find the solutions of initial value problem $x' = |x|^{1/2}$ with $x(0) = 0$ Show that the initial value problem
\begin{align}
&x' = |x|^{1/2}\\
&x_0 = 0
\end{align}
has four different solutions through the point $(0,0)$. 
I found three solutions which are
$$x_t= \frac14 t^2,$$
$$x_t= -\frac14 t^2,$$ and
$$x_t= 0$$
but I couldn't figure out the fourth one. 
 A: A solution to the differential equation is certainly non decreasing; suppose $f(t)$ is a nonconstant solution in the interval $[0,\infty)$ and set
$$
t_+=\sup\{t: t\ge0, f(t)=0\}
$$
Then $f$ satisfies $f(t_+)=0$; moreover, for $t>t_+$, $f(t)>0$ and $f'(t)=\sqrt{f(t)}$. Therefore
$$
\int_{t_+}^{t}\frac{f'(u)}{\sqrt{f(u)}}\,du=t-t_+
$$
that is,
$$
\sqrt{f(t)}=\frac{1}{2}(t-t_+)
$$
or
$$
f(t)=\frac{1}{4}(t-t_+)^2
$$
Similarly, if $f$ is nonconstant in the interval $(-\infty,0]$, there will be $t_-\le 0$ such that, for $t\le t_-$, $f(t)=-\frac{1}{4}(t-t_-)^2$, $f(t_-)=0$ and $f(t)=0$ for $t_-\le t\le 0$.
Conversely, it's easy to see that all functions
$$
f_{t_-,t_+}(t)=\begin{cases}
-\dfrac{1}{4}(t-t_-)^2 & \text{for $t\le t_-$}\\[4px]
0 & \text{for $t_-\le t\le t_+$}\\[4px]
\dfrac{1}{4}(t-t_+)^2 & \text{for $t\ge t_+$}
\end{cases}
$$
are solutions, where $t_-\le0$ and $t_+\ge0$ are arbitrary.
Also the functions
$$
f_{t_-,\infty}(t)=\begin{cases}
-\dfrac{1}{4}(t-t_-)^2 & \text{for $t\le t_-$}\\[4px]
0 & \text{for $t\ge t_-$}
\end{cases}
\qquad
f_{-\infty,t_+}(t)=\begin{cases}
0 & \text{for $t\le t_+$}\\[4px]
\dfrac{1}{4}(t-t_+)^2 & \text{for $t\ge t_+$}
\end{cases}
$$
are solutions, where, again, $t_-\le0$ and $t_+\ge0$ are arbitrary.
Possibly your book was referring to $f_{0,0}$, $f_{-\infty,0}$, $f_{0,\infty}$ and $f_{-\infty,\infty}$ (the constant solution).
