Solve $y'=2 \sqrt{|y|}$, $y(0)=0$ 
Solve $y'=2 \sqrt{|y|}$, $y(0)=0$.

$$\frac{dy}{2\sqrt{|y|}}=dx$$
The left side seems to be similiar to the derivative of $\sqrt{y}$ but the initial condition is $y(0)=0$ so it makes the denominator disappear. Any help?
 A: A solution of the equation must be a non decreasing function. Suppose $f$ is a solution and consider $k=\sup\{x\in\mathbb{R}:x\ge0, f(x)=0\}$. If $k=\infty$, then the solution is identically zero on $[0,\infty)$.
Otherwise it satisfies $f'(x)=2\sqrt{f(x)}$, $f(k)=0$ and $f(x)>0$ for $x>k$. Integrating over $(k,\infty)$
$$
\int\frac{f'(x)}{2\sqrt{f(x)}}\,dx =\int dx
$$
we get
$$
\sqrt{f(x)}=x+c
$$
so
$$
f(x)=(x+c)^2
$$
The initial condition $f(k)=0$ gives $c=-k$. 
Similarly, set $h=\inf\{x\in\mathbb{R}:x\le0, f(x)=0\}$. If $h=-\infty$ the function is identically zero on $(-\infty,0]$. Otherwise $f(x)<0$ for $x<h$ and $f$ satisfies $f'(x)=2\sqrt{-f(x)}$, $f(h)=0$ and $f(x)<0$ on $(-\infty,h)$. Integrating over $(-\infty,h)$ we get
$$
\int\frac{f'(x)}{2\sqrt{-f(x)}}\,dx=\int dx
$$
so
$$
-\sqrt{-f(x)}=x+c
$$
and therefore
$$
-f(x)=(x+c)^2
$$
The initial condition $f(h)=0$ gives $c=-h$.
Thus there are infinitely many solutions of one of the following forms:
\begin{gather}
f(x)=0\\
f(x)=\begin{cases}
(x-k)^2 & \text{if $x>k$}\\
0 & \text{if $x\le k$}
\end{cases}\\
f(x)=\begin{cases}
0 & \text{if $x\ge h$}\\
-(x-h)^2 & \text{if $x<h$}
\end{cases}\\
f(x)=\begin{cases}
(x-k)^2 & \text{if $x>k$}\\
0 & \text{if $h\le x\le k$}\\
-(x-h)^2 & \text{if $x<h$}
\end{cases}
\end{gather}
where $h\le 0$ and $k\ge 0$ are arbitrary.
A: By comparing
$$
\frac{d}{dx}(x+C)^2=2(x+C)
$$
and
$$
2\sqrt{|(x+C)^2|}=2\,|x+C|
$$
one finds that $y(x)=(x+C)^2$ is only a solution where $x+C>0$. The continuation to the left is by $y\equiv 0$ or $y(x)=-(x+D)^2$ for $x+D<0$.
This gives the general solutions as
$$
y(x)=(x-C)_+{}^2-(x-D)_{-}{}^2
$$
with $D\le C$ and $(x-C)_+=\max(0,x-C)$.
The IVP $y(0)=0$ has the infinity of solutions
$$
y(x)=(x-C)_+^2
$$
for any $C\ge 0$.
A: $$\frac { 1 }{ 2 } \int { \frac { dy }{ \sqrt { |y| }  }  } =\int { dx } \\ \sqrt { |y| } =x+C\\ y={ \left( x+C \right)  }^{ 2 }\\ y\left( 0 \right) =0\quad \Rightarrow \quad C=0\\$$ so the final answer is 
$$ y={ x }^{ 2 }$$
