Answer
$y_p(x)=A \cos 4x +B \sin 4x+x(C \cos 2x+D \sin 2x)$
Work Step by Step
The General equation for the polynomial has the form of $G(x)=e^{kx} P(x) \sin mx $ or $G(x)=e^{kx} P(x) \cos mx $
The trial solution for the method of undetermined coefficients can be determined as:
$y_p(x)=e^{kx} Q(x) \sin mx +e^{kx} R(x) \cos mx$
In the given problem, we have $m=k=1$ and so, the degree of the $B(x) $ and $C(x)$ is $1$.
Hence, the trial solution for the method of undetermined coefficients is as follows:
$y_p(x)=A \cos 4x +B \sin 4x+x(C \cos 2x+D \sin 2x)$
