If $y=e^{-x}(A \cos x+B \sin x)$, then y satisfies

$\frac{d^2 y}{d x^2}+2 \frac{d y}{d x}+2 y=0$

$y=e^{-x}(A \cos x+B \sin x)$                .......(1)

$\Rightarrow \frac{d y}{d x}=e^{-x}(-A \sin x+B \cos x)-e^{-x}(A \cos x+B \sin x)$

$\Rightarrow \frac{d y}{d x}=e^{-x}(-A \sin x+B \cos x)-y$                .......(2)

$\Rightarrow \frac{d^2 y}{d x^2}=e^{-x}(-A \cos x-B \sin x)-e^{-x}(-A \sin x+B \cos x)\left(-\frac{d y}{d x}\right)$

Using (1) and (2), we get

$\frac{d^2 y}{d x^2}+\frac{d y}{d x}=-y-y-\frac{d y}{d x}$

Hence (3) is the correct answer.