Answer
$y_p(x)=x(A x^3+Bx^+Cx+D)e^{x}$
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:
$y_p(x)=e^{ax} B(x) \sin mx +e^{ax} C(x) \cos mx$
when the sum of the coefficients of a differential equation is zero.
Then, we have $y_p(x)=x e^{ax} B(x)$
Given: $y''+3y'-4y=(x^3+x)e^x$
Here, the sum of the coefficients of a differential equation is $1+3-4=0$.
Thus, the trial solution for the method of undetermined coefficients is:
$y_p(x)=x(A x^3+Bx^+Cx+D)e^{x}$
