Answer
See below
Work Step by Step
We are given $y''-y=4e^{x}$
Solve the auxiliary equation for the differential equation.
$r^2-1=0$
Factor and solve for the roots.
$(r+1)(r-1)=0$
Roots are: $r_1=1$, as a multiplicity of 1 and $r_2=-1$, as a multiplicity of 1
The general solution is
$y(x)=c_1e^{x}+c_2e^{-x}$
The trial solution for $y_p=A_0e^{x}+Be^{-x}$ can be computed as by plugging back into the given differential equation.
Then we will have
$A_0=2x\\
B_0=-e^{2x}$
Hence,
$y(x)=c_1e^{x}+c_2e^{-x}+2xe^x-e^{x}$
