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. **Rewrite the equation:** Factor out $$\frac{1}{x}$$: $$y' = \frac{1}{x}(y^2 + y - 2)$$ 3. **Check the given solution:** Substitute $$y=1$$: $$y' = \frac{1}{x}(1 + 1 - 2) = \frac{0}{x} = 0$$ This confirms $$y=1$$ is a constant solution. 4. **Use substitution to reduce order:** Let $$y = 1 + \frac{1}{v}$$, so that we transform the equation to solve for $$v$$. 5. **Find $$y'$$ in terms of $$v$$:** $$y' = -\frac{1}{v^2}v'$$ 6. **Substitute into the original equation:** $$-\frac{1}{v^2}v' = \frac{1}{x} \left( \left(1 + \frac{1}{v}\right)^2 + \left(1 + \frac{1}{v}\right) - 2 \right)$$ 7. **Simplify the right side:** $$\left(1 + \frac{1}{v}\right)^2 + \left(1 + \frac{1}{v}\right) - 2 = 1 + \frac{2}{v} + \frac{1}{v^2} + 1 + \frac{1}{v} - 2 = \frac{3}{v} + \frac{1}{v^2}$$ 8. **Rewrite the equation:** $$-\frac{1}{v^2}v' = \frac{1}{x} \left( \frac{3}{v} + \frac{1}{v^2} \right)$$ 9. **Multiply both sides by $$v^2$$:** $$-v' = \frac{1}{x}(3v + 1)$$ 10. **Rewrite as:** $$v' + \frac{3}{x}v = -\frac{1}{x}$$ 11. **This is a linear first-order ODE in $$v$$.** The integrating factor is: $$\mu(x) = e^{\int \frac{3}{x} dx} = e^{3 \ln |x|} = |x|^3$$ 12. **Multiply both sides by $$\mu(x)$$:** $$x^3 v' + 3x^2 v = -x^2$$ 13. **Left side is derivative:** $$\frac{d}{dx}(x^3 v) = -x^2$$ 14. **Integrate both sides:** $$x^3 v = -\int x^2 dx + C = -\frac{x^3}{3} + C$$ 15. **Solve for $$v$$:** $$v = \frac{C}{x^3} - \frac{1}{3}$$ 16. **Recall substitution:** $$y = 1 + \frac{1}{v} = 1 + \frac{1}{\frac{C}{x^3} - \frac{1}{3}} = 1 + \frac{1}{\frac{C - \frac{x^3}{3}}{x^3}} = 1 + \frac{x^3}{C - \frac{x^3}{3}}$$ 17. **Simplify denominator:** $$C - \frac{x^3}{3} = \frac{3C - x^3}{3}$$ 18. **Final solution:** $$y = 1 + \frac{x^3}{\frac{3C - x^3}{3}} = 1 + \frac{3x^3}{3C - x^3}$$ 19. **Rewrite neatly:** $$\boxed{y = 1 + \frac{3x^3}{3C - x^3}}$$ This is the general solution to the differential equation given the particular solution $$y=1$$.