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Math Help - Differential Equation (Oscilation Problem)

  1. #1
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    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}
    <br />
\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.
    Brad
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  2. #2
    MHF Contributor chisigma's Avatar
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    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
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  3. #3
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    May be you need not solve this DE.
    You need to test that

    <br />
\displaystyle<br />
{<br />
y=const \; A \;  e^{i A^n t}<br />
}<br />

    where

    <br />
| \: A^n \; t \: | <<1<br />

    satisfies this DE.
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