Difficult exercise on unicity of solutions for an IVP Suppose $f$ and $g$ are continuous and $g$ is odd and  strictly increasing function. I have to prove that the IVP $$y'=f(x)g(y)$$ $$y(0)=1$$ has a unique solution if and only if $$\lim \limits_{u \to 0} \left [ \int_{u}^{1} \frac {1}{g(y)} dy \right] = + \infty$$
Does someone have a hint on what I could use? I have absolutely no idea from where the result follows. As $g$ is odd (and so $g(0)=0$), I would expect the limit to be always infinite. 
I know that any solution of the ODE will satisfy $$\int \frac {1}{g(y)} dy = \int f(x) dx$$ but I don't know how I could use this result... 
 A: I would comment on the original post, but I don't have the rep yet.
As is, the problem does not seem correctly posed.  Consider the initial value problem $$y'=y^{1/3}, \hspace{1cm} y(0)=1.$$
Here, $f(x)=1$ and $g(y)=y^{1/3}$.  $g$ is odd and strictly increasing.  As $g$ is differentiable on the positive real line, it is Lipschitz continuous in any finite interval of the form $[\alpha,1+\beta]$ for $0<\alpha<1$ and $\beta>0$.  Therefore, by the Picard-Lindelöf theorem, a unique solution $y(t)$ of the IVP exists for $t\in(1-\epsilon,1+\epsilon)$ and some $\epsilon>0$.
(EDIT: I forgot to mention, of course, we have $$\int_u^1 \frac{1}{y^{1/3}}dy=\frac{3}{2}\left(1-u^{2/3}\right)\rightarrow \frac{3}{2}$$ as $u\rightarrow 0^+$, so that the limit exists and is not infinite.)
The "correct" version is probably more along the lines of:
" Suppose $f$ and $g$ are continuous and $g$ is odd and strictly increasing.  The initial value problem $$y′=f(x)g(y), \hspace{1cm} \mathbf{y(0)=0}$$ has a unique solution if and only if ...limit relation... holds "
