 So solving exact differential equation seems to be pretty easy. So we might try to solve this differential equation, and as before, we assume the solutions are level curves f of x, y equals c, where the total derivative of f has this form. Comparing our two equations tells us the partial derivative of f with respect to y has to be x, y, and the partial derivative of f with respect to x must be x squared. And so merrily we anti-differentiate. And remember that our constants are in fact functions of the other variable. And since our two versions of f of x, y have to be the same function, we can compare them directly and we find... And there's a problem here. c1 of x has to be a function of x only. Now the problem isn't that we have minus one-half x, y squared here, and maybe we can get rid of that with whatever this other function is. But since c2 y can only be a function of y, it will be impossible to eliminate y from the right-hand side. And that means we can't actually find these functions that will make our differential equation exact. We don't have an exact differential equation. Now you might look at this and say, gee, it would have been nice to know before doing all the work whether the equation was exact. And in fact, we can test for exactness. This relies on an important theorem for multivariable calculus. Let f be a sufficiently nice function of x and y. In practice this means it has all the derivatives we need it to have and they're all as continuous as they need to be. Then the second partial derivative of f with respect to x and then y should be equal to the second partial derivative of f with respect to y and then x. This is referred to as the equality of mixed partial derivatives. And this suggests the following approach. If our solutions corresponds to the level curves of f of x, y equals c, then a differential equation is of the form partial of f with respect to y times dy dx plus partial of f with respect to x. And this means we already have the first partial derivatives. And what that means is we can check for exactness by comparing the mixed partials. If we take this and find the partial with respect to x and this and find the partial with respect to y, they should be equal. And if they're not, we don't have an exact equation. So let's test for exactness. So a useful way to remember which derivative to take is to separate the differentials. And our standard disclaimer, dy over dx is not a fraction, although we do treat it like one. If we treat it like one, we can separate our differentials to get the equation xy dy plus x squared dx equal to zero. Now, this suggests that xy is the result of having differentiated with respect to y. And so our derivative with respect to y is xy. Likewise, x squared appears to have been the result of having differentiated with respect to x. And so our partial with respect to x is x squared. So now we find the mixed partials. Here, we've already differentiated with respect to y. So let's differentiate with respect to x, and that gives us y. And here, we've already differentiated with respect to x. So we should differentiate with respect to y, and that gives us zero. And since they're not equal, they couldn't have come from a single function f of xy. And so this is not an exact differential equation. Since this equation isn't exact, we can't solve it by assuming it's solutions to be level curves with some f of xy. Or can we? To answer that question, let's solve another ordinary differential equation. So again, let's rewrite this by splitting the differential. So again, the way to read this is that 2y appears to have come from taking the partial derivative with respect to y. So our partial with respect to y is 2y. Likewise, 2x appears to have come from taking the partial derivative with respect to x. And so our partial of f with respect to x is 2x. So now we find the mixed partial derivatives. Here, we've already differentiated with respect to y. So we'll differentiate with respect to x, and get... Over here, we've already differentiated with respect to x. So we'll differentiate with respect to y, and get... And because they're equal, the equation is exact, and it's plausible that we have a solution in the form of a level curve. Of course, we still have to find that solution. Since this is exact, we know the solutions will be level curves of f of xy equals c. And again, our total derivative has this form. And so comparing our equations tells us that we must have the partial of f with respect to y has to be 2y. And the partial of f with respect to x has to be 2x. So finding the corresponding antiderivatives, we get our functions f of xy is y squared plus a function of x, or x squared plus a function of y. Again, our two expressions for f of xy must represent the same function, so we can compare them. And if we look at this, in order for the two sides to be equal, this function of x, c1 of x, has to be x squared. And that tells us c2 of y has to be y squared plus some constant, but we can include our constant as part of the definition of the level curve. And so this means that our general solutions are going to be of the form x squared plus y squared equals c. Finally, we can incorporate our initial conditions to find the specific solution. If y of 3 is equal to 4, then x is equal to 3, y is equal to 4, and we can find c by substituting these into the general equation. So c is equal to 25, and the particular solution corresponding to our initial condition is x squared plus y squared equals 25. It's worth making the following observation. Notice that if we took the exact differential equation, which we were able to solve, and multiplied by x over 2, we got the non-exact differential equation, which was the one we weren't able to solve. And this suggests we can start with a non-exact differential equation and multiply by some function to make it exact. In other words, we might be able to find an integrating factor. We'll take a look at that next.