# Exact equations example 1

First example of solving an exact differential equation. Created by Sal Khan.
Video transcript
OK, I filled your brain with a bunch of partial derivatives and psi's, with respect to x's and y's. I think now it's time to actually do it with a real differential equation, and make things a little bit more concrete. So let's say I have the differential, y, the differential equation, y cosine of x, plus 2xe to the y, plus sine of x, plus-- I'm already running out of space-- x squared, e to the y, minus 1, times y prime, is equal to 0. Well, your brain is already, hopefully, in exact differential equations mode. But if you were to see this pattern in general, where you see a function of x and y, here-- this is just some function of x and y-- and then you have another function of x and y, times y prime, or times dy, d of x, your brain should immediately say if this is inseparable. And I'm not going to try to make it separable, just because that'll take a lot of time. But if it's not separable, your brain said, oh, maybe this is an exact equation. And, you say, let me test whether this is an exact equation. So if this is an exact equation, this is our function M, which is a function of x and y. And this is our function N, which is a function of x and y. Now, the test is to see if the partial of this, with respect to y, is equal to the partial of this, with respect to x. So let's see. The partial of M, with respect to y, is equal to-- let's see, y is-- so this cosine of x is just a constant, so it's just cosine of x. Cosine of x plus-- now, what's the derivative? Well, 2x is just a constant, what's the derivative of e to the y, with respect to y? Well, it's just e to the y, right? So we have the constant on the outside, 2x times the derivative, with respect to y, so it's 2xe to the y. Fair enough. Now, what is the partial derivative of this, with respect to x? So N sub x, or the partial of N, with respect to x-- so what's the derivative of sine of x, with respect to x? Well, that's easy, that's cosine of x, plus 2x times e to the y, right? e and y is just a constant, because y is constant when we're taking the partial, with respect to x. So plus 2xe to the y. And then minus 1, the derivative of a constant, with respect to anything is going to be 0. So the derivative of N-- the partial of N, with respect to x, is cosine of x, plus 2xe to the y, which, lo and behold, is the same thing as the derivative, the partial of M, with respect to y. So there we have it. We've shown that M of y is equal to-- or the partial of M, with respect to y-- is equal to the partial of N, with respect to x, which tells us that this is an exact equation. Now, given that this is an exact equation-- oh, my wife snuck up behind me, I was wondering. I thought there was some critter in my house, or something. Anyway, so we know that this is an exact equation, so what does that tell us? Well, that tells us that there's some psi, where the partial derivative of psi, with respect to x, is equal to M, and the partial derivative of psi, with respect to y, is equal to N. And if we know that psi, then we can rewrite our differential equation as the derivative of psi, with respect to x, is equal to 0. So let's solve for psi. So we know that the partial of psi, with respect to x, is equal to M. So we could write that. We could write the partial of psi, with respect to x, is equal to M, which is y cosine of x, plus 2xe to the y. That's just here. That's my M of x. We could have done it the other way. We could have said the partial of y-- the partial of psi, with respect to y, is this thing over here. But let's just do it with x. Now, to at least get kind of a first approximation of what size-- not an approximation, but to start to get a sense of it-- let's take the derivative of both sides, with respect to-- sorry, take the antiderivative-- take the integral of both sides, with respect to x. So if you take the derivative of this, with respect to x, if you integrate-- sorry, if you were to take the antiderivative of this, with respect to x. So let me just write that down. The partial, with respect to x. We're going to [? integrate ?] it, with respect to x. That is going to be equal to the integral of this whole thing, with respect to x. Cosine of x plus 2xe to the y. We're integrating with respect to x. And normally when you integrate with respect to x, you'd say, OK, plus c, right? But it actually could be a plus-- since this is a partial, with respect to x, we could have had some function of y here in general, because y, we treat it as a constant, right? And that makes sense, because if you were to take the partial of both sides of this, with respect to x, if you were to take the partial of a function that is only a function of y, with respect to x, you would have gotten a 0 here. So when you take the antiderivative, you were, like, oh well, there might have been some function of y here that we lost when we took the partial, with respect to x. So anyway, this will simplify to psi. psi is going to be equal to the integral, with respect to x, or the antiderivative, with respect to x, here, plus some function of y that we might have lost when we took the partial, with respect to x. So let's do that. Let's figure out this integral. I'll do it in blue So y is just a constant. So the antiderivative of y cosine of x, is just y sine of x, plus-- either the y is constant, so 2x. The antiderivative of 2x, with respect to x, is x squared, so it's x squared e to the y. And then plus some function of y. And if you want to verify this, you should take the partial of this, with respect to x. If you take the partial of this, with respect to x, you're going to get this in here, which is our function, M, up here. And then when you take the partial of this, with respect to x, you'll get 0, and it'll get lost. OK, so we're almost there. We've almost figured out our psi, but we still need to figure out this function of y. Well, we know that if we take the partial of this, with respect to y, since this is an exact equation, we should get this. We should get our N function. So let's do that. So the partial-- I'll switch notation, just to expose you to it-- the partial psi, with respect to y, is going to be equal to-- so here, y sine of x, sine of x is just a constant. y is just y, so the derivative of this, with respect to y, is just sine of x. Plus the derivative of e to the y is e to the y. x squared is just a constant. So it's just x squared e to the y, plus-- what's the partial of f of y, with respect to y? It's going to be f prime of y. Well, what did we do? We took M, we integrated with respect to x, and we said, well, we might have lost some function of y, so we added that to it. And then we took the partial of that side that we've almost constructed, and we took the partial of that, with respect to y. Now, we know, since this is exact, that that is going to equal our N. So our N is up there. Cosine of x plus-- So that's going to be equal to-- I want to make sure I can read it up there-- to our N, right? Oh no, sorry. N is up here. Our N is up here. Sine of x-- let me write that-- sine of x plus x squared, e to the y, minus 1. So sine of x plus x squared, e to the y, minus 1. That was just our N, from our original differential equation. And now we can solve for f prime of y. So let's see, we get sine of x plus x squared, e to the y, plus f prime of y, is equal to sine of x plus x squared, e to the y, minus 1. So let's see, we can delete sine of x from both sides. We can delete x squared e to the y from both sides. And then what are we left with? We're left with f prime of y is equal to 1. And then we're left with f of y is equal to-- well, it equals y plus some constant, c, right? So what is our psi now? We wrote our psi up here, and we had this f of y here, so we can rewrite it now. So psi is a function of x and y-- we're actually pretty much almost done solving it-- psi is a function of x and y is equal to y sine of x, plus x squared, e to the y, plus y-- oh, sorry, this is f prime of y, minus 1. So this is a minus 1. So this is a minus y plus c. So this is going to be a minus y plus c. So we've solved for psi. And so what does that tell us? Well, we said that original differential equation, up here, using the partial derivative chain rule, that original differential equation, can be rewritten now as the derivative dx of psi is equal to-- psi is a function of x and y-- is equal to 0. Or if you were to integrate both sides of this, you would get that psi of xy is equal to c is a solution of that differential equation. So if we were to set this is equal to c, that's the differential equation. So we could say, y sine of x plus x squared, e to the y, minus y-- now we could say, plus this c-- plus this c, you call it c1, is equal to c2. Well, you could subtract the c's from both sides, and just be left with a c at the end. But anyway, we have solved this exact equation, one, first, by recognizing it was exact, by taking the partial of this, with respect to y, and seeing if that was equal to the partial of N, with respect to x. Once we saw that they were equal, we're like, OK, this is going to be exact. So let's figure out psi. Since this is exact, M is going to be the partial of psi, with respect to x. N is the partial of psi, with respect to y. Then to figure out y, we integrated M, with respect to x, and we got this. But since we said, oh, well, instead of a plus c, it could have been a function of y there, because we took the partial, with respect to x, so this might have been lost. To figure out the function of y, we then took our psi that we figured out, took the partial of that, with respect to y, got this. And we said, this was an exact equation, so this is going to equal our N of x y. We set those equal to each other, and then we solved for f of y. And then we had our final psi. Our final psi was this. And then the differential equation, because of the chain rule of partial derivatives, we could rewrite the differential equation as this. The solution is this, and so this is the solution to our differential equation. See you in the next video.