### Video Transcript

Find a relation between π’ and π‘
given that dπ’ by dπ‘ equals one plus π‘ to the fourth power over π’π‘ squared plus
π’ to the fourth power times π‘ squared.

Now, this differential equation
might look really nasty. But it is, in fact, a separable
differential equation. In this case, thatβs one where an
expression for dπ’ by dπ‘ can be written as some function of π’ times some function
of π‘. So how are we going to achieve
that? Well, we begin by factoring the
denominator by π‘ squared. And we find that dπ’ by dπ‘ is
equal to one plus π‘ to the fourth power over π‘ squared times π’ plus π’ to the
fourth power. We can now write this as some
function of π‘ times some function of π’. Itβs one plus π‘ to the fourth
power over π‘ squared times one over π’ plus π’ to the fourth power.

Letβs begin by multiplying both
sides of this equation by π’ plus π’ to the fourth power. Then, we recall that whilst dπ’ by
dπ‘ is not a fraction, we treat it a little like one. And we can say that π’ plus π’ to
the fourth power dπ’ is equal to one plus π‘ to the fourth power over π‘ squared
dπ‘. And thatβs great because weβre now
ready to integrate both sides of our equation. The left-hand side is quite
straightforward to integrate. The integral of π’ is π’ squared
over two. And the integral of π’ to the
fourth power is π’ to the fifth power over five. Donβt forget, since this is an
indefinite integral, we need that constant of integration π.

On the right-hand side, weβre going
to rewrite our integrand as one over π‘ squared plus π‘ to the fourth power over π‘
squared. And that simplifies to one over π‘
squared plus π‘ squared. But, of course, one over π‘ squared
is the same as π‘ to the power of negative two. And we can now integrate as
normal. When we integrate π‘ to the power
of negative two, we add one to the exponent to get negative one and then we divide
by that number. π‘ squared becomes π‘ cubed over
three. And we have a second constant of
integration; Iβve called that π. Now, of course, we can rewrite π‘
to the power of negative one over negative one as negative one over π‘.

Our final step is to subtract our
constant π from both sides of the equation. That gives us a new constant
π. Thatβs just the difference between
π and π. So weβve found a relation between
π’ and π‘. Itβs π’ to the fifth power over
five plus π’ squared over two equals negative one over π‘ plus two cubed over three
plus π.