Differential Equation $dy/dx=y^{1/3}$ and condition $y(x_0)=y_0$ has infinite solutions Prove that the differential equation 
  $$\frac{dy}{dx}= y^{1/3}$$
  with the initial value of $y(x_0)=y_0$ has infinite solutions.


I don't really understand the problem if I have to show that there are infinite solutions depending on the initial conditions or if it is something like if I proposed $(x_0,y_0)=(0,0)$ and prove that for that case the equation has infinite solutions. if you think is the first one could you explain how to do it.
If $y_0 &gt; 0$ then $f(y) = y^{1/3}$ is Lipschitz in a neighbourhood of $y_0$ (the derivative is bounded). So by Picard-Lindelöf there is a unique solution.

On the other hand, for $y_0 = 0$, $f$ is no longer Lipschitz and so we can no longer expect a unique solution. In fact two solutions are found readily:

$$ y(x) = 0 \text{ and } y(x) = \left[\frac{2}{3}(x - x_0)\right]^{3/2}. $$

But as Wikipedia alludes to in its article on the Peano existence theorem, the transition between the two solutions can happen at any point (not just at $x_0$). So the general solution is

$$ y_a(x) = \begin{cases}
0 &amp; x \le a, \\
\left[\frac{2}{3}(x - a)\right]^{3/2} &amp; x &gt; a,
\end{cases} $$

for any parameter $a \ge x_0$.
With $$ \frac{dy}{dx} = \sqrt[3]{y} \quad y(x_{0}) = y_{0}$$ then 
\begin{align}
y^{-1/3} \, dy &amp;= dx \\
\frac{3}{2} \, y^{2/3} &amp;= x + c_{0} \\
y(x) &amp;= \left[ \frac{2}{3} \, (x + c_{0}) \right]^{3/2}.
\end{align}

When $y(x_{0}) = y_{0}$ then
$$c_{0} = \frac{3}{2} \, y_{0}^{2/3} - x_{0}$$
and
$$ y(x) = \left[ \frac{2}{3} \, (x - x_{0}) + \sqrt[3]{y_{0}} \right]^{3/2}. $$