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, using the Riccati method. 2. **Recall the Riccati equation form:** A Riccati equation is of the form $$y' = a(x)y^2 + b(x)y + c(x)$$. Here, $$a(x) = \frac{1}{x}$$, $$b(x) = \frac{1}{x}$$, and $$c(x) = -\frac{2}{x}$$. 3. **Use the known solution:** Since $$y_1 = 1$$ is a particular solution, substitute $$y = y_1 + \frac{1}{v} = 1 + \frac{1}{v}$$ where $$v = v(x)$$ is a new function to find. 4. **Find $$y'$$ in terms of $$v$$:** $$y' = \frac{d}{dx}\left(1 + \frac{1}{v}\right) = 0 - \frac{v'}{v^2} = -\frac{v'}{v^2}$$. 5. **Substitute $$y$$ and $$y'$$ into the original DE:** $$-\frac{v'}{v^2} = \frac{1}{x}\left(1 + \frac{1}{v}\right)^2 + \frac{1}{x}\left(1 + \frac{1}{v}\right) - \frac{2}{x}$$. 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 $$\frac{3}{xv} + \frac{1}{xv^2}$$. 8. **Rewrite the equation:** $$-\frac{v'}{v^2} = \frac{3}{xv} + \frac{1}{xv^2}$$ Multiply both sides by $$v^2$$: $$-v' = \frac{3}{x}v + \frac{1}{x}$$ 9. **Rewrite as a linear ODE for $$v$$:** $$v' + \frac{3}{x}v = -\frac{1}{x}$$ 10. **Solve the linear ODE:** The integrating factor is $$\mu(x) = e^{\int \frac{3}{x} dx} = e^{3 \ln|x|} = |x|^3$$ Multiply both sides by $$x^3$$: $$x^3 v' + 3x^2 v = -x^2$$ This is $$\frac{d}{dx}(x^3 v) = -x^2$$ 11. **Integrate both sides:** $$x^3 v = -\int x^2 dx + C = -\frac{x^3}{3} + C$$ 12. **Solve for $$v$$:** $$v = \frac{C - \frac{x^3}{3}}{x^3} = \frac{C}{x^3} - \frac{1}{3}$$ 13. **Recall substitution:** $$y = 1 + \frac{1}{v} = 1 + \frac{1}{\frac{C}{x^3} - \frac{1}{3}} = 1 + \frac{1}{\frac{C - \frac{x^3}{3}}{x^3}} = 1 + \frac{x^3}{C - \frac{x^3}{3}}$$ 14. **Simplify denominator:** $$C - \frac{x^3}{3} = \frac{3C - x^3}{3}$$ So $$y = 1 + \frac{x^3}{\frac{3C - x^3}{3}} = 1 + \frac{3x^3}{3C - x^3}$$ 15. **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 substitution method.