Answer
The general solution is of the form: $y=c_1cos2x+c_2sin2x-\frac{1}{4}xcos2x$
Work Step by Step
The characteristic equation for $y''+4y=sin2x$ is $t^2+4=0$ with solutions $t=\pm 2i$
For particular solution, we have $Axcos2x+Bxsin2x$ gives $A=-1/4$ and $B=0$
The general solution is of the form: $y=c_1cos2x+c_2sin2x-\frac{1}{4}xcos2x$
