Kepler's third law for linear motion I do not know how to solve second order differential equations but am interested in seeing that Kepler's third law is valid for linear motion too. 
This can be proven using: 
$$\frac{d^2r}{dt^2}= -\frac{GM}{r^2}$$ 
After solving this differential equation, $r$
 should be made the subject of the formula and it would be seen that 
$$T^2 \varpropto r^3$$
I plugged this into wolfram alpha but it doesn't make $r$ the subject of the formula. 
Can someone prove Kepler's third law for linear motion using the differential equation I provided? I am not equipped with the mathematical tools to do it otherwise I'd have done it myself. 
Thanks. 
 A: If $AB$ is parallel to $CD$ then the areas of $\triangle CBD = \triangle CAD$ 

Linear motion with no external forces applied.
If motion is constant in traveling in the direction $B\to F$ in equal intervals of time, $BC = CD = DE = EF$

And we trace out equal areas in equal times.
Planetary motion with a periodic application of force.

We travel in our first unit of time from $A$ to $B.$  We then are perturbed hit by a force in the direction of O.  Without this kick we would have continued to $C$ but instead we continue to $C'$
$\triangle ABO = \triangle BCO$  because $AB = BC$
$CC'$ is parallel to $BO$
$\triangle BC'O = \triangle BCO = \triangle ABO$ 
And the same and be said for $\triangle C'D'O$, etc.
giving us equal areas in equal time.
These kicks of force to not have to be uniform magnitude.  They could even be negative. 
Since you seem to be familiar with differential equations, I will let you work out what happens a the time intervals becomes arbitrarily small.
A: In one dimension, the particle of mass $m$ moving along the $x$ axis, with $r=|x|$, has a potential energy $V(x)=-GMm/r$. To have a periodic orbit, the particle has to have negative energy (so it has two turning points) and it has to cross the origin of coordinates, where the force acting on the particle and the speed of the particle become infinite. Letting aside how physical this situation is, one still can compute the period of the orbit using the same general techniques one uses in 1-D for motion under conservative forces. From conservation of energy one gets:
$$\frac{1}{2}mv^2-\frac{GMm}{r} = E \Rightarrow v =\sqrt{\frac{2}{m}\left(E+\frac{GMm}{r}\right)}\,.$$ 
For a periodic orbit with turning point at $r=r_0$ (where the particle is at rest), the energy is:
$$E=-\frac{GMm}{r_0}\,,$$
so the speed of the particle as a function of $r$ and the turning point $r_0$ is:
$$v=\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)}$$
Because of the symmetry of the potential, the period of the orbit is four times the time it takes to go from the origin to $r=r_0$:
$$T=4\int_0^{r_0}\frac{dr}{v} = 4\int_0^{r_0}\frac{dr}{\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)}}\,.$$
Making the change of variables $r=r_0u$, we get:
$$T = r_0^{3/2}4\int_0^1\frac{du}{\sqrt{2GM\left(\frac{1}{u}-1\right)}}=
r_0^{3/2}4\int_0^1\sqrt{\frac{u}{2GM(1-u)}}\,du=\frac{2\pi}{\sqrt{2GM}}r_0^{3/2}\,.$$
The factor multiplying $r_0^{3/2}$ is a constant, so $T^2$ is proportional to $r_0^3$, i.e. Kepler's third law is also valid in 1-D. 
