Separable differential equations examples

- [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.