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Math Help - solve non linear ODE.

  1. #1
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    solve non linear ODE.

    hello,
    I have an ODE and am unable to solve it can somebody help:

    dy/dt = -5*t*y^2 + 5/t - 1/t^2 , initial condition is y(1) = 1

    I've look at various ways to solve but can't variable separable benoulli, integrating factor, exact etc but just can't seem to do it anyone can help.
    Thank you.
    Last edited by mr fantastic; April 3rd 2011 at 03:22 AM. Reason: Title.
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  2. #2
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bubuer View Post
    hello,
    I have an ODE and am unable to solve it can somebody help:

    dy/dt = -5*t*y^2 + 5/t - 1/t^2 , initial condition is y(t) = 1

    I've look at various ways to solve but can't variable separable benoulli, integrating factor, exact etc but just can't seem to do it anyone can help.
    Thank you.
    I believe this expression is a Ricatti DE. See Riccati equation - Wikipedia, the free encyclopedia

    I think they are a little tricky to solve and unfortunately I do not have experience with them. One of the math giants here might reply to you
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  3. #3
    A Plied Mathematician
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    It's not Ricatti right now, because there's no y term. Question: where is the initial condition? You have y(t) = 1. Is that supposed to be y(0) = 1?

    [EDIT]: See running-gag's post for a correction.
    Last edited by Ackbeet; April 2nd 2011 at 11:17 AM.
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  4. #4
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    sorry, initial condition is y(1) = 1.
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  5. #5
    MHF Contributor
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    Hi

    I confirm this is a Riccati equation

    If you can find a particular solution y_1 then letting y = y_1 + u, you can find a Bernoulli equation for u
    Équation de Riccati - Wikipédia

    Here y_1(t)=\frac{1}{t} is a particular solution

    At the end I can find y(t) = \frac{1}{Ke^{10t}-\frac{t}{2}-\frac{1}{20}}+\frac{1}{t}

    K can be found using the initial condition
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  6. #6
    A Plied Mathematician
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    Oh, right. You can have the coefficient of the y term be zero. Forgot about that.
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  7. #7
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    so, it's a special case of the ricatti eqations i'll have a closer look at these.
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