Answer
See below
Work Step by Step
We can notice that: $F(x)=y''+6y'+9y=4e^{-2x}$
Obtain: $(D^2+6D+9)y=4e^{-2x}$
Solve the auxiliary equation for the differential equation.
$(r+3)^2=0$
Roots are: $r_1=-3$, as a multiplicity of 2.
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{-3x}+c_2xe^{-3x}$
We can see that $(D+2)$ is the annihilator of $G(x)=4e^{-2x}$
Multiply $(D+3)$ to both sides of the given equation we get
$(D+3)^2(D+2)=0$
Solve the auxiliary equation for the differential equation.
$(r+2)(r+3)^2=0$
Roots are: $r_1=-3$, as a multiplicity of 2 and $r_2=-2$ as a multiplicity of 1.
The general solution to $(D+2)(D+3)^2=0$ is
$y(x)=c_1e^{-3x}+c_2xe^{-3x}+c_3e^{-2x}$
The trial solution to initial solution is $y_p=c_3e^{-2x}$.
We obtain the same trial solution with the use of annihilators.
