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Current time:0:00Total duration:5:37

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.