I'm struggling to figure out a differential equation problem. I'm trying to solve
$$x'' + x' + x = \sin2t$$
where I'm supposed to find a particular solution. I'm not completely sure how to break it up into $x(t) = A\cos(t) + B\sin(t)$
I substitute in for x(t), taking the derivative for $x''$ and $x'$, and I get: $$ (-Acos(t) -Bsin(t)) + (-Asin(t) + Bcos(t)) + (Acos(t) + Bsin(t)) = sin(2t) $$ which simplifies to: $$-Asin(t) + Bcos(t) = sin(2t)$$
At this point I get a bit lost. In other solutions, I'd find the values of A and B at $x=0$ and $x= pi/2$, but if I do that here I find the solutions: A= 0, B=0 This in turn, when substituted into the original $Acos(t) + Bsin(t) =$ particular solution just yields 0, which is not the answer.
What did I do wrong? How can I correctly solve this question? Thank you