Consider the DE $$y'=2\sqrt{|y-1|}$$ To which of the following conditions do we have solutions?

1. $y(0)=0$ and $y(1)=2$
2. $y(0)=0$ and $y(2)=2$
3. $y(0)=0$ and $y(3)=2$

Now, everywhere but on $y=1$ the function $f(y)=2\sqrt{|y-1|}$ is continuous and differentiable, so the solution to any IVP outside of $y=1$ must be unique. However, all of the listed conditions do cross the $1$ and I'm not sure what this means now. By separation of variables I found a general solution $$y(x)=\begin{cases}1-(x+\frac{C}{2})^2,& y\leq 1\\(x+\frac{D}{2})^2+1, & y\geq 1\end{cases}$$with constants $C,D$. This confuses me for two reasons:

1. The function $y(x)$ consists of two cases for different $y$ (instead of $x$ as it usually is).
2. Can't I now just plugin the initial values? For the first condition that would mean $C=\pm 2$ and $D=0$. For the second one $C=\pm 2,D=-2$ and for the last one $C=\pm 2, D=-4$.

This doesn't seem right. My initial guess was that only one of those is solvable. But does it even matter where the solution $y$ takes the value $2$ in terms of solving the DE? I'm a little lost here.