How are second order nonlinear ordinary differential equations solved? I conceived the following second order nonlinear ordinary differential equation:
$$\frac{d^2y(x)}{dx^2}=\frac{k}{(y(x))^2}$$
I can tell it's nonlinear because of the $\frac{k}{(y(x))^2}$ term and second order because of the second order derivative. 
Also, I did some research and concluded that it is of the type "missing x". In this category we use the relation, according to SOS math :  $$\frac{d^2y(x)}{dx^2}=v\frac{dv}{dy} (1)$$  by making v equal to $$\frac{dy(x)}{dx} (2)$$ and then, using the chain rule, simplify to obtain equation (1). 
Despite all that, I can't seem to find a logical solution to my ODE. I would appreciate any help or clue! Thanks! 
 A: Typically, if your equation has a second derivative and a zeroth derivative but no first derivative, you can reduce the order by multiplying both sides by the first derivative and integrating. This works mathematically because the $y''$ gives you the "du" you need to integrate the left side, while the $y'$ gives you the "du" you need to integrate the right side. In many physical situations, this amounts to identifying a conservation law.
In this case you wind up with:
$$\frac{1}{2} y'^2=-\frac{k}{y}+C.$$
This equation is separable, provided you can unambiguously choose a sign for the square root. 
A: You wrote that $\frac{d^2y}{dx^2}=v\frac{dv}{dy}$, so substitute that into your equation and you get $$v\frac{dv}{dy}=\frac{k}{y^2}.$$Now, $$v\ dv=\frac{k}{y^2}dy$$ $$-\frac{k}{y}+C_1=\frac{v^2}{2}$$ $$-\frac{2k}{y}+C_2=v^2$$ $$v=\pm\sqrt{-\frac{2k}{y}+C_2}$$Then, integrate and you get$$y = \pm\left(\sqrt{C_2-\dfrac{2k}{y}}y+\dfrac{k\ln\left(\frac{\left|\sqrt{C_2-\frac{2k}{y}}-\sqrt{C_2}\right|}{\sqrt{C_2-\frac{2k}{y}}+\sqrt{C_2}}\right)}{\sqrt{C_2}}\right)+C_3.$$Haven't checked yet, though... (Link here to the complicated integral, and click the "Go!" button.)
