## Calculus 8th Edition

$y_p(x)=A \cos 4x +B \sin 4x+x(C \cos 2x+D \sin 2x)$
We are give that $y''+4y=\cos 4x +\cos 2x$ 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: $y_p(x)=e^{\alpha x} B(x) \sin mx +e^{\beta x} C(x) \cos mx$ Here, we have $m=4, k=2$ Thus, the trial solution for the method of undetermined coefficients is: $y_{p_1}(x)=A \cos 4x +B \sin 4x; y_{p_2}(x)=C \cos 2x+D \sin 2x$ Hence, $y_p(x)=A \cos 4x +B \sin 4x+x(C \cos 2x+D \sin 2x)$