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Modeling population as an exponential function First order differential equations Khan Academy.mp3 | Let me just move the n over a little bit. Let me write it this way. N of t is going to be equal to this. This was our solution to this differential equation. Once again, this is just going to grow forever. If we know the initial conditions, let's say that we knew that n of zero, when time is equal to zero, let's just s... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | This was our solution to this differential equation. Once again, this is just going to grow forever. If we know the initial conditions, let's say that we knew that n of zero, when time is equal to zero, let's just say that's n sub naught. What would c be? Well, n of zero is going to be equal to c, c times e to the zero... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | What would c be? Well, n of zero is going to be equal to c, c times e to the zero power. E to the zero power is just one, so it's just going to be equal to c. C is equal to n sub naught. Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be careful, n naught, our initial population, times e to the rt. Now once again, this is an exponential. Essentially our population is going to look like this. If I were to graph it, it's g... |
Modeling population as an exponential function First order differential equations Khan Academy.mp3 | If I were to graph it, it's going to look, if that's my time axis, if that's my n axis right over here, I could say it's y equals n axis, however I want to denote it. That would be n naught, and it's going to grow exponentially from there. The rate of this exponential function is going to be dictated by this constant r... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | And we just tried out the solution y is equal to e to the rx. And we figured out that if you try that out, that it works for particular r's. And those r's we figured out in the last one were minus 2 and minus 3. But it came out of factoring this characteristic equation. And watch the last video if you forgot how we got... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | But it came out of factoring this characteristic equation. And watch the last video if you forgot how we got that characteristic equation. And we ended up with this general solution for this differential equation. And you could try it out if you don't believe me that it works. But what if we don't want the general solu... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | And you could try it out if you don't believe me that it works. But what if we don't want the general solution, we want to find the particular solution? Well, then we need initial conditions. So let's do this differential equation with some initial conditions. So let's say the initial conditions are, let me scroll this... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | So let's do this differential equation with some initial conditions. So let's say the initial conditions are, let me scroll this down, we have the solution that we figured out in the last video. And let me rewrite the differential equation. So it was the second derivative plus 5 times the first derivative plus 6 times ... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | So it was the second derivative plus 5 times the first derivative plus 6 times the function is equal to 0. And the initial conditions we're given is that y of 0 is equal to 2. And the first derivative at 0, or y prime at 0, is equal to 3. So they're giving us, what does y equal at the point 0? And what is the slope at ... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | So they're giving us, what does y equal at the point 0? And what is the slope at 0, at x is equal to 0? And the slope is 3. So how do we use these to solve for c1 and c2? Well, let's just use the first initial condition. y of 0 is equal to 2. So y of 0 is equal to 2, which is equal to, essentially, just substitute 0 in... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | So how do we use these to solve for c1 and c2? Well, let's just use the first initial condition. y of 0 is equal to 2. So y of 0 is equal to 2, which is equal to, essentially, just substitute 0 into this equation. So it's c1 times e to the minus 2 times 0. What's e to the, that's essentially e to the 0, right? So that'... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | So y of 0 is equal to 2, which is equal to, essentially, just substitute 0 into this equation. So it's c1 times e to the minus 2 times 0. What's e to the, that's essentially e to the 0, right? So that's just 1. So it's c1 times 1, which is just c1. Plus c2 times e to the minus 3 times 0. This is e to the 0, so it's jus... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | So that's just 1. So it's c1 times 1, which is just c1. Plus c2 times e to the minus 3 times 0. This is e to the 0, so it's just 1. So plus c2. So the first equation we get when we substitute our first initial condition is essentially c1 plus c2 is equal to 2. Now let's apply our second initial condition that tells us ... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | This is e to the 0, so it's just 1. So plus c2. So the first equation we get when we substitute our first initial condition is essentially c1 plus c2 is equal to 2. Now let's apply our second initial condition that tells us the slope at x is equal to 0. So y prime of 0. So this is our general solution. Let's take its d... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | Now let's apply our second initial condition that tells us the slope at x is equal to 0. So y prime of 0. So this is our general solution. Let's take its derivative. And then we can use this. So y prime of x is equal to what? The derivative of this is equal to minus 2 c1 times e to the minus 2x. |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | Let's take its derivative. And then we can use this. So y prime of x is equal to what? The derivative of this is equal to minus 2 c1 times e to the minus 2x. And what's the derivative of this? It's minus 3 c2 times e to the minus 3x. And now we can use our initial condition. |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | The derivative of this is equal to minus 2 c1 times e to the minus 2x. And what's the derivative of this? It's minus 3 c2 times e to the minus 3x. And now we can use our initial condition. y prime at 0. So when x is equal to 0, what's the right-hand side equal? It's minus 2 times c1. |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | And now we can use our initial condition. y prime at 0. So when x is equal to 0, what's the right-hand side equal? It's minus 2 times c1. And then e to the minus 0, e to the 0, so that's just 1. Minus 3 c2. And then once again, x is 0. |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | It's minus 2 times c1. And then e to the minus 0, e to the 0, so that's just 1. Minus 3 c2. And then once again, x is 0. So e to the minus 3 times 0, that's just 1. So it's just 1 times minus 3 c2. And it tells us that when x is equal to 0, what does this whole derivative equal? |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | And then once again, x is 0. So e to the minus 3 times 0, that's just 1. So it's just 1 times minus 3 c2. And it tells us that when x is equal to 0, what does this whole derivative equal? Well, that equals 3. y prime of 0 is equal to 3. So now we go back into your first year of algebra. We have two equations, two linea... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | And it tells us that when x is equal to 0, what does this whole derivative equal? Well, that equals 3. y prime of 0 is equal to 3. So now we go back into your first year of algebra. We have two equations, two linear equations with two unknowns. And we could solve. Let me write them in a form that you're probably more u... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | We have two equations, two linear equations with two unknowns. And we could solve. Let me write them in a form that you're probably more used to. So the first one is c1 plus c2 is equal to 2. And the second one is minus 2 c1 minus 3 c2 is equal to 3. So what can we do? Let's multiply this top equation by 2. |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | So the first one is c1 plus c2 is equal to 2. And the second one is minus 2 c1 minus 3 c2 is equal to 3. So what can we do? Let's multiply this top equation by 2. There's a ton of ways to solve this. But if you multiply the top equation times 2, you'll get, and I'll do this in a different color, just so that it's chang... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | Let's multiply this top equation by 2. There's a ton of ways to solve this. But if you multiply the top equation times 2, you'll get, and I'll do this in a different color, just so that it's changed. I'm just multiplying the top one by 2. You get 2 c1 plus 2 c2 is equal to 4. And now we can add these two equations. Let... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | I'm just multiplying the top one by 2. You get 2 c1 plus 2 c2 is equal to 4. And now we can add these two equations. Let's see, minus 2 c1 plus 2, those cancel out. So minus 3 plus 2, you get minus c2 is equal to 7. Or we could say that c2 is equal to minus 7. And now we could substitute back in here. |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | Let's see, minus 2 c1 plus 2, those cancel out. So minus 3 plus 2, you get minus c2 is equal to 7. Or we could say that c2 is equal to minus 7. And now we could substitute back in here. We have c1 plus c2, c2 is minus 7, so minus 7 is equal to 9. Or we know that c1 plus 2 c2, that's minus 7, is equal to 2. I'm just sub... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | And now we could substitute back in here. We have c1 plus c2, c2 is minus 7, so minus 7 is equal to 9. Or we know that c1 plus 2 c2, that's minus 7, is equal to 2. I'm just substituting back to this differential equation. Not a differential, it's just a simple linear equation. And then we get c1 is equal to 9. And now ... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | I'm just substituting back to this differential equation. Not a differential, it's just a simple linear equation. And then we get c1 is equal to 9. And now we have our particular solution to the differential equation. So this was our general solution. We can just substitute our c1's and our c2's back in. We have our pa... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | And now we have our particular solution to the differential equation. So this was our general solution. We can just substitute our c1's and our c2's back in. We have our particular solution for those initial conditions, and I think that warrants a different color. So our particular solution is y of x is equal to c1, wh... |
2nd order linear homogeneous differential equations 3 Khan Academy.mp3 | We have our particular solution for those initial conditions, and I think that warrants a different color. So our particular solution is y of x is equal to c1, which we figured out is 9e to the minus 2x plus c2. Well, c2 is minus 7. Minus 7e to the minus 3x. That is the particular solution to our original differential ... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | Now that we know a little bit about the convolution integral and how it applies to the Laplace transform, let's actually try to solve an actual differential equation using what we know. So I have this equation here, this initial value problem, where it says that the second derivative of y plus 2 times the first derivat... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | And that's nice and convenient that those initial conditions tend to make the problem pretty clean. But let's get to the problem. So the first thing we do is we take the Laplace transform of both sides of this equation. The Laplace transform of the second derivative of y is just s squared. This should be a bit of secon... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | The Laplace transform of the second derivative of y is just s squared. This should be a bit of second nature to you by now. It's s squared times the Laplace transform of y, which I'll just write as capital Y of s, minus s. So we start with the same degree as the number of derivatives we're taking, and then we decrement... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | Minus s times y of 0. You kind of think of this as the integral, and you take the derivative 1, so this isn't exactly the derivative of that, minus, you decrement that 1, you just have a 1 there, y prime of 0. And that's the Laplace transform of the second derivative. Now we have to do the Laplace transform of 2 times ... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | Now we have to do the Laplace transform of 2 times the first derivative. That's just going to be equal to plus 2 times s y of s, s times the Laplace transform of y, that's that there, minus y of 0, and we just have 1 left. The Laplace transform of 2y, that's just equal to plus 2 times the Laplace transform of y, and th... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | That's just alpha over s squared plus alpha squared. Now the next thing we want to do is we want to separate out the Laplace transform of y terms, or the y of s terms. Actually, even better, let's get rid of these initial conditions. y of 0 and y prime of 0 is 0, so this term is 0, that term is 0, and that term is 0. S... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | y of 0 and y prime of 0 is 0, so this term is 0, that term is 0, and that term is 0. So our whole expression, and I can get rid of the colors now, is just becomes, let me pick a nice color here, becomes s squared times y of s plus 2s y of s, that's that term right there, plus 2y of s is equal to the right-hand side, is... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | Now we can divide both sides of this equation by this thing right here, by that right there, and we get y of s, the Laplace transform of y, is equal to this thing, alpha over s squared plus alpha squared, times 1 over s squared plus 2s plus 2. I could just say divided by this, but it works out the same either way. Now ... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | Remember, I was doing this in the context of convolution, so I want to look for a Laplace transform that looks like the product of two Laplace transforms. I know what the inverse Laplace transform of this is. In fact, I just took it. It's sine of alpha t. So if I could figure out the inverse Laplace transform of this, ... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | It's sine of alpha t. So if I could figure out the inverse Laplace transform of this, I could at least express our function y of t at least as a convolution integral, even if I don't necessarily solve the integral. From there, it's just calculus, or if it's an unsolvable integral, we can just use a computer or somethin... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | But anyway, let's just try to get this in terms of a convolution integral. So what can I do with this? Let's see, this isn't a perfect square. So if this isn't a perfect square, the next best thing is to try to complete the square here. So let's try to write this as a s squared plus 2s plus something plus 2. I just rew... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | So if this isn't a perfect square, the next best thing is to try to complete the square here. So let's try to write this as a s squared plus 2s plus something plus 2. I just rewrote it like this. And if I wrote this as s squared plus 2s plus 1, that becomes s plus 1 squared. But if I add a 1, I have to also subtract a ... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | And if I wrote this as s squared plus 2s plus 1, that becomes s plus 1 squared. But if I add a 1, I have to also subtract a 1. I can't just add 1s arbitrarily to things. If I add 1, I have to subtract 1 to cancel out with that 1. So I really haven't changed this at all. I just rewrote it like this. But this now, I can ... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | If I add 1, I have to subtract 1 to cancel out with that 1. So I really haven't changed this at all. I just rewrote it like this. But this now, I can rewrite this term right here as s plus 1 squared, and then this becomes plus 1. That's this term right here, is the plus 1. So I could rewrite my whole y of s is now equa... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | But this now, I can rewrite this term right here as s plus 1 squared, and then this becomes plus 1. That's this term right here, is the plus 1. So I could rewrite my whole y of s is now equal to alpha over s squared plus alpha squared times 1 over this thing, s plus 1 squared plus 1. Now, I already said I know what the... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | Now, I already said I know what the inverse Laplace transform of this thing is. Now I just have to figure out what the inverse Laplace transform of this thing is, of this blue thing in the blue box, and then I can express it as a convolution integral. And how do I do that? I could just do it right now. I could just imm... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | I could just do it right now. I could just immediately say that y of t, let me write this down, y of t. So the inverse is equal to the inverse Laplace transform of, obviously, of y of s. Let me write that down. y of s, which is equal to the inverse Laplace transform of these two things, the inverse Laplace transform of... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | And now the convolution theorem tells us that this is going to be equal to the inverse Laplace transform of this first term in the product, so the inverse Laplace transform of that first term, alpha over s squared plus alpha squared convoluted with, I'll do the little convolution sign there, I was about to say convulsi... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | I have these two things, I recognize these independently, I can independently take the inverse of each of these things, so the inverse Laplace transform of their product is going to be the convolution of each of their inverse transforms. Now what's this over here? Why does this at the beginning of the problem? The inve... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | The inverse Laplace transform of this right here is sine of alpha t. Now, and then we're going to convolute that with the inverse Laplace transform of this right here. Let's do a little bit of work on the side just to make sure we get this right. So the Laplace transform of sine of t is equal to 1 over s squared plus 1... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | That looks like this, but I was shifted. I was shifted by minus 1. You might remember that the Laplace transform of e to the at sine of t, when you multiply e to the at times anything, you're shifting its Laplace transform. So that will be equal to 1 over s minus a squared plus 1. And now we have something that looks j... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | So that will be equal to 1 over s minus a squared plus 1. And now we have something that looks just, we essentially shifted it by a. So now we have something that looks very similar to this. If we just set our a to be equal to negative 1, then it fits this pattern. This is s minus negative 1. So the inverse Laplace tra... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | If we just set our a to be equal to negative 1, then it fits this pattern. This is s minus negative 1. So the inverse Laplace transform of this thing right here is just e to the a, which is minus 1, so minus 1t times sine of t. So this is the solution to our differential equation, even though it's not in a pleasant for... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | And we can, if we want to, express it as an integral. I'm not going to actually solve the integral in this problem, because it gets hairy and it's not even clear that, well, I won't even attempt to do it. But I just want to get into a form. And from there, it's just integral calculus. Or maybe a computer. Let's see. So... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | And from there, it's just integral calculus. Or maybe a computer. Let's see. So this is equal to, what's the convolution of these two things? It's the integral from 0 to t of sine of the first function of t minus tau. So I could write this. Well, I can actually switch, and I haven't shown you this, but we can switch th... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | So this is equal to, what's the convolution of these two things? It's the integral from 0 to t of sine of the first function of t minus tau. So I could write this. Well, I can actually switch, and I haven't shown you this, but we can switch the order either way. But actually, let me just do it this way. So it's sine, I... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | Well, I can actually switch, and I haven't shown you this, but we can switch the order either way. But actually, let me just do it this way. So it's sine, I could write this as sine of t minus tau times alpha. I'm taking the sine of all of those things. Times e to the minus tau sine of tau d tau. That's one way that if... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | I'm taking the sine of all of those things. Times e to the minus tau sine of tau d tau. That's one way that if I wanted to express the solution of this differential equation as an integral, I could write it like that. And it actually should be kind of obvious to you that this could go either way. Because when I wrote i... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | And it actually should be kind of obvious to you that this could go either way. Because when I wrote it, when it was a product up here, obviously, a product is, the order does not matter. I could write this term first, or I could write that term first. So regardless of which term is written first, I could take the same... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | So regardless of which term is written first, I could take the same principle would apply. And I'll formally prove it in a future video. So we could have also done it the other way. We could have written this expression as e to the minus t sine of t convoluted with sine of alpha t. And that would be equal to the integr... |
Using the convolution theorem to solve an initial value prob Laplace transform Khan Academy.mp3 | We could have written this expression as e to the minus t sine of t convoluted with sine of alpha t. And that would be equal to the integral from 0 to t of e to the minus t minus tau sine of t minus tau times sine of alpha tau d tau. So these are equivalent. Either of these would be an acceptable answer. If you have a ... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | Because I think you know how to solve essentially the homogenous version. So same problem as we did in the last video. The second derivative of y minus 3 times the first derivative of y minus 4 times the function. And now in the last example, we had the non-homogenous part was 3e to the 2x, but we're tired of dealing w... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And now in the last example, we had the non-homogenous part was 3e to the 2x, but we're tired of dealing with exponent functions, so let's make it a trigonometric function. So let's say it equals 2 sine of x. So the first step you do is what we've been doing, is you essentially solve the homogenous equation. So this le... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So this left-hand side is equal to 0. You do that by getting the characteristic equation, r squared minus 3r minus 4 is equal to 0. You get the solutions r is equal to 4, r is equal to minus 1. And then you get that general solution. And we did this in the last video. You get the general solution of the homogenous, may... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And then you get that general solution. And we did this in the last video. You get the general solution of the homogenous, maybe we'll call this the homogenous solution, y homogenous. We got c1e to the 4x plus c2e to the minus x. And that's all in good. But in order to get the general solution of this non-homogenous eq... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | We got c1e to the 4x plus c2e to the minus x. And that's all in good. But in order to get the general solution of this non-homogenous equation, I have to take the solution of the homogenous equation, if this were equal to 0, and then add that to a particular solution that satisfies this equation, that satisfies that wh... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And here, once again, we'll use undetermined coefficients. And undetermined coefficients, just think to yourself, what function, when I take its second and first derivatives and add and subtract multiples of them to each other, will I get sine of x? Well, two functions end up with sine of x when you take the first and ... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So it's a good guess. And that's really what you're doing in the method of undetermined coefficients. You take a guess of a particular solution, and then you solve for the undetermined coefficients. So let's say that our guess is y is equal to, I don't know, some coefficient times sine of x. And if this was sine of 2x,... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So let's say that our guess is y is equal to, I don't know, some coefficient times sine of x. And if this was sine of 2x, I'd put a times sine of 2x here, just because I want its derivatives to, I still want, no matter what happens here, I want the sine of 2x's or maybe cosine of 2x's to still exist. If this was a sine... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So whatever's here, I want here. Plus cosine of 2x. Plus b, some coefficient, undetermined coefficient, times cosine of x. And once again, if this was sine of 2x, I'd want a cosine of 2x here. So let's figure out its first and second derivatives. So the first derivative of this, y prime, is equal to a cosine of x. Cosi... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And once again, if this was sine of 2x, I'd want a cosine of 2x here. So let's figure out its first and second derivatives. So the first derivative of this, y prime, is equal to a cosine of x. Cosine derivative is minus sine, so minus b sine of x. And then the second derivative, I'll write down here, the second derivat... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And then the second derivative, I'll write down here, the second derivative is equal to what? Derivative of cosine is minus sine, so minus a sine of x minus b cosine of x. I think you're starting to see the hardest thing in most differential equations problems is not making careless mistakes. It's a lot of algebra and ... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | Every time I say that, I tend to make one. So I'm going to focus extra right now. So anyway, let's take these and substitute them back into this non-homogenous equation. Let's see if I can solve for a and b. So the second derivative is that. From that, I'm going to just rewrite it, just so that you see what I'm doing. ... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | Let's see if I can solve for a and b. So the second derivative is that. From that, I'm going to just rewrite it, just so that you see what I'm doing. So I'm going to take the second derivative, y prime prime, so that's minus a sine of x minus b cosine of x. I'm going to add minus 3 times the first derivative to that, a... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So I'm going to take the second derivative, y prime prime, so that's minus a sine of x minus b cosine of x. I'm going to add minus 3 times the first derivative to that, and I'm going to write the sines under the sines and the cosines under the cosines. So minus 3 times this. So the sine is plus 3b sine of x. So 3b sine... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So 3b sine of x minus 3 times this. So minus 3a cosine of x. And then minus 4 times our original function. So minus 4a sine of x minus 4 times this. Minus 4b cosine of x. When I take the sum of all of those, that's essentially the left-hand side of this equation. When I take the sum of all of that, that is equal to 2 s... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So minus 4a sine of x minus 4 times this. Minus 4b cosine of x. When I take the sum of all of those, that's essentially the left-hand side of this equation. When I take the sum of all of that, that is equal to 2 sine of x. I could have written them out in a line, but it would have just been more confusing. And now this... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | When I take the sum of all of that, that is equal to 2 sine of x. I could have written them out in a line, but it would have just been more confusing. And now this makes it easy to add up the sine of x's and the cosine of x's. So if I add up all the coefficients on the sine of x, I get minus a plus 3b minus 4a. So that... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So that looks like minus 5a plus 3b sine of x. Plus, and now what are the coefficients here? What are all the coefficients here? I have minus b and then I have another minus 4b. So minus 5b and then minus 3a. So minus 3a minus 5b. Ran out of space. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | I have minus b and then I have another minus 4b. So minus 5b and then minus 3a. So minus 3a minus 5b. Ran out of space. Cosine of x. This is the cosine of x should go right here. So anyway, how do I solve for a and b? |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | Ran out of space. Cosine of x. This is the cosine of x should go right here. So anyway, how do I solve for a and b? Well, I have minus 5a plus 3b is equal to whatever coefficients in front of sine of x here. So minus 5a plus 3b must be equal to 2. And then minus 3a minus 5b is a coefficient on cosine of x, although I k... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So anyway, how do I solve for a and b? Well, I have minus 5a plus 3b is equal to whatever coefficients in front of sine of x here. So minus 5a plus 3b must be equal to 2. And then minus 3a minus 5b is a coefficient on cosine of x, although I kind of squeezed in the cosine of x here. So this must be equal to whatever th... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And then minus 3a minus 5b is a coefficient on cosine of x, although I kind of squeezed in the cosine of x here. So this must be equal to whatever the coefficient on cosine of x is on the right-hand side. Well, the coefficient of cosine of x on the right-hand side is 0. So that sets up a system of two unknowns with two... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So that sets up a system of two unknowns with two equations, a linear system. So we get minus 5a plus 3b is equal to 2. And we get minus 3a minus 5b is equal to 0. And let's see if I can simplify this a little bit. See, this is a system of two unknowns, two equations. If I multiply the top equation by 5 thirds, I get m... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And let's see if I can simplify this a little bit. See, this is a system of two unknowns, two equations. If I multiply the top equation by 5 thirds, I get minus 25 over 3a plus 5b is equal to 5 thirds times this. 5 thirds times 2 is 10 thirds. And the bottom equation is I have minus 3a minus 5b is equal to 0. Let's add... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | 5 thirds times 2 is 10 thirds. And the bottom equation is I have minus 3a minus 5b is equal to 0. Let's add the two equations. I get 10 thirds is equal to these cancel out. And let's see. That's minus 25 over 3 minus 9 over 3a is equal to 10 thirds. It's getting a little bit messier than I like, but we'll soldier on. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | I get 10 thirds is equal to these cancel out. And let's see. That's minus 25 over 3 minus 9 over 3a is equal to 10 thirds. It's getting a little bit messier than I like, but we'll soldier on. So let's see. Minus 25 minus 9. What's minus 25 minus 9? |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | It's getting a little bit messier than I like, but we'll soldier on. So let's see. Minus 25 minus 9. What's minus 25 minus 9? So that is 34. So we get 34 over 3a is equal to 10 over 3. Or let's see. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | What's minus 25 minus 9? So that is 34. So we get 34 over 3a is equal to 10 over 3. Or let's see. We can multiply both sides by 3, divide both sides by 34. a is equal to 10 over 34, which is equal to 5 over 17. Nice, ugly number. 5 over 17. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | Or let's see. We can multiply both sides by 3, divide both sides by 34. a is equal to 10 over 34, which is equal to 5 over 17. Nice, ugly number. 5 over 17. And now we can solve for b. So let's see. minus 3 times a. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | 5 over 17. And now we can solve for b. So let's see. minus 3 times a. This color is nauseating me. Minus 3 times a. 5 over 17 minus 5b is equal to 0. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | minus 3 times a. This color is nauseating me. Minus 3 times a. 5 over 17 minus 5b is equal to 0. So that's what? Minus 15 over 17 is equal to plus 5b. I just took this, put it on the right hand side. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | 5 over 17 minus 5b is equal to 0. So that's what? Minus 15 over 17 is equal to plus 5b. I just took this, put it on the right hand side. And then divide both sides by 5. Actually, let me make sure. Oh, you know what? |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | I just took this, put it on the right hand side. And then divide both sides by 5. Actually, let me make sure. Oh, you know what? I realize I made a close mistake here. Minus 25 minus 9. That's minus 34 over 3. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | Oh, you know what? I realize I made a close mistake here. Minus 25 minus 9. That's minus 34 over 3. So minus 34a is equal to 10. a is equal to minus 10 over 34, or minus 5 over 17. So minus 3 times minus 5 over 17. So that's equal to plus 15 over 17 is equal to plus 5b. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | That's minus 34 over 3. So minus 34a is equal to 10. a is equal to minus 10 over 34, or minus 5 over 17. So minus 3 times minus 5 over 17. So that's equal to plus 15 over 17 is equal to plus 5b. And then we get b is equal to 3 over 17. That was hairy. And notice, the hard part was not losing your negative signs. |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | So that's equal to plus 15 over 17 is equal to plus 5b. And then we get b is equal to 3 over 17. That was hairy. And notice, the hard part was not losing your negative signs. But anyway, we now have our particular solution to this. Our particular solution is, let me try to write it in a non-nauseating color, although I... |
Undetermined coefficients 2 Second order differential equations Khan Academy.mp3 | And notice, the hard part was not losing your negative signs. But anyway, we now have our particular solution to this. Our particular solution is, let me try to write it in a non-nauseating color, although I think I picked a nauseating one. The particular solution is a minus 5 over 17 sine of x, right, that was a coeff... |
Logistic function application First order differential equations Khan Academy.mp3 | We saw that if we wanted to solve this, and we didn't want one of the constant solutions of n of t equals zero or n of t is equal to k, and we did this in the last few videos, we got the solution that n of t, n of t is equal to, is equal to our initial n, n naught, times our maximum population, all of that over, all of... |
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