Answer
$y(x)=C_1e^{-2x}+C_2xe^{-2x}+C_3e^{3x}$
Work Step by Step
Given $(D^2+4D+4)(D-3)y=0$
Solve the auxiliary equation for the differential equation.
$(r^2+4r+4)(r-3)=0$
Roots are: $r_1=-2$ as a multiplicity of 2 and $r_2=3$ as a multiplicity of 1.
This implies that there are two independent solutions to the differential equation
$y_1(x)=e^{-2x}$
$y_2=xe^{-2x}$ and
$y_3(x)=e^{3x}$
Therefore, the general equation is equal to $y(x)=C_1e^{-2x}+C_2xe^{-2x}+C_3e^{3x}$
