## Calculus: Early Transcendentals 8th Edition

$y_{p_1}(x)=A e^{3x}$ and $y_{p_2}(x)=x(Bx+C) \cos 2x+x(Dx+E) \sin 2 x$ We can see that there is no term of $y_{p_2}$ is a solution of the complimentary equation.
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 $1$. Here, the sum of the coefficients of a differential equation is $1+3-4=0$. When the sum of the coefficients of a differential equation is zero then, we have $y_p(x)=x e^{ax} Q(x)$ We get $y_{p_1}(x)=A e^{3x}$ and $y_{p_2}(x)=x(Bx+C) \cos 2x+x(Dx+E) \sin 2 x$ We can see that there is no term of $y_{p_2}$ is a solution of the complimentary equation.