Free-fall according to Newton's gravitation law Most analysis of free-fall assume that bodies fall with constant acceleration. If however one analyses  free-fall according to Newton's gravitation law, one is lead to a differential equation which I don't know how to solve. 
The differential equation one is lead to[1] is $\frac{d^2x}{dt^2}=-\frac{1}{x^2}$. Does this equation have a closed form solution with given initial conditions $x_0,v_0$?
[1] If one picks the units so that $MG=1$, fixes one mass at $x=0$ and places the falling mass at some $x>0$.
 A: Multiply the equation by $dx/dt$ and integrate once to get
$$
\frac{1}{2}\Bigl(\frac{dx}{dt}\Bigr)^2=\frac{1}{x}+\frac{v_0^2}{2}-\frac{1}{x_0}.
$$
Then
$$
\frac{dx}{dt}=\pm\sqrt{\frac{2}{x}+v_0^2-\frac{2}{x_0}}\,,
$$
a first order equation in separated variables whose solution is
$$
\int_{x_0}^x\frac{dz}{\sqrt{\frac{2}{z}+v_0^2-\frac{2}{x_0}}}=\pm t.
$$
The integral can be computed explicitly, although I do not think you can find a closed form for $x$ in terms of $t$.
A: Your Question is about radial trajectories that satisfy a central force inverse square law with unit "gravitational constant" $MG$.  The ordinary differential is equation is autonomous, meaning that time t does not explicitly appear and that time-translation of one solution gives another.
In this case you have fixed the (central) location of one (point) mass.  The "falling mass" is the only point in motion.  Depending on the initial conditions, the bodies may actually be moving apart rather than falling together (even assuming that the force is attractive rather than repellant).  Even though the motion is one-dimensional, terminology is borrowed from conic sections to describe the three cases:
Radial elliptic trajectory:  The moving point has less than escape velocity at all times and a period motion, alternately falling towards and moving away from the central mass.  At the transition times, velocity drops to zero.
Radial parabolic trajectory:  The moving point has exactly the escape velocity at all times, so eventually the distance between it and the central point increases without limit.  This special case has a rather simple expression for radial distance $x$ as a function of $t$, $x(t) = \left(\frac{9t^2}{2}\right)^{\frac{1}{3}}$, although this makes the initial velocity infinite where $x(0) = 0$.  Solutions with a positive $x(0)$ can be obtained by time translation (and by reversing time as necessary depending on the initial direction of motion).
Radial hyperbolic trajectory:  The moving point's initial velocity exceeds escape velocity, also eventually giving unbounded increasing distances between it and the central point.
The time-coordinate transformation $p(t) = \left(\frac{9t^2}{2}\right)^{\frac{1}{3}}$ that appears in the parabolic case above can be used to give a series solution in the "universal trajectory" cases.  Let $p$ be this transfomed time, the separation the moving point would have had if it were on a parabolic trajectory, and let $w = \frac{1}{x_0} - \frac{v_0^2}{2}$.  Then:
$$ x(t) = p - \frac{1}{5}wp^2 - \frac{3}{175}w^2p^3 - \frac{23}{7875}w^3p^4
 - \frac{1894}{3931875}w^4p^5 - \ldots $$
As @JuliánAguirre points out, it is simpler to obtain time as a function of radial distance $t(x)$ in the general case.
A: I managed to do it
NewtonGravitySolution
The final solution I had was:
$t=\sqrt{\frac {R}{2Gm} } \biggl( x \sqrt{\frac {R}{x} -1} +tan^{-1}\sqrt{\frac{R} {x}-1 }  \biggr)$
$R$ is the initial position where $v=0$
If you plug in values for Earth with Mass = Earth's Mass you can plot it.
