Finding the internal boundary condition $y(0)$ for $y'' = 2y-2\sqrt{y}$

Question

I have the second order ODE: $$y'' = 2y-2\sqrt{y}, \quad \quad y'(0)=y(C)=0, \quad \quad x \in (0,C)$$ and I would like to know the value $y(0)$ via a shooting-type method. Note that $C$ remains unknown at this point. So far, I used the substitution $v=y'$ to rewrite as the following IVP problem: $$y'=v, \quad \quad v' = 2y-2\sqrt{y}, \quad \quad y(0) = U,\quad v(0) = 0 , \quad \quad x \in (0,C),$$ where I know that $U \in (0,1)$. So my thinking was this: I plot the solution $y(x)$ for lots of different values of $U$, I then see which one respects the boundary value $y'(0)=0$ 'the most', and wherever $y=0$ will be the value of $C$. In the figure is my plots for $y$, but as you can see they are all very similar, how do I know which is the correct value for $U$? How do I even know if there is a unique value?