### Video Transcript

Find the solution of the
differential equation dπ¦ by dπ₯ equals π₯ times π to the power of π¦ that
satisfies the initial condition π¦ of zero equals zero.

A separable differential 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 dπ¦ by dπ₯ equals π of π₯ times π of π¦. Now, itβs quite clear that our
equation dπ¦ by dπ₯ equals π₯ times π to the power of π¦ is of this form. So, how do we solve it?

Well, weβre going to write it in
differential form so that all of our π¦s are on one side and all of our π₯s are on
the other. Weβre going to divide through by π
to the power of π¦. And weβre going to split up our dπ¦
by dπ₯. Now, remember, this isnβt a
fraction. But in this process, we do treat it
a little bit like one. So, we find that one over π to the
power of π¦ dπ¦ equals π₯ dπ₯. Now, in fact, weβre going to write
one over π to the power of π¦ as π to the power of negative π¦ just to make the
next step easier.

And then, our next step is to
integrate both sides of our equation with respect to their defined variable. So, on the left, we have the
indefinite integral of π to the power of negative π¦ with respect to π¦. And on the right, we have the
indefinite integral of π₯ with respect to π₯. Now, if we recall the indefinite
integral of π to the power of ππ₯ with respect to π₯ for some constant values of
π is one over π times π to the power of ππ₯ plus some constant of integration
π. So, when we integrate π to the
power of negative π¦ with respect to π¦, we get one over negative one times π to
the power of negative π¦ plus a constant π΄. And, of course, thatβs just
negative π to the power of negative π¦ plus π΄. The integral of π₯ is fairly
straightforward. We think of it as π₯ to the power
of one. We then increase the exponent by
one to give us two and divide by that new value. So, we get π₯ squared over two plus
some other constant of integration π΅.

Letβs combine our constants of
integration by subtracting π΄ from both sides. On the left-hand side, weβre simply
left with negative π to the power of negative π¦. And then on the right, we have π₯
squared over two plus a new constant πΆ. Letβs multiply through by negative
one. And we get negative π₯ squared over
two plus a different constant. Remember, itβs essentially negative
πΆ. But letβs call that plus π·. And then, weβre gonna find the
natural log of both sides of our equation. And the reason we do that is
because the natural log of π to the power of negative π¦ is just negative π¦.

And so, weβre almost there with a
general solution to our differential equation. To make π¦ the subject, we are
gonna multiply through by negative one one more time. And so, we find the general
solution to our differential equation to be π¦ equals negative the natural log of
negative π₯ squared over two plus a constant π·.

Well, there is one bit of
information we havenβt used yet. We can find the particular solution
to our differential equation by using the initial condition that π¦ of zero equals
zero. In other words, when π₯ is equal to
zero, π¦ is equal to zero. We replace π₯ and π¦ with zero. And we get zero equals negative the
natural log of negative zero squared over two plus π·. That gives us zero equals the
negative natural log of π·. But of course, we can multiply
through by negative one. And we find that zero is equal to
the natural log of π·.

Now, we can observe a solution
here. But weβre going to look at the full
method. Weβre going to raise both sides of
this equation as a power of π. That gives us π to the power of
zero equals π to the power of the natural log of π·. Well, anything to the power of zero
is one. And we know that π to the natural
log of π· is simply π·. So, we find π· to be equal to
one. And then, we can replace π· with
one in our general solution.

And we now have a particular
solution to the differential equation. Itβs π¦ equals negative the natural
log of negative π₯ squared over two plus one. And you might see this
alternatively written as negative the natural log of negative one-half π₯ squared
plus one.