The question states
The Initial value problem $y'=2\sqrt{y},y(0)=a$ has
(a) unique solution if $a<0$
(b) no solutions if $a<0$
(c)infintely many solutions for $a=0$
(d)unique solution if $a\geq 0$
It is given option (c) as correct..
My opinion is for $a=0$, clearly will have trival solution $y=0$ and after solving this i.e.
$\begin{equation}\begin{aligned}\frac{dy}{dx}&=2\sqrt{y}\\\frac{dy}{\sqrt{y}}&=2dx\\2\sqrt{y}&=2x+c\ \text{integrating both sides}\\\sqrt{y}&=x+C\\y&=(x+C)^2\end{aligned}\end{equation}$
Using the initial condition for $a=0$, $y(0)=0$ gives $c=0$, so we got the solution as $y=x^2$, and the other is the trivial solution so there are only two solutions...
So, why it is concluded as ininitely many solutions for $a=0$....
Thanks in advance...