1. **State the problem:** Solve the first-order linear differential equation $$y' + 2xy = x$$. 2. **Identify the type and formula:** This is a linear 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 the entire 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 $$x$$:** $$\int \frac{d}{dx} \left(e^{x^2} y\right) dx = \int x e^{x^2} dx$$ 7. **Simplify the left side:** $$e^{x^2} y = \int x e^{x^2} dx + C$$ 8. **Evaluate the integral on the right:** Let $$u = x^2$$, then $$du = 2x dx$$ or $$\frac{du}{2} = x dx$$. So, $$\int x e^{x^2} dx = \int e^u \frac{du}{2} = \frac{1}{2} \int e^u du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C$$ 9. **Substitute back:** $$e^{x^2} y = \frac{1}{2} e^{x^2} + C$$ 10. **Solve for $$y$$:** $$y = \frac{1}{2} + C e^{-x^2}$$ **Final answer:** $$y = \frac{1}{2} + C e^{-x^2}$$ where $$C$$ is an arbitrary constant.