Question:

Step 1: Recognize the differential equation structure. We are given the differential equation: $$f(x) + f'(x) = xe^{-x}$$ Rearranging this into standard form: $$f'(x) + f(x) = xe^{-x}$$ This is a first-order linear differential equation of the form $f'(x) + P(x)f(x) = Q(x)$, where $P(x) = 1$ and $Q(x) = xe^{-x}$. Step 2: Find and apply the integrating factor. For a linear differential equation $f'(x) + P(x)f(x) = Q(x)$, the integrating factor is: $$\mu(x) = e^{\int P(x)\,dx} = e^{\int 1\,dx} = e^x$$ Multiply both sides of the differential equation by $e^x$: $$e^x f'(x) + e^x f(x) = xe^{-x} \cdot e^x$$ $$e^x f'(x) + e^x f(x) = x$$ Step 3: Recognize the left side as a derivative of a product. The left side is the derivative of the product $e^x f(x)$: $$\frac{d}{dx}[e^x f(x)] = x$$ Step 4: Integrate both sides. Integrating both sides with respect to $x$: $$e^x f(x) = \int x\,dx = \frac{x^2}{2} + C$$ where $C$ is the constant of integration. Step 5: Apply the initial condition to find the constant. Using the given condition $f(0) = 0$: $$e^0 \cdot f(0) = \frac{0^2}{2} + C$$ $$1 \cdot 0 = 0 + C$$ $$C = 0$$ Step 6: Solve for $f(x)$. Substituting $C = 0$ back into the equation: $$e^x f(x) = \frac{x^2}{2}$$ $$f(x) = \frac{x^2}{2}e^{-x}$$ Step 7: Calculate $f(1)$. Substituting $x = 1$ into the expression for $f(x)$: $$f(1) = \frac{1^2}{2}e^{-1} = \frac{1}{2e}$$ **Final Answer:** The value of $f(1) = \dfrac{1}{2e}$, which corresponds to **Option 1**.