Answer
See below
Work Step by Step
We can notice that: $F(x)=y''+6y'+9y=4e^{-3x}$
Obtain: $(D^2+6D+9)y=4e^{-3x}$
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 $(D+3)$ is the annihilator of $G(x)=4e^{-3x}$
Multiply $(D+3)$ to both sides of the given equation we get
$(D+3)^3=0$
Solve the auxiliary equation for the differential equation.
$(r+3)^3=0$
Roots are: $r_1=-3$, as a multiplicity of 3.
The general solution to $(D+3)^3=0$ is
$y(x)=c_1e^{-3x}+c_2xe^{-3x}+c_3x^2e^{-3x}$
The trial solution to initial solution is $y_p=c_3x^2e^{-3x}$.
We obtain the same trial solution with the use of annihilators.
