Separable equations example (old)

I think it's reasonable to do one more separable differential equations problem, so let's do it. The derivative of y with respect to x is equal to y cosine of x divided by 1 plus 2y squared, and they give us an initial condition that y of 0 is equal to 1. Or when x is equal to 0, y is equal to 1. And I know we did a couple already, but another way to think about separable differential equations is really, all you're doing is implicit differentiation in reverse. Or another way to think about it is whenever you took an implicit derivative, the end product was a separable differential equation. And so, this hopefully forms a little bit of a connection. Anyway, let's just do this. We have to separate the y's from the x's. Let's multiply both sides times 1 plus 2y squared. We get 1 plus 2y squared times dy dx is equal to y cosine of x. We still haven't fully separated the y's and the x's. Let's divide both sides of this by y, and then let's see. We get 1 over y plus 2y squared divided by y, that's just 2y, times dy dx is equal to cosine of x. I can just multiply both sides by dx. 1 over y plus 2y times dy is equal to cosine of x dx. And now we can integrate both sides. So what's the integral of 1 over y with respect to y? I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function than that, whose derivative is actually 1 over y, and that's the natural log of the absolute value of y. And this is just a slightly broader function, because it's domain includes positive and negative numbers, it just excludes 0. While natural log of y only includes numbers larger than 0. So natural log of absolute value of y is nice, and it's actually true that at all points other than 0, its derivative is 1 over y. It's just a slightly broader function. So that's the antiderivative of 1 over y, and we proved that, or at least we proved that the derivative of natural log of y is 1 over y. Plus, what's the antiderivative of 2y with respect to y? Well, it's y squared, is equal to-- I'll do the plus c on this side. Whose derivative is cosine of x? Well, it's sine of x. And then we could add the plus c. We could add that plus c there. And what was our initial condition? y of 0 is equal to 1. So when x is equal to 0, y is equal to 1. So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of one, e to the what power is 1? Well, 0, plus 1 is-- sine of 0 is 0 --is equal to C. So we get c is equal to 1. So the solution to this differential equation up here is, I don't even have to rewrite it, we figured out c is equal to 1, so we can just scratch this out, and we could put a 1. The natural log of the absolute value of y plus y squared is equal to sine of x plus 1. And actually, if you were to graph this, you would see that y never actually dips below or even hits the x-axis. So you can actually get rid of that absolute value function there. But anyway, that's just a little technicality. But this is the implicit form of the solution to this differential equation. That makes sense, because the separable differential equations are really just implicit derivatives backwards. And in general, one thing that's kind of fun about differential equations, but kind of not as satisfying about differential equations, is it really is just a whole hodgepodge of tools to solve different types of equations. There isn't just one tool or one theory that will solve all differential equations. There are few that will solve a certain class of differential equations, but there's not just one consistent way to solve all of them. And even today, there are unsolved differential equations, where the only way that we know how to get solutions is using a computer numerically. And one day I'll do videos on that. And actually, you'll find that in most applications, that's what you end up doing anyway, because most differential equations you encounter in science or with any kind of science, whether it's economics, or physics, or engineering, that they often are unsolveable, because they might have a second or third derivative involved, and they're going to multiply. I mean, they're just going to be really complicated, very hard to solve analytically. And actually, you are going to solve them numerically, which is often much easier. But anyway, hopefully at this point you have a pretty good sense of separable equations. They're just implicit differentiation backwards, and it's really nothing new. Our next thing we'll learn is exact differential equations, and then we'll go off into more and more methods. And then hopefully, by the end of this playlist, you'll have a nice toolkit of all the different ways to solve at least the solvable differential equations. See you in the next video.