Is it possible to solve this second order autonomous differential equation? $$x''(t) = \frac{1}{x^2(t)}$$
I'm interested in this differential equation because it mimics the motion of an object subject to gravity. The solution of this differential equation will be an algebraic expression of the position  of such an object, which would be useful. 
 A: Multiply with $2x'$ and integrate
$$
x'^2=C-\frac{2}{x}\implies x'=\pm\sqrt{C-\frac{2}{x}}
$$
This now are two separable first order ODE. 
Set $u=x'$ and re-insert into the original equation eliminating $x$, $\frac1x=\frac{C-u^2}2$ so that
$$
u'=x''=\frac1{x^2}=\frac{(C-u^2)^2}4
$$
which should be solvable using separation and partial fraction decomposition to at least an implicit solution form.
A: $$x''(t)=\frac{1}{x^2(t)}$$
$$2x'x''=\frac{2x'}{x^2}$$
$$\left(\frac{dx}{dt}\right)^2=-\frac{2}{x}+c_1$$
With condition $-\frac{2}{x}+c_1\geq 0$ for all the following.
$$\pm\sqrt{\frac{1}{-\frac{2}{x}+c_1}}dx=dt$$
$$\pm\int\sqrt{\frac{x}{x-2c_1}}dx=t+\text{constant}$$
I suppose that you can integrate for $t(x)$.
The inverse function $x(t)$ cannot be expressed with a finite number of elementary functions. For approximate solution use series or numerical calculus. 
If some conditions where specified allowing to determine the constants of integration some simplification might occur.
A: Primes are differentiation with respect to time $t$
$$x''(t)=\frac{1}{x^2(t)}$$
$$2x'x''=\frac{2x'}{x^2}$$
$$x^{'2} =-\frac{2}{x}+\frac{2}{a}$$
where $a$ is a convenient arbitrary constant
$$\frac{dx}{dt}= \sqrt{2\frac{(x-a)}{a\,x}}$$
Integrate again by hand or taking Mathematica help,
$$ \sqrt{x(x-a)}+a \,\ln\big(\sqrt x+ \sqrt{x-a}\big)= \sqrt\frac{2}{a}\,t+b $$
where $b$ is another arbitrary constant.
Due square root/cosh type combination the function is transcendental in nature, i.e.,  $x$ cannot be expressed in terms of $t$ easily. When boundary conditions are given (two for second order DE) a numerical solution can be obtained by a CAS.
