1. **Stating the problem:** Solve the differential equation $$2(y')^2 = (y-1)y''$$ where $y' = \frac{dy}{dx}$ and $y'' = \frac{d^2y}{dx^2}$. 2. **Rewrite the equation:** Let $p = y'$, so $y'' = \frac{dp}{dx} = \frac{dp}{dy} \cdot \frac{dy}{dx} = p \frac{dp}{dy}$. Substitute into the equation: $$2p^2 = (y-1) p \frac{dp}{dy}$$ 3. **Simplify:** Divide both sides by $p$ (assuming $p \neq 0$): $$2p = (y-1) \frac{dp}{dy}$$ 4. **Separate variables:** $$\frac{dp}{dy} = \frac{2p}{y-1}$$ Rewrite as $$\frac{1}{p} dp = \frac{2}{y-1} dy$$ 5. **Integrate both sides:** $$\int \frac{1}{p} dp = \int \frac{2}{y-1} dy$$ This gives $$\ln|p| = 2 \ln|y-1| + C$$ 6. **Exponentiate:** $$|p| = e^C |y-1|^2$$ Let $A = e^C > 0$, so $$p = y' = A (y-1)^2$$ 7. **Separate variables again:** $$\frac{dy}{dx} = A (y-1)^2$$ Rewrite as $$\frac{dy}{(y-1)^2} = A dx$$ 8. **Integrate:** $$\int (y-1)^{-2} dy = \int A dx$$ The integral on the left is $$\int (y-1)^{-2} dy = - (y-1)^{-1} + C_1$$ So $$- \frac{1}{y-1} = A x + C_2$$ 9. **Solve for $y$:** $$\frac{1}{y-1} = - A x - C_2$$ Let $B = -C_2$, then $$\frac{1}{y-1} = - A x + B$$ Invert both sides: $$y - 1 = \frac{1}{- A x + B}$$ 10. **Final solution:** $$\boxed{y = 1 + \frac{1}{B - A x}}$$ where $A$ and $B$ are arbitrary constants determined by initial conditions.