 In this video, we provide the solution to question number 11 for practice exam number 3 for math 12-20, in which case we are given a separable differential equation. In this case, it is dy over dx is equal to e to the x over e to the y. And so since it's a separable differential equation, we can solve this by separating the variables. Part A asks us to first separate the variables. So you want a function of y, dy on the left-hand side, and some function of x, dx on the right-hand side. Now with this one, given that, well, of course, the derivative dy over dx is always a ratio. The right-hand side is also a ratio where you have e to the x on top, e to the y on the bottom. The simplest way to do that would just be to cross-multiply, in which case we end up with e to the y dy is equal to e to the x dx. And so very quickly we're able to separate the variables on this one. Sometimes it takes a little bit more effort than that, but that's all we need for this one. That's part A. Part B, then we want to take the separation of the variables and solve for y in that situation. So we're going to integrate both sides of the equation. So we integrate e to the y dy, and then that's equal to the integral of e to the x dx. So taking the antiderivatives on the left-hand side, we get the antiderivative of e to the y with respect to y is e to the y. On the right-hand side, the antiderivative of e to the x with respect to x is e to the x. We do need to have a constant in here because there could be an arbitrary constant. So let's put it on the right-hand side, you don't need it on both sides. So we get e to the y is equal to e to the x plus a constant. To solve for y, we're going to take the natural log of both sides. So y equals the natural log of e to the x plus a constant. Now because we don't know what that constant is, we really cannot simplify the right-hand side. So the general solution to the separable differential equation will be y equals the natural log of e to the x plus c.