Number of differentiable functions satisfying $y' = 2 \sqrt{y}$

Question

The following question was asked in KVPY $2021$ held on $22$nd May $2022$:

The number of differentiable functions $y:(-\infty, +\infty) \to [0, \infty)$ satisfying $y' = 2\sqrt{y}$, $y(0) = 0$ is

(A) $1$ (B) $2$ (C) finite but more than $2$ (D) infinite

Upon integrating and using $y(0) = 0$ we get $\sqrt{y} = x$. But since $\sqrt{y}$ is always positive but the domain of $x$ includes negative numbers, we square both sides, giving $y = x^2$ as a function which satisfies the given condition. I am unable to come up with any other function hence I think the answer should be (A), but the given answer is (D).

Could someone explain why there are infinite functions satisfying the given conditions? Or if my answer is correct and there is only one, please provide a convincing argument/proof.

Answer 1

Any function which is zero on $[0,a]$ for some $a>0$ and equals $(x-a)^2$ on $[a,\infty)$ satisfies the property. Since $a$ is any positive number, you have infinitely many solutions.

This is a standard example of non-uniqueness of solutions to initial value problems for ODEs. The same happens for $y'=y^\alpha$ for any $\alpha<1$.