Question

I'm trying to solve this DE using "variation of parameters" . The problem is: 4y"+y = cos x. The following is what I've managed to get so far, but I think I'm missing something: V1' cos ((1/2)x) + V2' sin ((1/2)x) = 0 and -V1'(1/2)sin((1/2)x) + V2'(1/2) cos((1/2)x) = cos x therefore V1' = V2' tan ((1/2)x). Then we take that and substitute to solve for V2'. -(1/2)V2' tan ((1/2)x) sin ((1/2)x) + (1/2)V2' cos((1/2)x) = cos x leading to V2' to = -2cos(x) cos ((1/2)x) and then we substitute V2' to solve for V1' and get V1' = -2cos(x) sin((1/2)x) and this is followed by taking the derivative o