 So let's see if we can find a way of turning a non-exact differential equation into an exact differential equation through the use of some integrating factor. So let's work on an actual differential equation. Let's check this for exactness. So, again, dy over dx is not a fraction. Nevertheless, we'll separate our differentials as a reminder that if this equation is exact, the derivative with respect to y must be this thing, and the derivative with respect to x must be this thing, and our check is whether the mixed partials are the same. So here, I've already differentiated with respect to y, so I'll differentiate with respect to x and get. And here, I've already differentiated with respect to x, so I'll differentiate with respect to y and get. And since these are different, this can't be exact. So could we try to find an integrating factor? Now, since our differential equation involves both x and y, the natural question to ask is, could we find an integrating factor i of x, y that will make this an exact differential equation? Well, let's try it out. So we'll take our differential equation, we'll multiply through by our function i of x, y, and separating our differentials. So if this is exact, our coefficient of dy is going to be a partial derivative of some function with respect to y, and the coefficient of dx is going to be a partial derivative of the same thing with respect to x. And if there's any hope for this to work out, the mixed partial derivatives must be equal. So we've already differentiated f with respect to y, so if we differentiate with respect to x, we get. And likewise here, we've already differentiated with respect to x, so if we differentiate with respect to y, we get. And so we need our mixed partial derivatives to be equal. And so this gives us the equation. And so if we can solve this partial differential equation, we can find the integrating factor. Yeah, no. The problem is we haven't yet talked about how to solve partial differential equations. Tactically, that isn't really a problem because remember the problem exists independent of our ability to solve it. The real problem is that in general, we don't have any way of solving partial differential equations. So we can't use this method effectively. So what can we do? So again, a useful rule in life and in mathematics is that it's easier to obtain forgiveness than permission. Now, the important requirement here is that it has to work, otherwise you won't get forgiveness either. So rather than assume our integrating factor is a function of both x and y, let's see if we can find an integrating factor that's a function of x only that will make our differential equation exact. So again, we'll take our differential equation and multiply it by our integrating factor. We'll separate the differentials. And so this coefficient of dy must have been the partial with respect to y. The coefficient of dx must have been the partial with respect to x. We'll find our mixed partial derivatives. And again, the requirement is that we want our mixed partial derivatives to be equal. So we'll set them equal to each other. And we'll do a little algebra. Both sides have a factor of e to the y, which is never going to be zero, so we can divide by it. We can get all of our xi of x terms over onto the right-hand side. And so whatever our integrating factor is, the derivative must be minus 1 over x times the integrating factor itself. And this looks awfully familiar. Let's see what we can do with this. So the derivative of our integrating factor is minus 1 over x i of x, and so rearranging things a little bit, we get. And now, since both sides only rely on the variable x, we can find the anti-derivative with respect to x. And so we have log of our integrating factor is minus the log of x, and we'll do a little bit of algebra. And we find our integrating factor is 1 over x. So remember, the whole point of this was to find a function that we could multiply our differential equation by to get an exact differential equation. So with integrating factor 1 over x, we have our differential equation, and if we multiply this by 1 over x, we get... which is exact. And so that means the solutions are level curves of f of x, y equals c. Well, if the solutions are level curves, then our differential equation will have the form of a complete derivative. And so our partial of f with respect to y is the coefficient of dy dx. And the partial of f with respect to x is the leftovers. So we can find f by anti-differentiating with respect to y, which gives us our function up to some function of x. Likewise, we can anti-differentiate with respect to x and get our function up to some function of y. But they have to be the same function, so we can compare the two. And so c1 of x has to be x itself, and c2 of y has to be 0. And so putting everything together, we find these solutions are going to be level curves of f of x, y equals x plus x e to power y.