Existence/Uniqueness of solution of singular ODE Consider the following singular ODE
$$
(t^{N-1}g(y'))'=t^{N-1}f(y)
$$
with initial condition
$$y(0)=y_0$$
$$y'(0)=0$$.
where $g$ is and increasing function.
How one can prove that this problem has a solution?
 A: One cannot prove existence in the generality that you have stated it.  For example, if $g(p) = e^p$, there is no solution when $N>1$ and $f$ is differentiable.
To see this, observe that, if there were a solution $y(t)$ to your problem in a neighborhood of $t=0$, then the curve $(t,y,p) = (t,y(t),y'(t))$ would be an integral curve of the system
$$
dy - p\ dt = d(t^{N-1}g(p)) - t^{N-1}f(y)\ dt = 0
$$ 
that passes through the point $(t,y,p) = (0,y_0,0)$.
Now, off the hypersurface $t=0$ in $typ$-space, this is equivalent to the system
$$
dy - p\ dt = \bigl((N{-}1)g(p)-tf(y)\bigr)\ dt + tg'(p)\ dp = 0.
$$
As long as $g(0)\not=0$ and $N>1$, this system has rank $2$ at the point $(t,y,p) = (0,y_0,0)$, so that there is a unique integral curve of the system passing through this point.  However, inspection shows that the curve is $(t,y,p) = (0,y_0,p)$ is an integral curve that passes through the point in question.  By uniqueness, it is the only one.
Thus, there is no integral curve of the kind that you would get from a solution to your problem.
I suspect that you have left out some hypotheses or not made a correct symmetry reduction of the system that you are trying to study.
