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:** This is a Riccati equation of the form $$y' = a(x) y^2 + b(x) y + c(x)$$ with $$a(x) = \frac{1}{x}, b(x) = \frac{1}{x}, c(x) = -\frac{2}{x}$$. 3. **Use the known solution:** Since $$y_1 = 1$$ is a particular solution, substitute $$y = y_1 + \frac{1}{u} = 1 + \frac{1}{u}$$ to transform the Riccati equation into a linear equation in $$u$$. 4. **Compute $$y'$$:** $$y' = \frac{d}{dx} \left(1 + \frac{1}{u} \right) = -\frac{u'}{u^2}$$. 5. **Substitute into the original equation:** $$-\frac{u'}{u^2} = \frac{1}{x} \left(1 + \frac{1}{u} \right)^2 + \frac{1}{x} \left(1 + \frac{1}{u} \right) - \frac{2}{x}$$. 6. **Expand the right side:** $$\left(1 + \frac{1}{u} \right)^2 = 1 + \frac{2}{u} + \frac{1}{u^2}$$ So, $$\frac{1}{x} \left(1 + \frac{2}{u} + \frac{1}{u^2} \right) + \frac{1}{x} \left(1 + \frac{1}{u} \right) - \frac{2}{x} = \frac{1}{x} + \frac{2}{x u} + \frac{1}{x u^2} + \frac{1}{x} + \frac{1}{x u} - \frac{2}{x}$$ 7. **Simplify the right side:** $$\frac{1}{x} + \frac{1}{x} - \frac{2}{x} = 0$$ and $$\frac{2}{x u} + \frac{1}{x u} = \frac{3}{x u}$$ So the right side is $$\frac{1}{x u^2} + \frac{3}{x u}$$. 8. **Rewrite the equation:** $$-\frac{u'}{u^2} = \frac{1}{x u^2} + \frac{3}{x u}$$ Multiply both sides by $$u^2$$: $$-u' = \frac{1}{x} + \frac{3 u}{x}$$ 9. **Rearranged:** $$u' + \frac{3}{x} u = -\frac{1}{x}$$ 10. **Solve the linear ODE for $$u$$:** The integrating factor is $$\mu(x) = e^{\int \frac{3}{x} dx} = e^{3 \ln |x|} = x^3$$. 11. **Multiply both sides by $$x^3$$:** $$x^3 u' + 3 x^2 u = -x^2$$ which is $$(x^3 u)' = -x^2$$. 12. **Integrate both sides:** $$x^3 u = -\int x^2 dx + C = -\frac{x^3}{3} + C$$ 13. **Solve for $$u$$:** $$u = -\frac{x^3}{3 x^3} + \frac{C}{x^3} = -\frac{1}{3} + \frac{C}{x^3}$$ 14. **Recall substitution:** $$y = 1 + \frac{1}{u} = 1 + \frac{1}{-\frac{1}{3} + \frac{C}{x^3}} = 1 + \frac{1}{\frac{C}{x^3} - \frac{1}{3}}$$ 15. **Simplify denominator:** $$\frac{C}{x^3} - \frac{1}{3} = \frac{3 C - x^3}{3 x^3}$$ 16. **Final solution:** $$y = 1 + \frac{1}{\frac{3 C - x^3}{3 x^3}} = 1 + \frac{3 x^3}{3 C - x^3} = \frac{3 C - x^3 + 3 x^3}{3 C - x^3} = \frac{3 C + 2 x^3}{3 C - x^3}$$ **Answer:** $$\boxed{y = \frac{3 C + 2 x^3}{3 C - x^3}}$$ where $$C$$ is an arbitrary constant.