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Find constants $ A $ and $ B $ such that the function $ y = A \sin x + B \cos x $ satisfies the differential equation $ y" + y' - 2y = \sin x. $

$A=-\frac{3}{10}, \quad B=-\frac{1}{10}$

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Harvey Mudd College

Baylor University

University of Nottingham

Boston College

It's Claire's, the one you married here. So we have. Why is equal to a sign minus B sign. We're gonna differentiate. Get a co sign minus B Sign we differentiate again. We got negative A sign minus b co sign. So when we substitute, we get the second derivative. Thus, the first derivative minus two. Why? It's equal to sign the nine week put negative, eh? Sine x minus b co sign waas a co sign minus b sign minus two times a sign must be co sign and it's equal to sign when we simplify and we get okay minus three b. No sign. Plus negative three A minus B I'm sign, which is equal to zero. I'm school sign. It was one time sign. No, we look at both sides. We have the equations, which is a minus three b equals Sarah Negative. Three a minus B people's one. So we're gonna find be in a So you get three a my next 93 My next three a minus b equals one. We get b is equal to negative one time and A is equal to negative. Three time