Answer
See below
Work Step by Step
The general solution for the given differential equation is: $y(x)=c_1\cos 2x+c_2 \sin 2x+A_0x\sin 2x+B_0x\cos 2x$
The trial solution for $y_p= A_0x\sin 2x+B_0x\cos 2x$ can be computed as by plugging back into the given differential equation.
So, we have: $y''+4y=8\sin 2x\\(D^2+4)y_p(x)=8\sin 2x\\ (D^2+4)(A_0x\sin 2x+B_0x\cos 2x)=8\sin 2x\\ 4A_0\cos 2x-4B_0\sin 2x=8\sin 2x$
On comparing coefficients, we get: $A_0=0$ and $B_0=-2$
Therefore, the general solution for the given differential equation is: $y(x)=c_1\cos 2x+c_2 \sin 2x-2x\cos x$
