differential equation $y'=\sqrt{|y|(1-y)}$ Consider the differential equation
$y'=\sqrt{|y|(1-y)}$ with $y\le1$
I know that $y=0$ and $y=1$ are solutions.
The problem is to compute the general solution for $0<y<1$. On which interval is the solution defined?
I don't know where or how to start. 
$y$ is always positive, so the absolute bars can be left out, but that doesn't seem to make it easier.
Is there anyone who can give me a hint?

Given $$\displaystyle \frac{dy}{dx} = \sqrt{|y|\cdot (1-y)}=\sqrt{y(1-y)}\;,$$ bcz $\; 0<y<1$
So $$\displaystyle \frac{dy}{\sqrt{y(1-y)}} = dx\Rightarrow \int \frac{1}{\sqrt{y(1-y)}}dy = \int dx$$ 
Now Put $$\displaystyle y=\left(z+\frac{1}{2}\right)\;,$$ Then $dy=dz$
So we get $$\displaystyle \int\frac{1}{\sqrt{\left(\frac{1}{2}\right)^2-z^2}}dz = \int dx$$
Now Put $\displaystyle z=\frac{1}{2}\sin \phi\;,$ Then $\displaystyle dz =\frac{1}{2}\cos \phi d\phi $
So we get $$\displaystyle \int\frac{\cos\phi}{\cos \phi}d\phi = \int dx\Rightarrow \phi=x+\mathcal{C}$$
So we get $$\displaystyle \sin^{-1}\left(2z\right) =x+\mathcal{C}$$
So we get $$\displaystyle \sin^{-1}\left(4y-2\right) = x+\mathcal{C}\Rightarrow 4y-2 = \sin(x+\mathcal{C})$$
