# Thread: Differential Equation (Oscilation Problem)

1. ## Differential Equation (Oscilation Problem)

Hello,
I am trying to solve what I thought was going to be a simple differential equation.

$\displaystyle m\ddot{x}=-kx^{2n+1}:n \in \mathbb{Z}-\{0\} and k,m \in \Re$.

So I set
$\displaystyle y=\frac{dx}{dt}\Rightarrow \frac{dy}{dt}=-\frac{k}{m}x^{2n+1}$
$\displaystyle \frac{dy}{dt}\frac{dt}{dx}=-\frac{k}{m}x^{2n+1}\frac{1}{y}=-\frac{k}{m}\frac{x^{2n+1}}{y}=\frac{dy}{dx}$

Then separating variables and integrating I got
$\displaystyle y^2+\frac{k}{m(n+1)} x^{2n+2}=C$

Solving for y
$\displaystyle y=\underline{+}\sqrt{C-\frac{k}{m(n+1)} x^{2n+2}$

remembering $\displaystyle y=\frac{dx}{dt}$ we can again separate variables
$\displaystyle \frac{dx}{\underline{+}\sqrt{C-\frac{k}{m(n+1)} x^{2n+2}}}=dt$

This is where I get stuck. I know this should be integrable to a sinusoid, but I don't know how. The actual question asks you to prove that the solution oscillates with a period proportional to $\displaystyle A^{-n}$ where A is the amplitude.

Any help would be appreciated.

2. The solution is strongly related to the integral...

$\displaystyle \displaystyle \int \frac{dx}{\sqrt{1+a\ x^{2\ (n+1)}}}$ (1)

... that can be solved integrating 'term by term' the expansion...

$\displaystyle \displaystyle \frac{dx}{\sqrt{1+a\ x^{2 (n+1)}}}= 1 - \frac{a}{2}\ x^{2 (n+1)} + a^{2}\ \frac{1\ 3}{2\ 4}\ x^{4 (n+1)} - a^{3}\ \frac{1\ 3\ 5}{2\ 4\ 6}\ x^{6 (n+1)} +...$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. May be you need not solve this DE.
You need to test that

$\displaystyle \displaystyle { y=const \; A \; e^{i A^n t} }$

where

$\displaystyle | \: A^n \; t \: | <<1$

satisfies this DE.