$\displaystyle 2(y')^2=y''(y-1)$
$\displaystyle y(1)=2 y'(1)=-1$
again i tried
y'=z
$\displaystyle z'y=z^{2}$
$\displaystyle \frac{{dz}}{dx}=\frac{dz}{dy}\frac{dy}{dx}$
$\displaystyle zz'=dz/dx$
You have
$\displaystyle \displaystyle \frac{2}{y-1}\,\frac{dy}{dx} = \frac{1}{\frac{dy}{dx}}\,\frac{d^2y}{dx^2}$.
On the RHS make the substitution $\displaystyle \displaystyle u = \frac{dy}{dx}$ so that $\displaystyle \displaystyle \frac{du}{dx} = \frac{d^2y}{dx^2}$ so that your DE becomes
$\displaystyle \displaystyle \frac{2}{y-1}\,\frac{dy}{dx} = \frac{1}{u}\,\frac{du}{dx}$.
Now integrating both sides gives
$\displaystyle \displaystyle \int{\frac{2}{y-1}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{u}\,\frac{du}{dx}\,dx}$
$\displaystyle \displaystyle \int{\frac{2}{y-1}\,dy} = \int{\frac{1}{u}\,du}$
$\displaystyle \displaystyle 2\ln{|y-1|} + C_1 = \ln{|u|} + C_2$
$\displaystyle \displaystyle \ln{|y-1|^2} + C = \ln{|u|}$, where $\displaystyle \displaystyle C = C_1 - C_2$
$\displaystyle \displaystyle C = \ln{|u|} - \ln{|(y-1)^2|}$
$\displaystyle \displaystyle C = \ln{\left|\frac{u}{(y-1)^2}\right|}$
$\displaystyle \displaystyle C = \ln{\left|\frac{\frac{dy}{dx}}{(y-1)^2}\right|}$
$\displaystyle \displaystyle \pm e^C = \frac{dy}{dx}\,\frac{1}{(y-1)^2}$
$\displaystyle \displaystyle A = \frac{dy}{dx}\,\frac{1}{(y-1)^2}$, where $\displaystyle \displaystyle A = \pm e^C$.
You can integrate both sides with respect to $\displaystyle \displaystyle x$ now to find $\displaystyle \displaystyle y$. You will need to use partial fractions.
From my previous post:
$\displaystyle 2\ln|y-1|=\ln|y'|+C_{1}.$
On both sides, you see, the derivative of the denominator appears in the numerator. Whenever that happens, you can do a u substitution, and you'll get du/u, which integrates to ln|u|. See how that works?