How to solve the ODE $\ddot{x}=\frac{c}{x}$ How to solve the following ODE? $$\ddot{x}=\frac{c}{x}$$
I have no idea how to solve it since it is not linear. Is there a way of separation or something like that?
 A: Hint: Multiply both sides by $\dot x$ and integrate over $t$ on both sides.
Hint 2: $\dot x dt = dx$
A: $\textbf{hint}$
$$
\ddot{x} = \dot{x}\dfrac{d}{dx}\dot{x} = \dfrac{d}{dx}\frac{\dot{x}^2}{2} = \frac{c}{x}
$$
Given that 
$$
\dfrac{d}{dt} =\dfrac{dx}{dt}\frac{d}{dx} = \dot{x}\dfrac{d}{dx}\tag{*}
$$
Then
$$
\ddot{x} = \dfrac{d}{dt}\dot{x} 
$$
Take eq(*) and put in the above
$$
\ddot{x} = \dot{x}\dfrac{d}{dx}\dot{x} 
$$
A: $$\frac{d^2x}{dt^2}=\frac{c}{x}$$
$$2\frac{d^2x}{dt^2}\frac{dx}{dt}=\frac{2c}{x}\frac{dx}{dt}=$$
$$\left(\frac{dx}{dt}\right)^2=2c\ln(x)+C_1$$
$$\frac{dx}{dt}=\pm \sqrt{2c\ln(x)+C_1}$$
$$dt=\pm\frac{dx}{\sqrt{2c\ln(x)+C_1}}$$
$$t=\pm\int{\frac{dx}{\sqrt{2c\ln(x)+C_1}}+C_2}$$
The closed form of this integral involves special function of the erf kind. $x(t)$ is the inverse function.
A: it is easier if you write it as a system of first order differential equations. like this $$\frac{dx}{dt} = y, \frac{dy}{dt} = \frac{c}{x}.$$ you can turn this into $$\frac{dy}{dx} = \frac{c}{xy} $$ which can be separated as $$y\, dy =  \frac{c}{x}$$
you can take it from there.
