Differential Equation @ 2nd Order How to solve this? $$\dfrac{d^2x}{dt^2}=-kx^2$$
I need it in physics problem.https://physics.stackexchange.com/questions/60202/non-shm-oscillatory-motion/     Though solved without it but still it would be better to have this.
 A: The differential equation is not easy to solve. The usual trick is to let $v=\frac{dx}{dt}$. Multiply both sides by $2\frac{dx}{dt}$. 
On the left-hand side we have the derivative of $v^2$. On the right, we have the derivative of $-\frac{2k}{3}x^3$. So $v^2=-\frac{2k}{3}x^3+C$. We end up with a separable equation. Unfortunately, unless we are very lucky and $C=0$, we end up with something that cannot be integrated in elementary terms. 
A: $$\frac{d^2x}{dt^2}=-kx^2$$
$$\implies \frac{d}{dt}\left(\frac{dx}{dt}\right)=-kx^2$$
$$\implies d\left(\frac{dx}{dt}\right)\left(\frac{dx}{dt}\right) =-kx^2dx$$
Integrating either sides, $$ \frac{\left(\frac{dx}{dt}\right)^2}2=\frac{-kx^3}3+c $$
$$\implies \frac{dx}{dt}=\pm \sqrt{\frac{6c-2kx^3}3}$$
A: One solution will be of the form $x=at^n$ so substitute that in to get $an(n-1)t^{n-2}=-ka^2t^{2n}$.  This gives $2n=n-2, n=-2$  Then $an(n-1)=6a=-ka^2$ so $a=-\frac 6k$.  Unfortunately as the equation is not linear, you can't use this to get all solutions.
A: I think this might be part of the solution
let x(t) be a Taylor series
the co-efficients of the taylor series contain x(0), x'(0), x''(0) ... etc 
I think these can be derived from d2xdt2=−kx2 and some initial conditions for x(0) = a, and x'(0) = b
http://i.stack.imgur.com/nRi4k.jpg
The final recurrence relation can be used to compute the x(0), x'(0), x''(0) co-efficients which can then be used to construct the taylor series for x(t)
It may not be nice but I believe it will work ...  
