Extensions of the ODE solution We have differential equation $y'=y^{\frac{1}{2}}$ with the condition $y(0)=0$.
How do we check if the solution is extendable on $\mathbb{R}$? I didn't find anything that could possibly help me with this problem. Thank you!
 A: Thanks to a remark of @Lutzl, the most general solution is as given in the following figure :

Fig. 1 : The half parabola can be placed at any positive abscissa, or even be absent (in this case, the solution is function $y=0$).
Why that ? Either we have a solution which is $y=0$ in a vicinity of the origin (which can be bounded or not), or there exists an $x_0>0$ for which $y(x_0)>0$ ; due to the continuity of function $y$, there is an open interval centered in $x_0$ such that such that $y(x) \neq 0$ ; thus, in this interval, the differential equation can be transformed into $y^{-1/2}{y'}=1$ ; integrate it as $2 y^{1/2}=x+k$, where $k$ is a constant, hence :
$$y=\frac{1}{4}(x+k)^2 \tag{1}$$ 
Here comes a small case study for "piecing" the solutions : 


*

*either this constant $k$ is $\leq 0$, and we have the solution depicted on the figure (where we have taken $k=-1.5$), by "piecing" the solution $y=0$ and the solution given by (1), and this is the only way to do the "piecing" in this case,

*or this constant $k$ is $>0$, and we cannot fulfill condition $y(0)=0$. 
A: Apart from $y \equiv 0$, $y=\frac {x^{2}} 4$ for $x \geq 0$ and $y=0$ for $x<0$ is a solution on $\mathbb R$. 
