... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 ... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 homo... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 o... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 t... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 h... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 ... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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-han... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 clear... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 ... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 solutio... |
Undetermined coefficients 1 Second order differential equations Khan Academy.mp3 | 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 functio... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 differentia... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 accel... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 a... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 sec... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 tran... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 t... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 ... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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, wher... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 func... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 ora... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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, le... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 si... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 s... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 t... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 h... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 th... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 ... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 wou... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 b... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 wor... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 pa... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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. |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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. |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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. |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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. |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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.... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 w... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 co... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 ... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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. |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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. |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 p... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 step function differential equation Laplace transform Khan Academy.mp3 | 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 ... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 proble... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 i... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 ... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 g... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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. S... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 hav... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 tim... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 ste... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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 mi... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Laplace step function differential equation Laplace transform Khan Academy.mp3 | 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... |
Worked example range of solution curve from slope field AP Calculus AB Khan Academy.mp3 | 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... |
Worked example range of solution curve from slope field AP Calculus AB Khan Academy.mp3 | 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 gr... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 ... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 l... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 ... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 inform... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 temperatur... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 wr... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 gonn... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 c... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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. A... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 e... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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. O... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 ... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 s... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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 t... |
Applying Newton's Law of Cooling to warm oatmeal First order differential equations Khan Academy.mp3 | 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, ... |
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