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(D) Given equation: $y = (x + K)e^{-x}$ Differentiating both sides with respect to $x$ using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x + K) \c

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$\frac{dy}{dx} + y = e^{-x}$

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(D) Given equation: $y = (x + K)e^{-x}$ Differentiating both sides with respect to $x$ using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x + K) \cdot e^{-x} + (x + K) \cdot \frac{d}{dx}(e^{-x})$ $\frac{dy}{dx} = 1 \cdot e^{-x} + (x + K) \cdot (-e^{-x})$ $\frac{dy}{dx} = e^{-x} - (x + K)e^{-x}$ Since $y = (x + K)e^{-x}$,we substitute $y$ into the equation: $\frac{dy}{dx} = e^{-x} - y$ Rearranging the terms gives: $\frac{dy}{dx} + y = e^{-x}$

The differential equation found by the elimination of the arbitrary constant $K$ from the equation $y = (x + K)e^{-x}$ is

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