Answer
See below
Work Step by Step
We can notice that: $y'''+9y''+24y'+16y=8e^{-x}+1$
Obtain: $(D^3+9D^2+24D+16)y=8e^{-x}+1$
Solve the auxiliary equation for the differential equation.
$r^3+9r^2+24r+16=0\\
(r+1)(r+4)^2=0$
Roots are: $r_1=-4$ as a multiplicity of 2 and $r_2=-1$ as a multiplicity of 1.
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{-4x}+c_2xe^{-4x}+c_3e^{-x}$
We can see that $D+1$ is the annihilator of $G(x)=8e^{-x}$
and $D$ is the annihilator of $H(x)=1$
then $F(x)=G(x)+H(x)=8e^{-x}+1$
Multiply $D(D+1)$ to both sides of the given equation we get
$D(D+1)^2(D+4)^2=0$
Solve the auxiliary equation for the differential equation.
$r(r+1)^2(r+4)^2=0$
Roots are: $r_1=-4$ as a multiplicity of 2, $r_2=-1$ as a multiplicity of 2 and $r_3=0$ as a multiplicity of 1.
The general solution to $D(D+1)^2(D+4)^2=0$ is
$y(x)=c_1e^{-4x}+c_2xe^{-4x}+c_3e^{-x}+c_4xe^{-x}+c_5$
The trial solution to initial solution is $y_p=c_4xe^{-x}+c_5$.
We obtain the same trial solution with the use of annihilators.
