Numer of solutions for IVP Consider the initial value problem
$\dfrac{dy}{dx}=3y^{{2}/{3}}$ with initial condition $y(0)=0$.
How many solutions are there for this IVP?


*

*1

*2

*3

*4

*infinitely many.
Clearly, $f(x,y)=3y^{\dfrac{2}{3}}$ does not satisfy Lipschitz's condition &  so the solution of the IVP is not unique.
Solving the equation with initial condition we get $y=x^{3}$. Again $y=0$ is the trivial solution. So I get two solutions.Are there any other solution(/s)? I want to know all the solutions & how we find their?
 A: There are infinitely many solutions, of course.
Indeed, all the functions of the form
$$
y_s(x) = 
\left\{
\begin{aligned}
&0 &&\text{ if } x\leq s\\
&(x-s)^3 &&\text{ if } x > s
\end{aligned}
\right.
$$
are solutions of your Cauchy problem for any $s \geq 0$. 

It is easy to check. Zero is a solution everywhere (in particular, for $x \leq s$); and $(x-s)^3$ is a solution of the Cauchy problem
$$
\frac{dy}{dx} = 3 y^{2/3} \quad \text{with} \quad y(s) = 0
$$
for $x \geq s$.
Now, if we "glue" zero and $(x-s)^3$, then it will be solution of your problem, except, probably, the point $x = s$.  However, at $x = s$ the function $y_s(x)$ is continuously differentiable, and hence, $y_s(x)$ is the correct solution of 
$$
\frac{dy}{dx} = 3 y^{2/3} \quad \text{with} \quad y(0) = 0.
$$
Thus, there are infinitely many of solutions, since $s \geq 0$ is arbitrary.
A: There is a slight subtlety about the definition of $y^{2/3}$ when $y<0$. If it is left undefined then I opt for $2$ solutions, if it is defined as $\bigl({\root 3\of{-| y|}}\bigr)^2=\bigl(-{\root 3\of {|y|}}\bigr)^2=|y|^{2/3}$ I opt for $4$ of them.
My reason for this is that I would count two solutions $x\mapsto\phi_1(x)$ and $x\mapsto\phi_2(x)$ which agree in an open neighborhood of $x=0$ as representants of the same solution germ. Since we can choose between $x\mapsto0$ and $x\mapsto x^3$ independently for $x\leq0$ and $x\geq0$ there are $4$ different germs in all (resp., $2$ of them if we leave $y^{2/3}$ undefined for $y<0$).
A: If we allow $x$ and $y$ to be complex there are arguably 3 nonzero solutions, with $x$ equated to the 3 complex roots of $y^{\tfrac{1}{3}}$, bringing the total number of solutions to 4. 
