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(B) Given the general solution: $y = A e^{-x} + B \cos x$ ... $(i)$ Differentiating with respect to $x$: $\frac{dy}{dx} = -A e^{-x} - B \sin x$ ... $(

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$(\cos x-\sin x) \frac{d^2 y}{d x^2}+2 \cos x \frac{d y}{d x}+(\sin x+\cos x) y=0$

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(B) Given the general solution: $y = A e^{-x} + B \cos x$ ... $(i)$ Differentiating with respect to $x$: $\frac{dy}{dx} = -A e^{-x} - B \sin x$ ... $(ii)$ Differentiating again: $\frac{d^2y}{dx^2} = A e^{-x} - B \cos x$ ... $(iii)$ From $(i)$ and $(ii)$, we eliminate $B$: $y \sin x + \frac{dy}{dx} \cos x = A e^{-x} (\sin x - \cos x)$ ... $(iv)$ From $(i)$ and $(iii)$, we eliminate $B$: $y + \frac{d^2y}{dx^2} = 2 A e^{-x}$ ... $(v)$ From $(iv)$ and $(v)$, we eliminate $A$: $y \sin x + \frac{dy}{dx} \cos x = \frac{1}{2} (y + \frac{d^2y}{dx^2}) (\sin x - \cos x)$ Multiplying by $2$: $2y \sin x + 2 \frac{dy}{dx} \cos x = (y + \frac{d^2y}{dx^2}) (\sin x - \cos x)$ Rearranging terms: $(\cos x - \sin x) \frac{d^2y}{dx^2} + 2 \cos x \frac{dy}{dx} + (\sin x + \cos x) y = 0$.

If $A$ and $B$ are arbitrary constants, then the differential equation having $y=A e^{-x}+B \cos x$ as its general solution is

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