Answer
$y(x)=C_1e^{-2x}+C_2 e^{2x}+C_3xe^{2x}$
Work Step by Step
Solve the characteristic equation for the differential equation. $$r^3-2r^2-4r+8=0$$
Factor and solve for the roots. $$(r+2)(r-2)^2=0$$ $$r_1=-2$$ as a multiplicity $1$ and $r=2$ as as a multiplicity $2$.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-2x}$ and $y_2= e^{2x}$ and $y_3=xe^{2x}$
Therefore, the general equation is equal to $y(x)=C_1e^{-2x}+C_2 e^{2x}+C_3xe^{2x}$