Solving second order differential equation of unusual form

Question

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?

Answer 1

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 $$ x(t) = \left(\frac{1}{2}(1-\gamma)\sqrt{\frac{K'}{\gamma + 1}}t + C_1 \right)^{\frac{2}{1-\gamma}} $$

Answer 2

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 $$ x'(t) = \sqrt{C + \frac{2K}{1-\gamma}x(t)^{1 - \gamma}} \, . $$