Answer
$y_p(x)=A xe^x+B \cos x+C \sin x$
Work Step by Step
Consider $G(x)=e^{\alpha x} A(x) \sin mx $ or $G(x)=e^{\beta x} A(x) \cos mx $
The trial solution for the method of undetermined coefficients is defined as:
$y_p(x)=e^{\alpha x} B(x) \sin mx +e^{\beta x} C(x) \cos mx$
On comparing the above equation with the given equation, we get
$m=k=1$
Thus, the trial solution for the method of undetermined coefficients is:
$y_{p_1}(x)=A xe^x$ and $y_{p_2}(x)=B \cos x+C \sin x$
Hence, $y_p(x)=A xe^x+B \cos x+C \sin x$
