differential equations - solution curves Solve the differential equation $$~(y')^2= 4y~$$ to verify the general solution curves and singular solution curves. Determine the points $~(a,b)~$ in the plane for which the initial value problem $$~(y')^2= 4y,\quad y(a)= b~$$ has 
$(a)\quad$ no solution , 
$(b)\quad$  infinitely many solutions
that are defined for all values of $~x (c)~$  on some neighborhood of the point $~x=a~$ , only finitely many solutions. 
General solution that I am getting is $~y (x) = (x-c)^2~$  and singular solution is $~y(x)=0~$. 
I wish a clarification in the second part. 
The family of curves consists on parabolas with vertex varying on the $~x~$-axis. 
I need someone to explain how to proceed to find suitable choice of $~(a,b)~$ in each part. 
 A: The solutions are the ones you listed.    
The solutions all have shape $y=(x-c)^2$ or $y=0$. Thus if $b<0$, then none of the solutions curves pass through $(a,b)$.  So for all pairs $(a,b)$ such that $b<0$, there cannot be a solution satisfying $y(a)=b$.  We do not know (yet) whether these are all the pairs $(a,b)$ for which there is no solution, but soon we will.
For any $a$, if $b=0$ there are exactly two solutions satisfying $y(a)=b$, the singular solution and the solution $y=(x-a)^2$.
Finally, we look at pairs $(a,b)$ with $b$ positive.  We look for values of $c$ such that $y(a)=b$. 
The solution $y=(x-c)^2$ passes through $(a,b)$ if and only if $(a-c)^2=b$. This equation has exactly two solutions, $c=a\pm\sqrt{b}$.
Conclusion: (a) The pairs $(a,b)$ for which there is no solution satisfying $y(a)=b$ are all $(a,b)$ with $b<0$. (b)  There are no pairs $(a,b)$ for which there are infinitely many solutions with initial condition $y(a)=b$. (c) For all remaining pairs $(a,b)$, that is, all pairs with $b \ge 0$, there are finitely many solutions, indeed exactly two solutions that satisfy $y(a)=b$.  
The geometry: The conclusion can also be reached geometrically, by visualizing the family of parabolas. All of your parabolas are obtained by sliding the standard parabola $y=x^2$ along the $x$-axis.  For any $(a,b)$ with $b \gt 0$, there are exactly two such parabolas that pass through (a,b): one whose "left" half goes through $(a,b)$, and one whose "right" half goes through $(a,b)$.
Note: One could interpret the word "finite" to include the possibility of $0$ solutions: $0$ is certainly finite!  That is obviously not the intended interpretation here. But if we interpret "finite" as including $0$, the answer to (c) is all pairs $(a,b)$.
A: $y=0$ is a solution, which works for $b=0$ and any $a$.
Else,
$$\frac{y'}{2\sqrt y}=\pm1$$
has the solutions
$$\sqrt y=c\pm x.$$
Now
$$\sqrt b=c\pm a$$
requires that $b\ge0$, and there is no restriction on $a$. By every point $(a,b)$ such that $a\ne0$, there are two solutions.
