Solving second order differential equation of unusual form How would one go about solving a differential equation of the form
$$ \frac{d^2x}{dt^2} = K( \frac{1}{x^\gamma})$$
where both K and Gamma are constants?
 A: This ode is autonomous and can be transformed by first multiplying by
$$
\dot{x} = \frac{dx}{dt}
$$
leaving
$$
\dot{x}\ddot{x} = K\cdot x^{\gamma}\dot{x}
$$
The left hand side is
$$
\dot{x}\ddot{x} = \frac{1}{2}\frac{d}{dt}\dot{x}^2
$$
we have
$$
K\cdot x^{\gamma}\dot{x} = \frac{K}{\gamma + 1}\frac{d}{dt}x^{\gamma + 1}
$$
thus we have
$$
\frac{d}{dt}\left[\frac{1}{2}\dot{x}^2 - \frac{K}{\gamma + 1}x^{\gamma + 1}\right] = 0
$$
or
$$
\dot{x}^2 - \frac{2K}{\gamma + 1}x^{\gamma + 1} = C
$$
(I will transform $2K \to K'$ due to missing out the factor 2)
The constant is usually a parameter of the model, to fit the observed data.
I will integrate the simpler model i.e. $C=0$
$$
\dot{x} = \sqrt{\frac{K'}{\gamma + 1}}x^{\frac{1}{2}(\gamma + 1)} 
$$
so we have
$$
\int_x x^{-\frac{1}{2}(\gamma + 1)} dx = \sqrt{\frac{K'}{\gamma + 1}}t + C_1
$$
or
$$
\frac{1}{-\frac{1}{2}(\gamma + 1) + 1}x^{-\frac{1}{2}(\gamma + 1) + 1} = \frac{1}{\frac{1}{2}(1 -\gamma)}x^{-\frac{1}{2}(\gamma + 1) + 1}
$$
or
$$
x^{\frac{1}{2}(1 -\gamma) } = \frac{1}{2}(1-\gamma)\sqrt{\frac{K'}{\gamma + 1}}t + C_1
$$
or
$$
x(t) = \left(\frac{1}{2}(1-\gamma)\sqrt{\frac{K'}{\gamma + 1}}t + C_1 \right)^{\frac{2}{1-\gamma}}
$$
A: Multiply both sides with $\frac{d}{dt} x(t) = x'(t)$. The result is
$$
\frac{1}{2} \frac{d}{dt} |x'(t)|^2 = x''(t)x'(t) = Kx(t)^{-\gamma}x'(t) = \frac{K}{1 - \gamma} \frac{d}{dt} x(t)^{1-\gamma}
$$
unless $\gamma = 1$. Now integrate both side to obtain
$$
|x'(t)|^2 = C + \frac{2K}{1-\gamma}x(t)^{1 - \gamma}
$$ 
Thus solutions live on level sets of the function $F(x,y) = y^2 - \frac{2K}{1-\gamma}x^{1 - \gamma}$ .
Take the square root and try to solve the resulting first ode 
$$
x'(t) = \sqrt{C + \frac{2K}{1-\gamma}x(t)^{1 - \gamma}} \, .
$$
