Answer
See below
Work Step by Step
We are given $y''+6y'+9y=4e^{-2x}$
Solve the auxiliary equation for the differential equation.
$r^2+6r^2+9r=0$
Factor and solve for the roots.
$(r+3)^2=0$
Roots are: $r_1=-3$, as a multiplicity of 2.
The general solution is
$y(x)=c_1e^{-3x}+c_2xe^{-3x}+A_0e^{-2x}$
The trial solution for $y_p=A_0e^{-2x}$ can be computed as by plugging back into the given differential equation.
$4A_0e^{-2x}-12A_0e^{-2x}+9A_0e^{-2x}=4e^{-2x}$
then
$A_0e^{-2x}=4e^{-2x}$
We have $A_0=4$
Hence,
$y(x)=c_1e^{-3x}+c_2xe^{-3x}+4e^{-2x}$
