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 d𝑦/dπ‘₯ = 5 cos π‘₯ with the initial condition 𝑦(πœ‹/2) = 4? [A] 𝑦 = 5 sin π‘₯ βˆ’ 4 [B] 𝑦 = βˆ’5 sin π‘₯ + 9 [C] 𝑦 = 5 sin π‘₯ βˆ’ 1 [D] 𝑦 = βˆ’5 sin π‘₯ βˆ’ 9 [E] 𝑦 = 5 sin π‘₯ + 1

02:10

Video Transcript

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

A differential equation is an equation which contains derivatives. Our job then is to solve the differential equation d𝑦 by dπ‘₯ equals five cos π‘₯. And this is going to involve integration at some point. Here I have d𝑦 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 𝑦 with respect to π‘₯ with respect to π‘₯ is 𝑦.

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 𝑦 is equal to five sin π‘₯ plus 𝑐, that constant of integration.

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

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

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.