The initial value problem,$$y'=2\sqrt{y}\;\;\;\; , y(0)=a$$ has
(1) A unique solution if $a<0$
(2) No solution if $a>0$
(3) Infinitely many solutions if $a=0$
(4) A unique solution if $a \ge 0$
This is what I have done so far-
Here, $f(x,y)=2\sqrt{y}\;\;$ and consider the rectancular region
$R=\{ (x,y)\in \Bbb R^2:|x|\le c ,|y-a | \le b\}$
$\mathbf {If \;\;\ a>0}:$
Both $f(x,y)$ and $\dfrac{\partial f}{\partial y}(x,y)$ are continuous in some rectancular region $R$.
Thus there exists a unique solution for the IVP.
$\mathbf {If \;\;\ a<0}:$
Then, $f$ is not defined over the rectangular region. So we try to solve the DE.
On solving we get, $\;\;\;y^{1/2}=x+\dfrac{c}{2}$
From the initial conditions, $y(0)=a$ we have,
$a^{1/2}=0+\dfrac{c}{2}$ which is not defined since $a<0$.
Thus the IVP has no solution.
$\mathbf {If \;\;\ a=0}:$
$f$ is not defined on any rectangular region $R$ containing $(0,0)$, hence nothing can be said about the existence or the uniqueness of the solution. So we investigate further using initial conditions.
$y^{1/2}=x+\dfrac{c}{2}$ with $y(0)=0$,
$\implies c=0$.
So, $y^{1/2} = x$, which is valid for $x\in [-c,c]$ for come $c>0$.
But in any such interval, $x$ will take negative values and thus $y^{1/2} = x$ will not hold. Hence it cannot be the solution of the IVP.
Furthermore, $y(x)=0$ is clearly a solution in this case.
I am not sure about the cases, $a<0$ and $a=0$.
Can anyone please guide me towards a solution?
For $a < 0$, if you don't want to allow square roots of negative numbers (which would be complex numbers), there can't be a solution because the right side of the differential equation is undefined.
The interesting case is $a = 0$, because that's where the conditions for the uniqueness theorem don't hold. Obviously $y = 0$ is a solution, but so is $y = x^2$ for $x \ge 0$. In fact, there are infinitely many solutions of the form $$ y = \cases{0 & for $x \le x_0$\cr (x-x_0)^2 & for $x > x_0$}$$
where $x_0 \ge 0$.