My problem is the following: $y'' + 4y = 4\sin(2x)$
I tried making a particular solution to this by assuming that $y = A\sin(2x) + B\cos(2x)$
This gives me the derivate $y' = 2A\cos(2x) - 2B\sin(2x)$ and
The second derivate $y'' = -4A\sin(2x) - 4B\cos(2x)$
If I plug the second derivate and 4 times the equation into the starting problem:
I get: $-4A\sin(2x) - 4B\cos(2x) + 4[A\sin(2x) + B\cos(2x)] = 4\sin(2x) + 0cos(2x)$
As you can see, I get $0 = 4\sin(2x)$, which is impossible because $0$ is not $4$.
I can't find this problem anywhere, how do I go about solving this problem?