Answer
$y(x)=C_1e^{-4x}+C_2 e^{2x}+C_3e^{4x}$
Work Step by Step
Solve the auxiliary equation for the differential equation. $$(r-2)(r^2-16)=0$$
Factor and solve for the roots. $$(r-2)(r^2-16)^2=0$$ $$(r-2)(r-4)(r+4)=0$$
Roots are: $r_1=-4$ and $r_2=2$ and $r_3=4$
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-4x}$ and $y_2= e^{2x}$ and $y_3=e^{4x}$
Therefore, the general equation is equal to $y(x)=C_1e^{-4x}+C_2 e^{2x}+C_3e^{4x}$