Vtyjkyuk

I want to solve the differential equation$$begincasesx''=frac2xx^2-1\x'(0)=0\x(0)=x_0endcases$$

This is what I have done so far. I have not studied differential equation much, and introducing the function $s$ below is just a trick that I learned, but I'm not sure why/if it works.

Let $s:=x'$. Then
$$fracd^2xdt^2=fracdsdt=fracdsdxfracdxdt=fracdsdxs$$
so the above equation becomes $$sfracdsdx=frac2xx^2-1$$
$$s,ds=frac2xx^2-1,dx$$$$int s,ds=intfrac2xx^2-1,dx+C$$
$$s^2=loglvert x^2-1rvert+C$$$$x'=omegasqrtloglvert x^2-1rvert+C,,omega:mathbb R_+mapsto-1,1$$
and with $x'(0)=0$, $x(0)=x_0$ we have
$$x'=omegasqrtlogbigglvertfracx^2-1x_0^2-1biggrvert$$
How can I continue to solve for $x(t)$? And am I justified in introducing $s$ and treat the notation like I did to obtain $x'$?

Your calculus is correct.
$$fracdxdt=omegasqrtlnbigglvertfracx^2-1x_0^2-1biggrvert$$
with $omega=pm 1$ .
$$t=pmint fracdxsqrtlnbigglvertfracx^2-1x_0^2-1biggrvert$$
With condition $x(0)=x_0$ :
$$t=int_x_0^x fracdxisqrtlnbigglvertfracxi^2-1x_0^2-1biggrvert$$
One can show that the integral is convergent for $xito x_0$. That isn't the main trouble.

The integral cannot be expressed in terms of standard functions, a fortiori the inverse function $x(t)$ as well. Numerical calculus is required to fully solve the problem.

The procedure is correct. The last equation is separable, and its solution can be written as
$$
int_x_0^xfracdxsqrtlogBiglvertdfracx^2-1x_0^2-1Bigrvert=x_0+omega,t.
$$
However, the integral has no closed form in terms of elementary functions (or even special functions, as far as I can tell.)