Question:

Step 1: Recognize the differential equation structure. We are given the equation $f'(x) + f(x) = x$, which is a first-order linear differential equation of the form $f'(x) + P(x)f(x) = Q(x)$ where $P(x) = 1$ and $Q(x) = x$. Step 2: Find the integrating factor. For a first-order linear ODE, the integrating factor is $\mu(x) = e^{\int P(x)\,dx}$. Since $P(x) = 1$, we have: $$\mu(x) = e^{\int 1\,dx} = e^x$$ Step 3: Multiply both sides by the integrating factor. Multiplying the original equation by $e^x$: $$e^x f'(x) + e^x f(x) = xe^x$$ The left side is the derivative of a product: $$\frac{d}{dx}\left[e^x f(x)\right] = xe^x$$ Step 4: Integrate both sides. Integrating both sides with respect to $x$: $$e^x f(x) = \int xe^x\,dx$$ To evaluate the integral on the right, we use integration by parts with $u = x$ and $dv = e^x\,dx$: $$\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C$$ Therefore: $$e^x f(x) = e^x(x-1) + C$$ Step 5: Solve for $f(x)$. Dividing both sides by $e^x$: $$f(x) = x - 1 + Ce^{-x}$$ Step 6: Apply the polynomial constraint. Since $f(x)$ is given to be a polynomial function, it cannot contain exponential terms. The term $Ce^{-x}$ is exponential unless $C = 0$. Therefore, we must have $C = 0$, which gives: $$f(x) = x - 1$$ Step 7: Calculate $f(4)$. Substituting $x = 4$ into $f(x) = x - 1$: $$f(4) = 4 - 1 = 3$$ The value of $f(4)$ is equal to **3**, which corresponds to **Option 3**.