1. **State the problem:** Solve the differential equation $$y' = \frac{1}{x}y^2 + \frac{1}{x}y - \frac{2}{x}$$ given that $$y=1$$ is a solution. 2. **Identify the type of equation:** This is a Riccati differential equation of the form $$y' = a(x)y^2 + b(x)y + c(x)$$ where $$a(x) = \frac{1}{x}, b(x) = \frac{1}{x}, c(x) = -\frac{2}{x}$$. 3. **Use the known solution:** Since $$y_1 = 1$$ is a particular solution, substitute $$y = y_1 + \frac{1}{v} = 1 + \frac{1}{v}$$ to transform the Riccati equation into a linear equation in terms of $$v$$. 4. **Compute $$y'$$:** $$y' = \frac{d}{dx}\left(1 + \frac{1}{v}\right) = 0 - \frac{v'}{v^2} = -\frac{v'}{v^2}$$. 5. **Substitute $$y$$ and $$y'$$ into the original equation:** $$-\frac{v'}{v^2} = \frac{1}{x}\left(1 + \frac{1}{v}\right)^2 + \frac{1}{x}\left(1 + \frac{1}{v}\right) - \frac{2}{x}$$. 6. **Expand the right side:** $$\left(1 + \frac{1}{v}\right)^2 = 1 + \frac{2}{v} + \frac{1}{v^2}$$ So, $$\frac{1}{x}\left(1 + \frac{2}{v} + \frac{1}{v^2}\right) + \frac{1}{x}\left(1 + \frac{1}{v}\right) - \frac{2}{x} = \frac{1}{x} + \frac{2}{xv} + \frac{1}{xv^2} + \frac{1}{x} + \frac{1}{xv} - \frac{2}{x}$$ 7. **Simplify the right side:** $$\frac{1}{x} + \frac{1}{x} - \frac{2}{x} = 0$$ and $$\frac{2}{xv} + \frac{1}{xv} = \frac{3}{xv}$$ So the right side is $$\frac{1}{xv^2} + \frac{3}{xv}$$. 8. **Rewrite the equation:** $$-\frac{v'}{v^2} = \frac{1}{xv^2} + \frac{3}{xv}$$ Multiply both sides by $$v^2$$: $$-v' = \frac{1}{x} + \frac{3v}{x}$$ 9. **Rewrite:** $$v' + \frac{3}{x}v = -\frac{1}{x}$$ This is a linear first-order ODE in $$v$$. 10. **Find integrating factor:** $$\mu(x) = e^{\int \frac{3}{x} dx} = e^{3 \ln|x|} = |x|^3$$ 11. **Multiply both sides by $$\mu(x)$$:** $$x^3 v' + 3x^2 v = -x^2$$ 12. **Recognize left side as derivative:** $$\frac{d}{dx}(x^3 v) = -x^2$$ 13. **Integrate both sides:** $$x^3 v = -\int x^2 dx + C = -\frac{x^3}{3} + C$$ 14. **Solve for $$v$$:** $$v = \frac{C}{x^3} - \frac{1}{3}$$ 15. **Recall substitution:** $$y = 1 + \frac{1}{v} = 1 + \frac{1}{\frac{C}{x^3} - \frac{1}{3}} = 1 + \frac{1}{\frac{C - \frac{x^3}{3}}{x^3}} = 1 + \frac{x^3}{C - \frac{x^3}{3}}$$ 16. **Simplify denominator:** $$C - \frac{x^3}{3} = \frac{3C - x^3}{3}$$ So, $$y = 1 + \frac{x^3}{\frac{3C - x^3}{3}} = 1 + \frac{3x^3}{3C - x^3}$$ 17. **Final solution:** $$\boxed{y = 1 + \frac{3x^3}{3C - x^3}}$$ This is the general solution to the given Riccati differential equation using the known particular solution and the Riccati method.