Answer
See below
Work Step by Step
$y''+4y'+4y=169\sin 3x$
The complementary function is
$y_c(x)=c_1xe^{-2x}+c_2e^{-2x}$
We consider the complex differential equation
$z''+4z'+4z=169e^{3ix}$
An appropriate complex-valued trial solution for this differential equation is:
$z_p(x)=A_0e^{3ix}$
where $A_0$ is a complex constant.
so that $z _p$ is a solution if and only if
$-5A_0e^{3ix}+12iA_0e^{3ix}=169A_0e^{3ix}\\
A_0=\frac{169}{12i-5}=-5+12i$
A particular solution is:
$y_p(x)=Re_{\{z_p\}}=-12\cos 3x -5\sin 3x$
so that the general solution is
$y(x)=c_1xe^{-2x}+c_2e^{-2x}-12\cos 3x -5\sin 3x$
