consider the initial value problem $ y'=y^{1/3}, y(0)=0$ consider the initial value problem $y'=y^{\frac{1}{3}} , y(0)=0 $ This is already asked
Here
I have a doubt in those answer, as that question is 6 years old, i ask this seperately.


*

*Its is written in comment that the function is ill-defined in negative axis. What does it means? Please explain this intuitively


*Also in that accepted answer how answerer got that idea of that function. It's not trivial

 A: The only reasonable definition of $y^{1/3}$ for $y<0$ in this context is
$$
y^{1/3} = -|y|^{1/3} \qquad \forall y<0.
$$
Whoever claims differently is just confusing people.
The non uniqueness of solutions is related to the non-Lipschitzianity of the function; it has nothing to do with it being "ill-defined in negative axis", whatever that means. You would have the same problem with the ODE $y'=|y|^{1/3}$. What happens on the negative $y$ axis does not matter at all in this exercise.
The example of a parametrized family of solutions starting at $y(0)=0$ is a standard exercise. The key insights are that:


*

*the ODE is autonomous, so you can translate solutions,

*$y=0$ is a solution,

*you can "join" the $y=0$ solution with any one departing from it.

A: *

*In general, $y \mapsto y^\beta$ with $\beta > 0$ real is only defined for $x\ge0$. How would you define $(-2)^\pi$? This is why the function is said to be ill-defined for $x<0$.

*To find the solutions of $y^\prime=y^{\frac{1}{3}} , y(0)=0$, you can notice that functions always vanishing on an interval are solutions. And for $y(t) \neq 0$, you can rewrite the equation as 
$$\frac{y^\prime(t)}{(y(t))^{1/3}} = 1$$ and integrate it on $(\alpha,x)$. You get the solution $y(x) = (2/3)^{3/2} (x-\alpha)^{3/2}$. You then see that the functions


$$y(x)=\begin{cases}
0 & \text{ for } x < \alpha \\
(2/3)^{3/2} (x-\alpha)^{3/2} & \text{ for } x \ge  \alpha
\end{cases}$$
are also solutions.
