### Video Transcript

Find a relation between π¦ and π₯
given that π₯π¦ times π¦ prime is equal to π₯ squared minus five.

Now, this first step isnβt entirely
necessary. But it can make it a little easier
to see what to do next. We write π¦ prime using Leibniz
notation. And we see that π₯π¦ dπ¦ by dπ₯ is
equal to π₯ squared minus five. And then we recall that a separable
differential equation is one in which the expression for dπ¦ by dπ₯ can be written
as some function of π₯ times some function of π¦. Now, in fact, if we divide both
sides of our equation by π₯π¦, we see that we can achieve this. We get dπ¦ by dπ₯ equals π₯ squared
minus five over π₯π¦ or π₯ squared minus five over π₯ times one over π¦. And thatβs great because π of π₯
is, therefore, π₯ squared minus five over π₯. And our function of π¦ is one over
π¦.

Now, in fact, we just performed
this to demonstrate that we did indeed have a separable differential equation. We couldβve kept π¦ on the
left-hand side as shown. And then, we perform this rather
strange step. dπ¦ by dπ₯ isnβt a fraction, but we do treat it a little like one. And we say that π¦ dπ¦ equals π₯
squared minus five over π₯ dπ₯. Our next step is to integrate both
sides of this equation. Remember, to integrate a polynomial
term whose exponent is not equal to negative one, we add one to the exponent and
divide by that new number. So the integral of π¦ is π¦ squared
over two plus some constant of integration π.

Then, it might look like we need to
perform some sort of substitution to evaluate this integral. But actually, if we separate our
fraction into π₯ squared over π₯ minus five over π₯, we find that we need to
integrate π₯ minus five over π₯ with respect to π₯. Then, we recall the general result
for the integral of one over π₯. Itβs the natural log of the
absolute value of π₯ plus π. And so, when we integrate the
right-hand side of our equation, we get π₯ squared over two minus five times the
natural log of the absolute value of π₯ plus some second constant of integration
which Iβve called π.

Our last step is to subtract π
from π and then multiply the entire equation by two. When we do, we find that π¦ squared
equals π₯ squared minus 10 times the natural log of the absolute value of π₯ plus
π. This π is a different constant
achieved by subtracting π from π and then multiplying by two. And so, given our differential
equation, weβve found a relation between π¦ and π₯. π¦ squared equals π₯ squared minus
10 times the natural log of the absolute value of π₯ plus π.