## Calculus: Early Transcendentals 8th Edition

$y_p(x)=A xe^x+B \cos x+C \sin x$
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 $1$. $y_{p_1}(x)=A xe^x$ and $y_{p_2}(x)=B \cos x+C \sin x$ Hence, the trial solution for the method of undetermined coefficients is as follows: $y_p(x)=A xe^x+B \cos x+C \sin x$