Uniqueness of second order ODE I am trying to determine whether or not 
$$ y'' = y^{1/3}, \; y(0) = 0, \; y'(0) = 1
$$
has a unique solution.  My gut feeling is that there is not, but apparently one can use Osgood's criterion to prove uniqueness.  I have attempted to do so, but was unable to come up with a function to satisfy the Osgood criterion.  Osgood's criterion can be found at https://mathoverflow.net/questions/55289/existence-uniqueness-of-solutions-to-quasi-lipschitz-odes .
My first step was to write this equation as
$$ x'(t) = f(x(t),t), \;\text{ where } f(x(t),t) = \begin{pmatrix} x_1 \\ x_0^{1/3} \end{pmatrix}
$$
where $x = \begin{pmatrix} x_0 \\ x_1 \end{pmatrix} = \begin{pmatrix} y(t) \\ y'(t) \end{pmatrix}$ in the initial problem statement.
Then 
$$ |f(x^{(1)},t) - f(x^{(2)},t)| \leq \begin{pmatrix} |x_1^{(1)} - x_1^{(2)}| \\ |x_0^{(1)^{1/3}}-x_0^{(2)^{1/3}}| \end{pmatrix} \leq \begin{pmatrix} |x_1^{(1)} - x_1^{(2)}| \\ |x_0^{(1)}-x_0^{(2)}|^{1/3} \end{pmatrix} 
$$
However, I do not think that this last string of inequalities provides anything useful for Osgood's criterion.  Any suggestions? Thanks in advance.
 A: When $\text{n}$ is a constant, multiply both sides by $y'(x)$ (when $y'(x)\ne0$): 
$$y''(x)=y(x)^{\frac{1}{\text{n}}}=\sqrt[\text{n}]{y(x)}\Longleftrightarrow y'(x)y''(x)=y'(x)\sqrt[\text{n}]{y(x)}$$
Now integrate both sides with respect to $x$:
$$\int y'(x)y''(x)\space\text{d}x=\int y'(x)\sqrt[\text{n}]{y(x)}\space\text{d}x$$
Now, use:


*

*Substitute $u=y'(x)$ and $\text{d}u=y''(x)\space\text{d}x$:
$$\int y'(x)y''(x)\space\text{d}x=\int u\space\text{d}u=\frac{u^2}{2}+\text{K}_1=\frac{y'(x)^2}{2}+\text{K}_1$$

*Substitute $s=y(x)$ and $\text{d}s=y'(x)\space\text{d}x$:
$$\int y'(x)\sqrt[\text{n}]{y(x)}\space\text{d}x=\int\sqrt[\text{n}]{s}\space\text{d}s=\frac{s^{1+\frac{1}{\text{n}}}}{1+\frac{1}{\text{n}}}+\text{K}_2=\frac{y(x)^{1+\frac{1}{\text{n}}}}{1+\frac{1}{\text{n}}}+\text{K}_2$$


So, we get:
$$\frac{y'(x)^2}{2}=\frac{y(x)^{1+\frac{1}{\text{n}}}}{1+\frac{1}{\text{n}}}+\text{K}\Longleftrightarrow\frac{y'(x)}{\sqrt{\frac{2y(x)^{1+\frac{1}{\text{n}}}}{1+\frac{1}{\text{n}}}+\text{K}}}=\pm1$$
Now integrate both sides with respect to $x$, again:
$$\int\frac{y'(x)}{\sqrt{\frac{2y(x)^{1+\frac{1}{\text{n}}}}{1+\frac{1}{\text{n}}}+\text{K}}}\space\text{d}x=\pm\int1\space\text{d}x$$
Now, use:


*

*$$\pm\int1\space\text{d}x=\text{C}\pm x$$

*Substitute $p=y(x)$ and $\text{d}p=y'(x)\space\text{d}x$:
$$\int\frac{y'(x)}{\sqrt{\frac{2y(x)^{1+\frac{1}{\text{n}}}}{1+\frac{1}{\text{n}}}+\text{K}}}\space\text{d}x=\int\frac{1}{\sqrt{\frac{2p^{1+\frac{1}{\text{n}}}}{1+\frac{1}{\text{n}}}+\text{K}}}\space\text{d}p$$


That will give a elliptic integral.
