Newtonian Mechanics - Differential equation If we combine Newton's second law of motion i.e. $F=m\ddot{x}$ and Newton's law of gravity i.e., 
$$
F=G\frac{mM}{x^2},
$$
where $x$ is distance, we obtain the following equation:

$$\ddot{x}=MG\frac{1}{x^2}$$ My question is how do we solve this differential equation for $x$ ? I see no ordinary way of doing it. I have tried using Laplace transformation but apparently there exists no transformation for $\frac{1}{x^2}$ so I'm left with nothing to use.
Hint:  Set $p(x)=x'(t)$ so you'll have $x''(t)=p(t)p'(t)$ so $$x''=\frac{MC}{x^2}\to pp'=\frac{MC}{x^2}$$ and so $$\int pdp=\int \frac{MC}{x^2}dx$$
(double dot χ)*(χ^2)= MG
But (double dot χ) equals the acceleration a, which equals ΔV/t.
This implies that;
(χ squered)= MGt/ΔV.
Hence
χ= Squere root(MGt/ΔV).
I hope this is helpful, as it is simple since it does not involve the trouble of integration constants.