1. **State the problem:** Solve the differential equation $y' + 2xy = x$. 2. **Identify the type:** This is a first-order linear ordinary differential equation of the form $y' + P(x)y = Q(x)$ where $P(x) = 2x$ and $Q(x) = x$. 3. **Find the integrating factor (IF):** The integrating factor is given by $$\mu(x) = e^{\int P(x) \, dx} = e^{\int 2x \, dx} = e^{x^2}.$$ 4. **Multiply both sides of the differential equation by the integrating factor:** $$e^{x^2} y' + 2x e^{x^2} y = x e^{x^2}.$$ 5. **Recognize the left side as the derivative of a product:** $$\frac{d}{dx} \left(e^{x^2} y \right) = x e^{x^2}.$$ 6. **Integrate both sides with respect to $x$:** $$e^{x^2} y = \int x e^{x^2} \, dx + C.$$ 7. **Evaluate the integral:** Let $u = x^2$, then $du = 2x \, dx$ or $x \, dx = \frac{du}{2}$. So, $$\int x e^{x^2} \, dx = \int e^u \frac{du}{2} = \frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C.$$ 8. **Substitute back:** $$e^{x^2} y = \frac{1}{2} e^{x^2} + C,$$ 9. **Solve for $y$:** $$y = \frac{1}{2} + C e^{-x^2}.$$ **Final answer:** $$\boxed{y = \frac{1}{2} + C e^{-x^2}}.$$