TITLE: An INITIAL VALUE PROBLEM
QUESTION [0 upvotes]: The initial value problem $y' = 2\times\sqrt{y}$ , $y(0)=a$ has

a unique solution for $a<0$
no solution for $a>0$
infinitely many solutions if $a=0$
a unique solution if $a\ge 0$.
My answer is 4.
As if we solve the ode by separating $y'$ ,then we have a general solution as $\sqrt{y}=x+c$.
Now using $y(0)=a$ , we have $c=\sqrt a$, which is valid if $a \ge 0$, and in that case we get a unique solution for that $a$.
Now the answer is given to be 3, if $a=0$ ,then we get $y=x²$ ,which is unique , then how is it 3 ?
Is my method correct and my answer also ? Or am I wrong ?
Hoping for a help.Thank you.

REPLY [0 votes]: You are wrong. For instace, for any $\beta\ge0$ the function $$f_\beta(x)=\begin{cases}(x-\beta)^2&\text{if }x\ge \beta\\ 0&\text{if }x<\beta\end{cases}$$
satisfies $f'_\beta=2\sqrt{f_\beta}$ and $f_\beta(0)=0$.
The problem with your argument is that, keen as it is on working only algebraically, it "conveniently" skips all the crucial details of a proper proof, basically:

the fact that the differential equation $y'=2\sqrt{\lvert y\rvert}$ fails to satisfy the hypothesis of Cauchy-Lipschitz theorem in any neighbourhood of the line $\Bbb R\times \{0\}$
the fact that, while for $y(0)>0$ you can rewrite the problem as an integral equation $$\int_0^x \frac{y'(t)}{2\sqrt{y(t)}}\,dt=\int_0^x1\,ds,$$
this cannot be done when $y$ is constantly $\equiv 0$ in a neighbourhood of $0$.
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