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?

The differential is separable, thus you simply need to compute
$$ \int \frac{ dy}{ \sqrt{ y-y^2}} = x + const$$ 
To solve the above, complete the square and use a trig substitution.
