1. The problem is to find the general solution of the differential equation $$y' + y = x$$. 2. This is a first-order linear differential equation of the form $$y' + p(x)y = q(x)$$ where $$p(x) = 1$$ and $$q(x) = x$$. 3. The integrating factor (IF) is given by $$\mu(x) = e^{\int p(x) dx} = e^{\int 1 dx} = e^x$$. 4. Multiply both sides of the differential equation by the integrating factor: $$e^x y' + e^x y = x e^x$$ 5. The left side is the derivative of $$y e^x$$: $$\frac{d}{dx}(y e^x) = x e^x$$ 6. Integrate both sides with respect to $$x$$: $$y e^x = \int x e^x dx + C$$ 7. To integrate $$\int x e^x dx$$, use integration by parts: Let $$u = x$$, $$dv = e^x dx$$, then $$du = dx$$, $$v = e^x$$. 8. Applying integration by parts: $$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C_1 = e^x (x - 1) + C_1$$ 9. Substitute back: $$y e^x = e^x (x - 1) + C$$ 10. Divide both sides by $$e^x$$: $$y = x - 1 + C e^{-x}$$ 11. Therefore, the general solution is: $$y = x - 1 + C e^{-x}$$ 12. Comparing with the options, the correct answer is (B).