# 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.

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

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

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

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

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 $A^{-n}$ where A is the amplitude.

Any help would be appreciated.

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

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

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

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

Kind regards

$\chi$ $\sigma$

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

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

where

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

satisfies this DE.