1. **Stating the problem:** We need to solve the differential equation $$y y'' = 2(y')^2 - 2 y'$$ where $y' = \frac{dy}{dx}$ and $y'' = \frac{d^2y}{dx^2}$. 2. **Rewrite the equation:** Let $p = y' = \frac{dy}{dx}$. Then $y'' = \frac{dp}{dx} = \frac{dp}{dy} \frac{dy}{dx} = p \frac{dp}{dy}$ by chain rule. 3. **Substitute into the equation:** $$y y'' = y p \frac{dp}{dy} = 2 p^2 - 2 p$$ This gives: $$y p \frac{dp}{dy} = 2 p^2 - 2 p$$ 4. **Simplify and separate variables:** Divide both sides by $p$ (assuming $p \neq 0$): $$y \frac{dp}{dy} = 2 p - 2$$ Rewrite as $$y \frac{dp}{dy} - 2 p = -2$$ 5. **Solve the linear differential equation in $p(y)$:** It's a first order linear ODE: $$\frac{dp}{dy} - \frac{2}{y} p = - \frac{2}{y}$$ 6. **Find integrating factor:** $$\mu(y) = e^{-\int \frac{2}{y} dy} = e^{-2 \ln y} = y^{-2}$$ 7. **Multiply both sides by integrating factor:** $$y^{-2} \frac{dp}{dy} - \frac{2}{y^{3}} p = - \frac{2}{y^{3}}$$ Which can be written as $$\frac{d}{dy} (p y^{-2}) = - \frac{2}{y^{3}}$$ 8. **Integrate both sides:** $$p y^{-2} = \int -2 y^{-3} dy + C = -2 \left(-\frac{1}{2 y^{2}} \right) + C = \frac{1}{y^{2}} + C$$ 9. **Solve for $p$:** $$p = y^{2} \left( \frac{1}{y^{2}} + C \right) = 1 + C y^{2}$$ Recall $p = \frac{dy}{dx}$ so: $$\frac{dy}{dx} = 1 + C y^{2}$$ 10. **Separate variables and integrate:** $$\frac{dy}{1 + C y^{2}} = dx$$ Case 1: $C = 0$ $$\int dy = \int dx$$ $$y = x + D$$ Case 2: $C \neq 0$ Use integral formula: $$\int \frac{dy}{1 + a y^{2}} = \frac{1}{\sqrt{a}} \arctan( y \sqrt{a}) + K$$ So, $$x + E = \frac{1}{\sqrt{C}} \arctan( y \sqrt{C})$$ or $$y = \frac{1}{\sqrt{C}} \tan\left(\sqrt{C} (x + E)\right)$$ 11. **Final solution:** $$y = x + D \quad \text{or} \quad y = \frac{1}{\sqrt{C}} \tan\left(\sqrt{C} (x + E)\right)$$ where $C, D, E$ are constants.