Where do these other initial conditions come from?

I have the problem $2\frac{dy}{dt}=y(y-2),y\left(0\right)=3$, which I get how to solve up until the intial conditions. The solution is $y\left(t\right)=\frac{2}{1-C{e}^{t}}$. The solution then plugs in three initial conditions and solves for C. They do $y\left(0\right)=1,y\left(0\right)=-1$, and $y\left(0\right)=3$ and then solve each of these. Where did the other two come from?

Also, the problem asks to determine the interval of existence. This one is straightforward, but there are others where the interval is something like $(-\mathrm{\infty},2)\cup (2,\mathrm{\infty})$ and they plug in $t=0$ and then take only the interval that solution lies in. Why is that?

Exact wording: Identify the equilibrium (constant) solutions for this nonlinear ODE. a) identify the equilibrium solutions for the nonlinear ODE. b) Solve the IVP for the initial condition, $y\left(0\right)=3$. Determine the existence interval and the limit of y(t) as t approaches the endpoints of the interval.