Answer
See below
Work Step by Step
$y''-4y=100xe^{x}\sin x$
The complementary function is
$y_c(x)=c_1 e^{2x}+c_2 e^{-2x}$
We consider the complex differential equation
$z''-4z=100xe^{(1+i)x}$
An appropriate complex-valued trial solution for this differential equation is:
$z_p(x)=(A_0+B_0x)e^{(1+i)x}$
where $A_0,B_0$ are complex constants.
so that $z _p$ is a solution if and only if
$2(1+i)B_0e^{(1+i)x}-2(2-i)B_0xe^{(1+i)x}-(4-2i)A_0e^{(1+i)x}=100xe^{(1+i)x}\\
\rightarrow A_0=-\frac{i}{2}$
Hence, $z=-\frac{i}{2}xe^{(-1+i)x}$
A particular solution is:
$y_p(x)=Re_{\{z_p\}}=-e^{-x}x\cos x$
so that the general solution is
$y(x)=c_1e^{2x}+c_2e^{-2x}-e^{-x}x\cos x$
