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Math Help - Non-equilibrium solutions of an ODE

  1. #1
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    Non-equilibrium solutions of an ODE



    So the equilibrium solutions are -1 and 1, but what are the non-equilibrium solutions?

    I started with:

    dy/(1-y^2) = dt

    -ln(1-y^2) = t + c

    Am I going in the right direction?
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  2. #2
    A Plied Mathematician
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    Your integration of the DE seems a bit off, to me. Try the y integration again.
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  3. #3
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    Unfortunately not. Remember, the integral of dU/U is the natural log of the absolute value of U, but you don't have the differential of your denominator, -(y^2-1) in the numerator. You'll have to use partial fraction decomposition.
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  4. #4
    MHF Contributor chisigma's Avatar
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    A simple preliminary analysis of the DE...

    y^{'} = 1 - y^{2} (1)

    ... permits You to conclude that is...

    a) y^{'} >0 for -1< y < 1

    b) y^{'} = 0 for y=1 or y=-1

    c) y^{'} < 0 for y=1 or y=-1

    ... so that is...

    a)  \lim_{t \rightarrow \infty} y(t) = 1 for y(0) >-1

    b)  \lim_{t \rightarrow \infty} y(t) = - 1 for y(0) = -1

    c)  \lim_{t \rightarrow \infty} y(t) = -\infty for y(0) <-1

    Kind regards

    \chi \sigma
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