Answer
See below
Work Step by Step
We can notice that: $F(x)=y'''-6y''+25y'=x^2$
Obtain: $(D^3-6D^2+25D)y=4e^{-2x}$
Solve the auxiliary equation for the differential equation.
$r^3-6r^2+25r=0\\
r(r^2-6r+25)^2=0$
Roots are: $r_1=0$, as a multiplicity of 1, $r_2=3-4i$ as a multiplicity of 1 and $r_3=3+4i$ as a multiplicity of 1.
Therefore, the general solution for the given differential equation is:
$y(x)=c_1+c_2e^{3x}\cos 4x+c_3e^{3x}\sin 4x$
We can see that $D^3$ is the annihilator of $G(x)=x^2$
Multiply $D^3$ to both sides of the given equation we get
$D^3(D^3-6D^2+25D)=0$
Solve the auxiliary equation for the differential equation.
$r^3(r^3-6r^2+25r)=0\\
r^4(r^2-6r+25)=0$
Roots are: $r_1=0$ as a multiplicity of 4, $r_2=3-4i$ as a multiplicity of 1 and $r_2=3+4i$ as a multiplicity of 1.
The general solution to $D^3(D^3-6D^2+25D)=0$ is
$y(x)=c_1+c_2e^{3x}\cos 4x+c_3e^{3x}\sin 4x+c_4x+c_5x^2$
The trial solution to initial solution is $y_p=c_4x+c_5x^2$.
We obtain the same trial solution with the use of annihilators.
