### Video Transcript

Solve the differential equation dπ¦
by dπ₯ plus π¦ equals one.

A separable equation is a first
order differential equation in which the expression for dπ¦ by dπ₯ can be factored
as a function of π₯ times a function of π¦. In other words, it can be written
in the form π of π₯ times π of π¦. So letβs rearrange our equation dπ¦
by dπ₯ plus π¦ equals one so that itβs in this form. To achieve this, weβre going to
subtract π¦ from both sides of the equation. And we obtain dπ¦ by dπ₯ to be
equal to one minus π¦. Now, it may not look like it, but
we have achieved our aim. Our function in π¦ is one minus π¦,
and our function in π₯ is simply one.

Then, to solve this equation, weβre
going to rewrite it using differentials. Now, remember, dπ¦ by dπ₯
absolutely isnβt a fraction. But we do treat it a little like
one for the purposes of this process. We begin by dividing both sides of
our equation by one minus π¦. And we see that this is equivalent
to saying one over one minus π¦ dπ¦ equals one dπ₯. And now, weβre ready to integrate
both sides of this equation. So how do we integrate one over one
minus π¦ with respect to π¦? Well, we begin by quoting the
general result for the integral of one over π₯ with respect to π₯. Itβs the natural log of the
absolute value of π₯ plus some constant of integration, π.

What weβre going to do is perform a
substitution for our integral. Weβre going to let π’ be equal to
one minus π¦ so that dπ’ by dπ¦ is equal to negative one. We could write this equivalently as
negative dπ’ equals dπ¦. And then weβre going to replace dπ¦
with negative dπ’ and one minus π¦ with π’. And we see that we now need to
integrate negative one over π’ with respect to π’. Well, thatβs negative the natural
log of the absolute value of π’ plus that constant of integration π. By replacing π’ with one minus π¦,
we find the integral of one over one minus π¦ to be the negative natural log of the
absolute value of one minus π¦ plus a constant of integration which Iβm going to
call π.

When we integrate one with respect
to π₯, itβs a little more straightforward. We get π₯ plus a second constant of
integration. Letβs call that π. Letβs subtract π from both sides
of our equation and multiply through by negative one. That gives us the natural log of
the absolute value of one minus π¦ equals negative π₯ plus π one. π one is a new constant, and itβs
achieved by subtracting π from π and multiplying by negative one. And then, we notice that we can
raise both sides of this equation as a power of π. So we obtain the absolute value of
one minus π¦ to be equal to π to the power of negative π₯ plus π one.

By using the laws of exponents,
though, we see that we can rewrite π to the power of negative π₯ plus π one as π
to the power of negative π₯ times π to the power of π one. But, of course, π to the power of
π one is itself a constant. Letβs call that π two. And we see we can rewrite the
right-hand side as π two times π to the power of negative π₯. Well, of course, weβre dealing with
the absolute value of one minus π¦. And we can say that this means that
one minus π¦ could be equal to positive π two times π to the power of negative π₯
or negative π two times π to the power of negative π₯. But π two is a constant, so we
donβt actually need to write that.

In our final step, weβre going to
add π¦ to both sides of the equation and then subtract π two π to the power of
negative π₯. That gives us π¦ equals one minus
π two times π to the power of negative π₯. But of course, once again, since π
two is a constant, we can change this to plus π. And we see that π¦ equals one plus
π times π to the power of negative π₯ is the solution to our separable
differential equation.