### Video Transcript

Which of the following is the
solution to the differential equation dπ¦ by dπ₯ equals five cos π₯ with the initial
condition π¦ at π by two is equal to four? Is it a) π¦ equals five sin π₯
minus four? Is it b) π¦ equals negative five
sin π₯ plus nine? Is it c) π¦ equals five sin π₯
minus one, d) π¦ equals negative five sin π₯ minus nine, or e) π¦ equals five sin π₯
plus one?

A differential equation is an
equation which contains derivatives. Our job then is to solve the
differential equation dπ¦ by dπ₯ equals five cos π₯. And this is going to involve
integration at some point. Here I have dπ¦ by dπ₯ as some
function of π₯. In this case, we can simply
integrate both sides of our equation with respect to π₯. The integral of the derivative of
π¦ with respect to π₯ with respect to π₯ is π¦.

On the right-hand side, we can take
the five outside of the integral sign and focus on integrating cos π₯ with respect
to π₯. We know that the integral of cos π₯
with respect to π₯ is sin π₯ plus that constant of integration. So we can see then that π¦ is equal
to five sin π₯ plus π, that constant of integration.

But we have the condition π¦ of π
by two is equal to four. In other words, when π₯ is equal to
π by two, π¦ is four. So we can substitute these values
into our solution to the differential equation to find the value of π. That gives us four equals five sin
of π by two plus π. And of course sin of π by two is
one.

So we have four equals five times
one plus π. Well, five times one is just
five. And we can solve for π by
subtracting five from both sides of this equation. And we see that π is equal to
negative one. We replace π with negative one in
the equation π¦ equals five sin π₯ plus π. And we see that the solution to the
differential equation dπ¦ by dπ₯ equals five cos π₯ with that initial condition is
π. Itβs π¦ equals five sin π₯ minus
one.