### Video Transcript

Find the solution for the following
differential equation for π¦ of zero is equal to two. One minus π to the power of
negative π₯ times dπ¦ by dπ₯ is equal to zero, where π¦ of zero is equal to two.

The question gives us a
differential equation and an initial condition, and it wants us to find the solution
to this differential equation. Itβs worth noting we call this an
initial condition because itβs when our variable π₯ is equal to zero. First, we need to know what it
means to find a solution of a differential equation. We need to find π¦ is equal to some
function π of π₯ so that this satisfies our differential equation and our initial
condition. π¦ equals π of π₯ satisfying our
differential equation means that one minus π to the power of negative π₯ times π
prime of π₯ is equal to zero. And π¦ being equal to π of π₯
satisfying our initial condition means that π of zero must be equal to two.

So how are we going to find our
function π of π₯? Well, we can rearrange our
differential equation to make π prime of π₯ the subject. First, we add π to the power of
negative π₯ times π prime of π₯ to both sides of the equation. We get one is equal to π to the
power of negative π₯ times π prime of π₯. Next, we want to divide both sides
of our equation through by π to the power of negative π₯. And we know this is the same as
just multiplying both sides of our equation by π to the power of π₯. This gives us that π prime of π₯
is equal to π to the power of π₯.

But this is an equation for our
derivative function π prime of π₯. We want an expression for our
function π of π₯. And to do this, we need to remember
that integration and differentiation are opposite processes. So weβll integrate both sides of
this equation with respect to π₯. So, because integration and
differentiation are opposite processes, integrating π prime of π₯ with respect to
π₯ will give us π of π₯. And itβs worth reiterating this
will only be up to our constant of integration.

Now, letβs evaluate our other
integral, the integral of π to the power of π₯ with respect to π₯. Well, we know that this is just
equal to π to the power of π₯ plus πΆ. So weβve now found an expression
for π of π₯ up to our constant of integration πΆ. Itβs equal to π to the power of π₯
plus πΆ. Now, to find our value of πΆ, weβre
going to use our initial condition, π¦ of zero is equal to two or, equivalently, π
evaluated at zero is equal to two. Substituting this information into
our expression for π of π₯. We get that two is equal to π to
the zeroth power plus πΆ. And of course, π to the zeroth
power is equal to one. So we can rearrange this equation
and find that πΆ is equal to one.

So weβve shown that πΆ is equal to
one. So weβll use this in our expression
for π of π₯. And this gives us that π of π₯ is
π to the power of π₯ plus one. And remember, π¦ is equal to π of
π₯. So this gives us the equation π¦ is
equal to π to the power of π₯ plus one. And this is our final answer, weβve
shown the solution to the differential equation one minus π to the power of
negative π₯ times dπ¦ by dπ₯ is equal to zero and π¦ at zero is equal to two is π¦
is equal to π to the power of π₯ plus one.