Answer
See below
Work Step by Step
We are given $y''-4y=5e^x$
Solve the auxiliary equation for the differential equation.
$r^2-4=0$
Factor and solve for the roots.
$(r-2)(r+2)=0$
Roots are: $r_1=2$, as a multiplicity of 1, $r_2=-2$ as a multiplicity of 1.
The general solution is
$y(x)=c_1e^{4x}+c_2e^{-4x}$
The trial solution for $y_p=A_0e^x$ can be computed as by plugging back into the given differential equation.
$A_0e^x-4A_0e^x=5e^x$
then
$A_0=-\frac{5}{3}$
Hence,
$y(x)=c_1e^{4x}+c_2e^{-4x}-\frac{5}{3}e^x$
