TITLE: Existence and uniqueness of soluctions of $y'=xy^{2/3}$
QUESTION [5 upvotes]: It is asked to analyze the existance and uniqueness of solutions of the ode at every point $(x_o, y_o)$
$$y' = 3y^{2/3}$$
My attempt: We consider the initial condition $ y(x_o)=y_o$. If we consider the subset of $R^2$ such that $y\neq 0$, then we can verify the function $F(x,y)=3y^{2/3}$ is continuous and lipschitzian, and then admit an unique solution y such that $y(x_o)=y_o$
Now, if $y = 0,$ clearly $y=0$ is a solution of this ODE. We could solve the ODE by separating the variables too and obtain $y=(x+c)^3$
But my doubt is: If $y=0,$ wasnt 0 the only solution or the solution obtained separating the vsrizbles is also a solution? Applying the initial condition, we should have $y=(x-x_0)^3$, but is it a solution?
Thanks in advanced!

REPLY [3 votes]: At $y = 0$, $F$ is not Lipschitz, so the existence and uniqueness theorem does not apply if the initial condition is $y_0 = 0$. That is, there is no guarantee that there are unique solutions (thus it is no surprise that you found two distinct solutions). 
Further, you can get even more solutions, for any $c > x_0$:
$$
y = \begin{cases} 
 0 & : x \leq c, \\
(x - c)^3 &: x > c.
\end{cases}
$$