Answer
$y_p(x)=A_0xe^x$
Work Step by Step
We are given $y''+2y'-3y=5e^{x}$
We have: $F(x)=5e^x$
Obtain: $5e^x(D-1)=0$
So, $D-1$ is the annihilator of $F(x)=5e^x$
Therefore, the general solution for the given differential equation is:
$y(x)=C_1e^x+C_2xe^x+A_0xe^x$
The trial solution is
$y_p(x)=A_0xe^x$
