### Video Transcript

Solve the differential equation dπ§
by dπ‘ plus π to the power of two π‘ plus two π§ equals zero.

Remember, a separable differential
equation is one for which the expression for dπ§ by dπ‘ can be expressed as some
function of π§ times some function of π‘. So how exactly are we going to
achieve that for our equation? Well, weβre going to begin by
subtracting π the power of two π‘ plus two π§ from both sides of our equation. We then recall that the laws of
exponents tell us that π₯ to the power of π plus π can be written as π₯ to the
power of π times π₯ to the power of π. So we can write negative π to the
power of two π‘ plus two π§ as negative π to the power of two π‘ times π to the
power of two π§.

Now, of course, dπ§ by dπ‘ isnβt a
fraction, but we treat it a little like one. And we can say that this is
equivalent to one over π to the power of two π§ dπ§ equals negative π to the two
π‘ dπ‘. And this is great because weβre now
ready to integrate both sides. It can be simpler to express one
over π to the power of two π§ as π to the power of negative two π§. And then, we quote the general
result for the integral of π to the power of ππ₯ for some constant π. Itβs π to the power of ππ₯ over
π plus π. So this means the integral of π to
the power of negative two π§ is negative π to the power of negative two π§ over
two. And the integral of negative π to
the two π‘ is negative π to the two π‘ over two.

Our next step is to subtract π one
from both sides of the equation and then multiply through by negative two. Remember, in solving our
differential equation, ideally, we want an equation for π§ in terms of π‘, So we
find that π to the power of negative two π§ is equal to π to the power of two π‘
plus π three. π three is a new constant obtained
by subtracting π one from π two and then multiplying by negative two. To solve for π§, we find the
natural log of both sides of this equation. But the natural log of π to the
power of negative two π§ is just negative two π§. So our final step is to divide
through by negative two. And weβve solved our differential
equation. π§ is equal to negative one-half
times the natural log of π to the power of two π‘ plus some constant; letβs call
that π.