1. Given the differential equation: $$xy' = y^2 + y$$ 2. Rewrite it as: $$xy' = y(y+1)$$ which implies $$y' = \frac{y(y+1)}{x}$$ 3. Separate variables: $$\frac{dy}{y(y+1)} = \frac{dx}{x}$$ 4. Use partial fractions on the left side: $$\frac{1}{y(y+1)} = \frac{A}{y} + \frac{B}{y+1}$$ 5. Solve for constants: multiplying both sides by $y(y+1)$ gives $$1 = A(y+1) + By$$ 6. Let $y=0$ then $1 = A(0+1) \Rightarrow A=1$ 7. Let $y=-1$ then $1 = B(-1) \Rightarrow B = -1$ 8. Thus: $$\frac{dy}{y(y+1)} = \left( \frac{1}{y} - \frac{1}{y+1} \right) dy$$ 9. Integrate both sides: $$\int \left( \frac{1}{y} - \frac{1}{y+1} \right) dy = \int \frac{1}{x} dx$$ 10. Integrals evaluate to: $$\ln|y| - \ln|y+1| = \ln|x| + C$$ 11. Combine logarithms on left side: $$\ln\left| \frac{y}{y+1} \right| = \ln|x| + C$$ 12. Exponentiate both sides: $$\left| \frac{y}{y+1} \right| = e^C |x|$$ 13. Let $K = \pm e^C$, so $$\frac{y}{y+1} = Kx$$ 14. Solve for $y$: $$y = Kx(y+1) = Kxy + Kx$$ 15. Rearrange terms: $$y - Kxy = Kx \Rightarrow y(1 - Kx) = Kx$$ 16. Hence the implicit general solution is: $$y = \frac{Kx}{1 - Kx}$$ Final answer: $$y = \frac{Cx}{1 - Cx}$$ where $C$ is an arbitrary constant determined by initial conditions.