Answer
$y(x)=C_1e^{-x}+C_2\cos 2x+C_3 \sin 2x+C_4 x\cos 2x+C_5x \sin 2x$
Work Step by Step
Solve the auxiliary equation for the differential equation. $$(r^2+4)^2(r+1)=0$$
Factor and solve for the roots. $r_1=-1$, as a multiplicity of $1$ and $r_2=-2i, r_3=2i$ as a multiplicity of $2$.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-x}$ and $y_2= \cos 2x $ and $y_3=\sin 2x$ and $y_4=x \cos 2x$ and $y_5=x \sin 2x$
Therefore, the general equation is equal to $y(x)=C_1e^{-x}+C_2\cos 2x+C_3 \sin 2x+C_4 x\cos 2x+C_5x \sin 2x$
