1. **State the problem:** Solve the differential equation $$y' + \frac{2y - 1}{x} = x + 1$$. 2. **Rewrite the equation:** Distribute the fraction: $$y' + \frac{2y}{x} - \frac{1}{x} = x + 1$$ Rearranged: $$y' + \frac{2}{x}y = x + 1 + \frac{1}{x}$$ 3. **Identify the type:** This is a linear first-order differential equation of the form $$y' + P(x)y = Q(x)$$ where $$P(x) = \frac{2}{x}$$ and $$Q(x) = x + 1 + \frac{1}{x}$$. 4. **Find the integrating factor (IF):** $$\mu(x) = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$$ 5. **Multiply both sides by the integrating factor:** $$x^2 y' + x^2 \cdot \frac{2}{x} y = x^2 \left(x + 1 + \frac{1}{x}\right)$$ Simplify: $$x^2 y' + 2x y = x^3 + x^2 + x$$ 6. **Recognize the left side as a derivative:** $$\frac{d}{dx} (x^2 y) = x^3 + x^2 + x$$ 7. **Integrate both sides:** $$x^2 y = \int (x^3 + x^2 + x) dx + C$$ Calculate the integral: $$\int x^3 dx = \frac{x^4}{4}, \quad \int x^2 dx = \frac{x^3}{3}, \quad \int x dx = \frac{x^2}{2}$$ So, $$x^2 y = \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + C$$ 8. **Solve for $$y$$:** $$y = \frac{1}{x^2} \left( \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + C \right) = \frac{x^2}{4} + \frac{x}{3} + \frac{1}{2} + \frac{C}{x^2}$$ **Final answer:** $$y = \frac{x^2}{4} + \frac{x}{3} + \frac{1}{2} + \frac{C}{x^2}$$