Worked example: separable differential equation (with taking log of both sides) | Khan Academy

Let's say we need to find a solution to the differential equation that the derivative of y with respect to x is equal to x squared over e to the y. Pause this video and see if you can have a go at it. I will give you a clue: it is a separable differential equation.

All right, now let's do this together. So whenever you see any differential equation, the first thing you should try to see is: is it separable? When I say separable, I mean I can get all the expressions that deal with y on the same side as the dy, and I can separate those from the expressions that deal with x, and they need to be on the same side as my differential dx.

So how can we do that? Well, if we multiply both sides by e to the y, then e to the y will go away right over here, so we will get rid of this y expression from the right-hand side. Then we can multiply both sides by dx.

So if we did that, let me move my screen over a bit to the left. I’m going to multiply both sides by e to the y, and I’m also going to multiply both sides by dx. Multiplying by dx gets rid of the dx on the left-hand side, and it sits on the right-hand side with the x squared. So all of this is now e to the y dy is equal to x squared dx.

Just the fact that we were able to do this shows that it is separable. Now what we can do now is integrate both sides of this equation.

So let's do that. What is the integral of e to the y dy? Well, one of the amazing things about the expression, or you could say the function, something is equal to e to the — and normally we say e to the x, but in this case it’s e to the y — is that the anti-derivative of this is just e to the y. We’ve learned that in multiple videos; I always express my fascination with it.

So this is just e to the y, and likewise, if you took the derivative of e to the y with respect to y, it would be e to the y. Remember this works because we are integrating with respect to y here. So the integral of e to the y with respect to y is e to the y, and that is going to be equal to the anti-derivative of x squared.

Well, that is, we increment the exponent, so that gets to x to the third power, and we divide by that incremented exponent. Since I took the indefinite integral of both sides, I have to put a constant on at least one of these sides. So let me throw it over here, plus c.

Just to finish up, especially on a lot of examinations like the AP exam, they might want you to write in a form where y is explicitly an explicit function of x. So to do that, we can take the natural log of both sides.

So we take the natural log of that side, and we take the natural log of that side. Well, the natural log of e to the y — what power do we have to raise e to get to e to the y? Well, that's why we took the natural log; this just simplifies as y.

And we get y is equal to the natural log of what we have right over here: x to the third over three plus c. And we are done.