Find the largest interval for solution If $y(x)$ is the solution of differential equation$\frac {dy}{dx}=2(1+y)(\sqrt y)$ satisfying $y(0)=0,y(\frac {\pi}{2})=1 $ then the largest interval (to the right of origin ) on which the solution exists is 
(a)$[0,\frac {3\pi}{4}]$
(b)$[0,\pi)$
(c)$[0,2\pi)$
(d) )$[0,\frac {2\pi}{3}]$
Exact image of question
This question was asked in GATE Mathematics exam
My try
I solved the givrn differential equation and the solution is
$${tan^{-1}}(\sqrt y)=x+c$$
Can anyone give me hint in these question?

The curve $$y=\tan^2 x\qquad(x\geq0)\tag{1}$$
and all its horizontal translates are solution curves of this ODE. To this family of curves its envelope $y(x)\equiv0$ has to be added, since it is a solution as well. But this is not all: Since the ODE does not satisfy a crucial assumption of the existence and uniqueness theorem in the points $(x,0)$ we have to accept that for such initial points there are several solutions. Intuitively: We may choose an arbitrary $p\in{\mathbb R}$, then follow the envelope as long as $x\leq p$,   and for $x\geq p$  go along a translate of the curve $(1)$.
Now you want the solution passing through the point $\bigl({\pi\over2},1\bigr)$. It so happens that the following "splined" function does the trick:
$$y(x)=\left\{\eqalign{0\qquad\qquad&\bigl(x\leq{\pi\over4}\bigr)\cr \tan^2\bigl(x-{\pi\over4}\bigr)\qquad&\bigl({\pi\over4}\leq x<{3\pi\over4}\bigr)\ .\cr}\right.$$
This solution lives in the $x$-interval $\ \bigl]-\infty,{3\pi\over4}\bigr[\ $. It follows that answer (a) comes nearest.
