If y = e-x (A cos x + B sin x), then y is a solution of

The correct option is: d²y/dx² + 2(dy/dx) + 2y = 0.

To understand why this is the correct differential equation, let us start with the given function:
$$y=e^{-x}(A\cos x+B\sin x)$$

First, we find the first derivative of y with respect to x, denoted as ^dy/_dx, using the product rule of differentiation:
$$\frac{dy}{dx}=\frac{d}{dx}\left[e^{-x}\right](A\cos x+B\sin x)+e^{-x}\frac{d}{dx}[A\cos x+B\sin x]$$

Differentiating each part yields:
$$\frac{dy}{dx}=-e^{-x}(A\cos x+B\sin x)+e^{-x}(-A\sin x+B\cos x)$$

Notice that the first term on the right-hand side is simply −y. Therefore, we can simplify this expression to:
$$\frac{dy}{dx}=-y+e^{-x}(-A\sin x+B\cos x)$$

Rearranging the terms, we get:
$$\frac{dy}{dx}+y=e^{-x}(-A\sin x+B\cos x)$$

Now, we differentiate this equation once more with respect to x to obtain the second derivative:
$$\frac{d}{dx}[\frac{dy}{dx}+y]=\frac{d}{dx}[e^{-x}(-A\sin x+B\cos x)]$$

Applying the sum rule on the left side and the product rule on the right side:
$$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-e^{-x}(-A\sin x+B\cos x)+e^{-x}(-A\cos x-B\sin x)$$

We can simplify this by factoring out negative signs. Notice that:
$$-e^{-x}(-A\sin x+B\cos x)=-(\frac{dy}{dx}+y)$$
and
$$e^{-x}(-A\cos x-B\sin x)=-e^{-x}(A\cos x+B\sin x)=-y$$

Substituting these two expressions back into the derivative equation gives:
$$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-(\frac{dy}{dx}+y)-y$$

Expanding the right-hand side yields:
$$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-\frac{dy}{dx}-2y$$

Finally, we move all terms to the left-hand side to write the differential equation in standard form:
$$\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+2y=0$$

This matches the correct option.