Answer
$y(x)=C_1e^{-2x}+C_2xe^{-2x}+C_3e^{2x}+C_4xe^{2x}$
Work Step by Step
Solve the auxiliary equation for the differential equation. $$r^4-8r^2+16=0$$
Factor and solve for the roots. $$(r-2)^2(r+2)^2=0$$
Roots are: $r_1=-2$, as a multiplicity of $2$ and $r_2=2$ as a multiplicity of $2$.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-2x}$ and $y_2=xe^{-2x}$ and $y_3=e^{2x}$ and $y_4=xe^{2x}$
Therefore, the general equation is equal to $y(x)=C_1e^{-2x}+C_2xe^{-2x}+C_3e^{2x}+C_4xe^{2x}$