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Math Help - Help with 2nd order non-linear DE

  1. #1
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    Help with 2nd order non-linear DE

    y''-(1+t^2)y'+1=1 when y(0)=1 y'(0)=0

    im not even sure where to begin... can someone please give me a hint
    Last edited by mr fantastic; March 23rd 2010 at 07:34 PM. Reason: Changed post title
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  2. #2
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    Are you SURE you've written it correctly?

    As it is written, I'm tempted to subtract 1 and set up a very simple system where f = y' and f' = y".
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  3. #3
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    sorry i messed it up

    y''-(1+t^2)y'+y=1 when y(0)=1 y'(0)=0
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  4. #4
    MHF Contributor chisigma's Avatar
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    We suppose that the solution of the second order linear DE...

    y^{''} - (1 + t^{2})\cdot y^{'} + y = 1 , y(0)=1 , y^{'} (0) = 0 (1)

    ... can be expressed as McLaurin series, so that is...

    y(t) = \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\cdot t^{n} (2)

    Now it is possible to find how many terms of the (2) You like, starting from the first two that are given as 'initial conditions'. From (1) we derive y^{''} (0) as ...

     y^{''} (0) = (1 + t^{2})\cdot y^{'} (0) - y(0) + 1 = 0 (3)

    From (3) we derive y^{(3)} (0) as...

    y^{(3)} (0) = (1 + t^{2})\cdot y^{''} (0) + (2 t -1)\cdot y^{'} (0) = 0 (4)

    From (4) we derive y^{(4)} (0) as...

    y^{(4)} (0) = (1 + t^{2})\cdot y^{(3)} (0) + (4 t +1) \cdot y^{''} (0) + 2\cdot y^{'}(0) = 0 (5)

    At this point it is evident that \forall n>0 is y^{(n)} (0) = 0 so that the solution of (1) is [simply] y(t)=1 ...

    Kind regards


    \chi \sigma
    Last edited by chisigma; March 25th 2010 at 05:26 AM. Reason: Previous 'solution' completely wrong... sorry!...
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