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Math Help - Solve a 2nd order linear ODE

  1. #1
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    Solve a 2nd order linear ODE

    How can one solve the following ODE:
    y''+(x^2-1)y=0

    Thank you very much.
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  2. #2
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    Re: Solve a 2nd order linear ODE

    That one's nasty. WolframAlpha gives parabolic cylinder functions. Mathematica gives hypergeometric functions. In any case, a series solution is, I think, the way to get where you need to go.
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  3. #3
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    Re: Solve a 2nd order linear ODE

    Quote Originally Posted by Ackbeet View Post
    That one's nasty. WolframAlpha gives parabolic cylinder functions. Mathematica gives hypergeometric functions. In any case, a series solution is, I think, the way to get where you need to go.
    Thank you very much for your hints.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Solve a 2nd order linear ODE

    Quote Originally Posted by xinglongdada View Post
    How can one solve the following ODE:
    y''+(x^2-1)y=0

    Thank you very much.
    If we write the ODE as...

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

    ... it is clear that the general solution is...

    y(x)=c_{1}\ y_{e} (x) + c_{2}\ y_{o} (x) (2)

    ... where...

    y_{e} (x) = \sum_{n=1}^{\infty} a_{2 n}\ x^{2n} (3)

    ... is an even function and...

    y_{o} (x) = \sum_{n=1}^{\infty} a_{2 n+1}\ x^{2n+1} (4)

    ... is an odd function. The a_{2 n} and a_{2 n +1} are found directly from (1) imposing initial conditions y(0)=y^{'}(0)=1 \implies a_{0}=a_{1}=1...

    y^{(2)}= (1-x^{2})\ y \implies y^{(2)}(0)=1 \implies a_{2}= \frac{1}{2} (5)

    y^{(3)}= (1-x^{2})\ y^{(1)} -2 x\ y \implies y^{(3)}(0)=1 \implies a_{3}= \frac{1}{6} (6)

    y^{(4)}= (1-x^{2})\ y^{(2)} -4 x\ y^{(1)} -2\ y \implies y^{(4)}(0)=-1 \implies a_{4}= - \frac{1}{24} (7)

    y^{(5)}= (1-x^{2})\ y^{(3)} -6 x\ y^{(2)} -6\ y^{(1)} \implies y^{(5)}(0)=-5 \implies a_{5}= - \frac{1}{24} (8)

    Of course You can proceed in the computation... in next post we will try to find a general expression for the a_{n}...

    Kind regards

    \chi \sigma
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: Solve a 2nd order linear ODE

    In the previous post it has been demonstrated that the second order linear ODE...

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

    ... has general solution...

    y(x)= c_{1}\ h_{e} (x) + c_{2}\ h_{o}(x) (2)

    ... where...

    h_{e} (x)= \sum_{n=0}^{\infty} a_{n}\ \frac{x^{2n}}{(2n)!} (3)

    ... is an even function and...

    h_{o} (x)= \sum_{n=0}^{\infty} b_{n}\ \frac{x^{2n+1}}{(2n+1)!} (4)

    ... is an odd function. In general the computation od the derivatives of y(*) from (1) can be done recursively as...

    y^{(n)} (x) = (1-x^{2})\ y^{(n-2)} (x) -2\ (n-2)\ x \ y^{(n-3)}(x) - (n-2)\ (n-3)\ y^{(n-4)}(x) (5)

    ... and from (5), setting x=0, we can derive the recursive relations...

    a_{n}= a_{n-1}- (2n-2)\ (2n-3)\ a_{n-2}\ ,\ a_{0}=1, a_{-1}=0 (6)

    b_{n}= b_{n-1}- (2n-1)\ (2n-2)\ b_{n-2}\ ,\ b_{0}=1, b_{-1}=0 (7)

    Deriving explicit solutions of the recursive relations (6) and (7) seems not very confortable... the recursive computation of the a_{n} and b_{n} is however very easy. The figure shows the values of h_{e}(x) and h_{o}(x) in -2<x<2 computed using 25 terms of (3) and (4)...



    Kind regards

    \chi \sigma
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  6. #6
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    Re: Solve a 2nd order linear ODE

    Thank you very much. And I know now it is solved by the method of series.
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