Answer
$y(x)=-2e^{-2x} -2 x e^{-2x}+2e^{2x}$
Work Step by Step
Solve the auxiliary equation for the differential equation. $$r^3+2r^2-4r-8=0$$
Roots are: $(r-2) (r+2)^2 =0 \implies r_1=-2; r_2=2$
This implies that there are $\bf{3}$ independent solutions to the differential equation and the general equation is equal to $y(x)=C_1 e^{-2x} +C_2 x e^{-2x}+C_3 e^{2x}$
Next, after applying the initial conditions we get: $C_1=-2; C_2=-2; C_3=2$
Therefore, $y(x)=-2e^{-2x} -2 x e^{-2x}+2e^{2x}$ is the solution to the initial-value problem.
