Answer
$y_{p_1}(x)=A e^{3x}$ and $y_{p_2}(x)=x(Bx+C) \cos 2x+x(Dx+E) \sin 2 x$
There is no term of $y_{p_2}$ that is a solution of the complimentary equation.
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: $y''+4y=e^{3x}+x\sin 2x$
Here, we have $m=k=1$
Thus, the trial solution for the method of undetermined coefficients is:
$y_{p_1}(x)=A e^{3x}$ and $y_{p_2}(x)=x(Bx+C) \cos 2x+x(Dx+E) \sin 2 x$
There is no term of $y_{p_2}$ that is a solution of the complimentary equation.
