Solve a nonlinear differential equation I am interested in the following differential equation :
$$ y(y-1)\ddot{y} - (2y-1)(\dot{y})^2 = 0. $$
In order to solve it, I notice that it is equivalent to 
$$ \frac{\ddot{y}}{\dot{y}} - \frac{\dot{y}}{1-y} + \frac{\dot{y}}{y} = 0 $$
which would integrate as :
$$ \ln(\vert \dot{y} \vert) + \ln(1-y) + \ln(y) = c, \; c \in \mathbb{R}. $$
But I do not know the sign of $\dot{y}$. By Cauchy-Lipschitz, I can show that, since $y \equiv 0$ and $y \equiv 1$ are solutions, $y \in ]0,1[$ for an initial condition $y(0) \in ]0,1[$. But I am wondering if a change of variables could help solve this differential equation since my method is not really neat.
 A: You have the signs strange, I see, for $y\ne 0$ (which is a constant solution)
$$
\frac{\ddot y}{\dot y}-\frac{\dot y}{y-1}-\frac{\dot y}{y}=0
$$
which integrates to
$$
\ln|\dot y|-\ln|y-1|-\ln|y|=c
$$
and this can be transformed to
$$
\dot y = C·y(y-1),
$$
$C=\pm e^c$, which again can be solved via separation and partial fractions.
A: Under the substitution, $y \rightarrow \dfrac{f}{1+f}$, the resulting equation is $\dfrac{f'^2 - f \cdot f''}{(1+f)^4}$, which should seem familiar and suggest the method of solution.  In fact, putting $y \rightarrow \dfrac{\mathrm{e}^{f}}{1+\mathrm{e}^{f}}$, we get $\dfrac{-\mathrm{e}^{2f} f''}{(1+\mathrm{e}^f)^4} = 0$, forcing $f''=0$.  Applying an initial condition at $x=0$ to capture one of the arbitrary constants, the resulting solutions are $y(x) = \dfrac{y(0)\exp(ax)}{1-y(0)+y(0)\exp(ax)}$ for arbitrary constant $a$.  The two solutions you found correspond to $y(0) = 1$ (for $f \equiv 1$) and $y(0)=0$ (for $f \equiv 0$).  The provided form also gives solutions with $y(0) >1$ or $y(0) < -1$.
