Answer
$y(x)=C_1e^{- x}\cos x+C_2 e^{-x}\sin x$
Work Step by Step
Solve the characteristic equation for the differential equation. $$r^2+2r+2=0$$
Factor and solve for the roots. $$r_1=-1-i ; r_2=-1+i$$ as roots.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{- x}\cos x$ and $y_2= e^{-x}\sin x$.
Therefore, the general equation is equal to $y(x)=C_1e^{- x}\cos x+C_2 e^{-x}\sin x$.