Answer
$y(x)=C_1e^{-3x}+C_2xe^{-3x}+C_3x^2e^{-2x}+C_4e^{-2x}\cos 3x+C_5e^{-2x}\sin 3x$
Work Step by Step
Given $(D+3)^3(D^2-4D+13)y=0$
Solve the auxiliary equation for the differential equation.
$(r+3)^3(r^2-4r+13)=0$
Roots are: $r_1=-3$, as a multiplicity of 3 and $r_2=-2 \pm 3i$ as a multiplicity of 2.
This implies that there are two independent solutions to the differential equation
$y_1(x)=e^{-3x}$
$y_2=xe^{-3x}$ and
$y_3(x)=x^2e^{-3x}\\
y_4(x)=e^{-2x}\cos 3x\\
y_5(x)=e^{-2x}\sin 3x$
Therefore, the general equation is equal to $y(x)=C_1e^{-3x}+C_2xe^{-3x}+C_3x^2e^{-3x}+C_4e^{-2x}\cos 3x+C_5e^{-2x}\sin 3x$
