1. The problem is to solve the differential equation $$xy' = y^2 + y$$ or equivalently $$x \frac{dy}{dx} = y^2 + y$$. 2. Divide both sides by $x$ (assuming $x \neq 0$) to rewrite the equation as $$\frac{dy}{dx} = \frac{y^2 + y}{x}$$. 3. Factor $y$ out of the numerator on the right side: $$\frac{dy}{dx} = \frac{y(y+1)}{x}$$. 4. This is a separable differential equation; rewrite as $$\frac{dy}{y(y+1)} = \frac{dx}{x}$$. 5. Use partial fractions to decompose $$\frac{1}{y(y+1)} = \frac{A}{y} + \frac{B}{y+1}$$ which yields $A=1$ and $B=-1$. 6. Integrate both sides: $$\int \left(\frac{1}{y} - \frac{1}{y+1} \right) dy = \int \frac{1}{x} dx$$. 7. The integrals become: $$\ln|y| - \ln|y+1| = \ln|x| + C$$. 8. Combine the logarithms: $$\ln \left| \frac{y}{y+1} \right| = \ln|x| + C$$. 9. Exponentiate both sides to solve for $y$: $$\frac{y}{y+1} = K x$$ where $K = e^C$ is an arbitrary constant. 10. Solve the algebraic equation for $y$: $$y = K x (y+1) = K x y + K x$$. 11. Rearranging: $$y - K x y = K x$$ which gives $$y (1 - K x) = K x$$. 12. Finally, $$y = \frac{K x}{1 - K x}$$, representing the implicit solution of the differential equation. The diagonal green line with points $(0,0),(1,1),(2,2),(3,3),(4,4),(5,5)$ corresponds to the particular solution for some specific $K$ and initial condition. Final answer: $$y = \frac{K x}{1 - K x}$$ with arbitrary constant $K$.