Worked example: separable differential equations | AP Calculus AB | Khan Academy

What we're going to do in this video is get some practice finding general solutions to separable differential equations.

So, let's say that I had the differential equation Dy/Dx, the derivative of y with respect to X, is equal to e^X over y. See if you can find the general solution to this differential equation. I'm giving you a huge hint: it is a separable differential equation.

All right, so when we're dealing with a separable differential equation, what we want to do is get the Y's and the Dy on one side and then the x's and the DXs on the other side. We really treat these differentials kind of like variables, which is a little handwavy with the mathematics, but that's what we will do.

So, let's see. If we multiply both sides by y, we're going to get y * derivative of y with respect to X is equal to e^X. Now, we can multiply both sides by the differential DX. If we multiply both of them by DX, those cancel out and we are left with y * Dy is equal to e^X DX.

Now we can take the integral of both sides, so let us do that. What is the integral of y Dy? Well, here we would just use the reverse power rule. We would increment the exponent, so it's y to the 1, but now when we take the anti-derivative, it will be y^2. Then we divide by that incremented exponent is equal to...

Well, the exciting thing about e^X is its anti-derivative is, and its derivative is e^X. So, we can say it is equal to e^X + C. We can leave it like this if we like; in fact, this right over here, this isn't an explicit function.

Y here isn't an explicit function of X. You could actually say Y is equal to the plus or minus square root of (2 * all of this business), but this would be a pretty general relationship which would satisfy this separable differential equation.

Let's do another example. So, let's say that we have the derivative of y with respect to X is equal to, let's say, it's equal to y^2 * sin(X). Pause the video and see if you can find the general solution here.

So, once again, we want to separate our y's and our x's. Let's see, we can multiply both sides by y to the -2 power. These become one, and then we could also multiply both sides by DX. So, if we multiply DX here, those cancel out and then we multiply DX here.

Now, we're left with y to the -2 power * Dy is equal to sin(X) DX. Now we just can integrate both sides. What is the anti-derivative of y to the -2? Well, once again, we use the reverse power rule.

We increment the exponent, so it's going to be y to the 1, and then we divide by that newly incremented exponent. Dividing by negative 1 would just make this thing negative, so that is going to be equal to...

So, what's the anti-derivative of sin(X)? Well, you might recognize it. If I put a negative there and a negative there, the anti-derivative of negative sin(X) well that's cosine of X. So, this whole thing is going to be negative cosine of X.

Another way to write this: I can multiply both sides by -1, and so these would both become positive. I could write 1/Y is equal to cosine of X.

Actually, let me write it this way: plus C. Don't want to forget the plus C's, plus C. Or, I can take the reciprocal of both sides. If I want to solve explicitly for y, I could get Y is equal to 1 over (sin(X) + C) as our general solution, and we're done.

That was strangely fun!