Answer
$y_{p_1}(x)=A e^{3x}$ 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.
Work Step by Step
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.
