DIfferential Equations Initial Value Problem Existence and Uniqueness

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I'm working on a textbook question, and it's the following:

Determine in terms of $a$ and $b$ how many different solutions the initial value problem $y'=4x\sqrt{y}$, $y(a)=b$ has, given that the solution is:
$$y(x) = \begin{cases} 0& \text{if $x^2 \leq c$} \\ (x^2-c)^2 & \text{if $x^2>c$} \end{cases}$$

Okay, so using the existence and uniqueness theorem:
I get that $f(x,y)=4x\sqrt{y}$ is continuous $\forall y\geq 0$ and,
$f_y(x,y)=\frac{2x}{\sqrt{y}}$ is continuous $\forall y>0$.
I believe I have two cases for these two functions: either some small enough open rectangle containing the point $(a,b)$ has all of its region above the $x$ axis, or some of the rectangular region sits below the $x$ axis.

So if $b>0$ then we are guaranteed existence of unique solutions, and
if $b \leq 0$ then we are not guaranteed existence and uniqueness because the rectangular region will have points below $y=0$, also having $b<0$ would have no solutions because we can't take square roots of negative numbers

But if $y \equiv 0$ then we have part of the piece wise function that satisfies that being a solution. So we can say that there exists unique solutions if $b \geq 0$

But here's the actual answer from the textbook:

The initial value problem $y'=4x \sqrt{y}$, $y(a)=b$ has infinitely many solutions (defined for all x ) if $b \geq 0 $; no solutions if $b < 0$. However, if $b > 0$ then it has a unique solution near $x = a$.

I'm actually quite confused at this point, and not quite sure what I'm doing. There aren't any set rules I'm following, and I'm used to having a mechanical way of doing things, but I'm just not sure what to do. Any help regarding the intuition of the answer or "instructions" of doing these types of questions would be appreciated.

Thanks.

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The solution formula is not general enough. You actually have $$ y(x)=\begin{cases} (x^2-c_1)^2&\text{for }x<{-}\sqrt{c_1},\\ 0&\text{for }{-}\sqrt{c_1}\le x\le \sqrt{c_2},\\ (x^2-c_2)^2&\text{for }\sqrt{c_2}<x \end{cases} $$ so that you have 2 branching points of which at least 1 is always independent of the initial condition.