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Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3 | 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 infinit... |
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3 | 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 te... |
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3 | 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 ... |
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3 | 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 tran... |
Laplace transform of t L{t} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 s... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 actu... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 goi... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 divid... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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.... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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. |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 cosi... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 o... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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. |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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. |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 ... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 solut... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 th... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 ... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 ... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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'... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 mi... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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? |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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. |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 fr... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 minu... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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,... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 ... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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, ... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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... |
Complex roots of the characteristic equations 3 Second order differential equations Khan Academy.mp3 | 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 ... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 ... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 tran... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 l... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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, m... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 ... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 ... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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,... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 i... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 e... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 w... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 t... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 ... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 equa... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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. |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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. |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 thi... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 th... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 trivi... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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 directl... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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, bu... |
Laplace transform of t^n L{t^n} Laplace transform Differential Equations Khan Academy.mp3 | 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, ... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 di... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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,... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 fig... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 co... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 i... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 e... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 tha... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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... |
Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3 | 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 plu... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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 t... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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 ... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. N... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. A... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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, whateve... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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 equ... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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 fi... |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
Undetermined coefficients 3 Second order differential equations Khan Academy.mp3 | 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. |
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