 So let's take a further look at exact differential equations. So remember this is based on the idea that if we have a differential equation, then if it's exact, then the solutions will be level curves of some function f of x, y. On the other hand, if it's not exact, we try to find an integrating factor that will make it exact. So for example, let's consider this differential equation. Now, maybe we're fantastically lucky and we've been given an exact differential equation. But I wouldn't count on it. If it's exact, then the differential equation can be expressed as a complete derivative of f of x, y. So we can set up our complete derivative of f of x, y, and compare it to the equation given. And if these things are really true, then a coefficient of dy dx has to be the partial of capital F with respect to y. And the leftover bits must be the partial derivative of capital F with respect to x. Well, keep in mind this is what we want, but you can't always get what you want. And so we should check the mixed partial derivatives. And since they're not equal, this equation is not exact. So let's see if we can find an integrating factor. Suppose i of x is an integrating factor. Then we can multiply our equation by i of x. And since we want our equation to be exact, we must have this expression equal to the total differential of some function capital F. And this requires that our coefficient of dy dx be our partial of f with respect to y. And the leftovers have to be the partial of f with respect to x. And because we want our equation to be exact, the mixed partial derivatives have to be equal. So let's find those mixed partials. And so whatever our integrating factor is, it has to make this mixed partial equal to this mixed partial. So let's see if we can solve that equation. So the unknown here is really i of x. So let's try and get all of our i of x terms together. We'll factor. And we have an i prime of x here on the left hand side and an i of x over on the right hand side. So let's at least try to get them together. We'll divide both sides by i of x. And so here we have the quotient of two functions of x, but it's expressed as a function of x and y. And so since the left hand side is a function of x only, we can't solve this equation. No such function exists. So that means we can't find an integrating factor of the form i of x. But why assume i is a function of x only? So suppose i of y is an integrating factor. Then if we multiply our differential equation by i of y we get. And since we want this equation to be exact, this must be the total differential of some function. So again, our coefficient of dy dx must be the partial derivative of f with respect to y. And the leftovers have to be the partial of f with respect to x. And our mixed partials must be equal. So in order for these two sides to be equal, then we can say something about i of y. So let's solve for i of y. And getting all of our i and i prime terms onto one side, we find that the quotient of our two functions of y, i of y and its derivative, is a function of y. And this at least is a possibility. So we can anti-differentiate both sides, do a little algebra, and find our integrating factor is 1 over y. And remember the reason we're doing all of this is to give ourselves an exact differential equation. So using our integrating factor gives us a new differential equation which is exact. Of course we still need to solve the differential equation. And so our expression must be the total derivative of some function capital F. So again the coefficient of dy dx must be the partial of f with respect to y. And so I can find capital F of xy by anti-differentiation where our constant is actually a function of x only. Similarly our leftover bits must be the partial of f with respect to x and so we can find a second expression for our function where our constant is a function of y only. And we compare our two expressions for f of xy and simplify. We see that our function of x only must be x cubed. Our function of y only must be a true constant so we'll omit it. And our solutions will be the level curves of f of xy equals 8xy plus log y plus x cubed. And again we can go back and write this as a general solution in the form function equal to some undetermined constant.