Answer
$y_p(x)=A xe^x+B \cos x+C \sin x$
Work Step by Step
Consider $G(x)=e^{ax} A(x) \sin mx $ or $G(x)=e^{ax} A(x) \cos mx $
The trial solution for the method of undetermined coefficients can be calculated as:
$y_p(x)=e^{ax} B(x) \sin mx +e^{ax} C(x) \cos mx$
Given: $y''-3y'+2y=e^x +\sin x$
Here, we have $m=k=1$
Thus, the trial solution for the method of undetermined coefficients is:
$y_{p_1}(x)=A xe^x$ and $y_{p_2}(x)=B \cos x+C \sin x$
or, $y_p(x)=A xe^x+B \cos x+C \sin x$
