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

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

The correct answer is Option (3) → $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$ ##

Given that, $y = e^{-x} (A \cos x + B \sin x)$

Solution can be obtained by eliminating $A$ and $B$.

On differentiating both sides w.r.t. $x$ we get

$\frac{dy}{dx} = -e^{-x} (A \cos x + B \sin x) + e^{-x} (-A \sin x + B \cos x)$

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

Again, differentiating both sides w.r.t. $x$, we get

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

$\Rightarrow \quad \frac{d^2y}{dx^2} = -\frac{dy}{dx} - y - \left[ \frac{dy}{dx} + y \right]$

$\Rightarrow \quad \frac{d^2y}{dx^2} = -\frac{dy}{dx} - y - \frac{dy}{dx} – y$

$\Rightarrow \quad \frac{d^2y}{dx^2} = -2\frac{dy}{dx} - 2y$

$\Rightarrow \quad \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$