Worked example: separable differential equations

- [Instructor] What we're gonna be doing 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 to the 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 wanna do is get the Ys and the DYs on one side, and then the Xs and the DXs on the other side. And we really treat these differentials kind of like variables, which is a little hand-wavy with the mathematics. But that's what we will do. So let's see. If we multiply both sides times Y, so we're gonna multiply both sides times Y, what are we going to get? We're gonna get Y times a derivative of Y, with respect to X, is equal to E to the X, and now we can multiply both sides by the differential, DX; multiply both of them by DX; those cancel out. And we are left with Y times DY is equal to E to the X, DX. And now we can take the integral of both sides. So let us do that. So 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 first, but so now when we take the anti-derivative, it will be Y squared, and then we divide by that incremented exponent, is equal to, well the exciting thing about E to the X is that it's anti-derivative, and its derivative, is E to the X, is equal to E to the X, plus is equal to E to the X plus C. And so we can leave it like this if we like. In fact this right over here is, this isn't an explicit function. Y here isn't an explicit function of X. We could actually say Y is equal to the plus or minus square root of two times 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 squared times sine of X. Pause the video and see if you can find the general solution here. So once again, we wanna separate our Ys and our Xs. So let's see, we can multiply both sides times Y to the negative two power, Y to the negative two, Y to the negative two, these become one, and then we could also multiply both sides times DX. So if we multiply DX here, those cancel out, and then we multiply DX here, and so we're left with Y to the negative two power times DY is equal to sine of X, DX, and now we just can integrate both sides. Now what is the anti-derivative of Y to the negative two? Well, once again we use the reverse power rule. We increment the exponents, so it's gonna be Y to the negative one, and then we divide by that newly incremented exponent. So we divide by negative one. Well that would just make this think negative. That is going to be equal to... So, what's the anti-derivative of sine of X? Well, it is, you might recognize it if I put a negative there, and a negative there. The anti-derivative of negative sine of X, well that's cosine of X. So this whole thing is gonna be negative cosine of X, or another way to write this: I could multiply both sides times a negative one, and so these would both become positive, and so I could write one over Y is equal to cosine of X, and actually let me write it this way, plus C; don't wanna forget my plus Cs. Plus C, or I could take the reciprocal of both sides if I wanna solve explicitly for Y, I could get Y is equal to one over cosine of X plus C as our general solution. And we're done. That was strangely fun.