# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 29

Which of the following is the solution to the differential equation dπ¦/dπ₯ = 5 cos π₯ with the initial condition π¦(π/2) = 4? [A] π¦ = 5 sin π₯ β 4 [B] π¦ = β5 sin π₯ + 9 [C] π¦ = 5 sin π₯ β 1 [D] π¦ = β5 sin π₯ β 9 [E] π¦ = 5 sin π₯ + 1

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### 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. Itβs π¦ equals five sin π₯ minus one.