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Math Help - Did I make a mistake with this ODE via Power Series?

  1. #1
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    Did I make a mistake with this ODE via Power Series?

    I need to solve the following ODE using Power Series:


    xy'' + y' + xy = 0


    Now, when I do it, this is what happens:


    y = \sum_{n=0}^{\infty} a_n x^n

    y' = \sum_{n=1}^{\infty} na_n x^{n - 1}

    y'' = \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2}

    x \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} + \sum_{n=1}^{\infty} na_n x^{n - 1} + x \sum_{n=0}^{\infty} a_n x^n = 0

    \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 1} + \sum_{n=1}^{\infty} na_n x^{n - 1} + \sum_{n=0}^{\infty} a_n x^{n + 1} = 0


    Respectively:

    k = n - 1
    k = n - 1
    k = n + 1

    \sum_{k=1}^{\infty} k(k + 1) a_{k + 1} x^k + \sum_{k=0}^{\infty} (k + 1)a_{k + 1} x^k + \sum_{k=1}^{\infty} a_{k - 1} x^k = 0

    \sum_{k=1}^{\infty} k(k + 1) a_{k + 1} x^k + \sum_{k=1}^{\infty} (k + 1)a_{k + 1} x^k + \sum_{k=1}^{\infty} a_{k - 1} x^k + a_1 = 0

    \sum_{k=1}^{\infty} [x^k[(k + 1)ka_{k + 1} + (k + 1)a_{k + 1} + a_{k - 1}]] + a_1 = 0

    a_1 = 0

    (k + 1)ka_{k + 1} + (k + 1)a_{k + 1} + a_{k - 1} = 0

    a_{k + 1} =  \frac{-a_{k - 1}}{(k + 1)k + (k + 1)}


    y = a_0(1 - \frac{1}{4}x^2 + \frac{1}{64}x^4 - \frac{1}{2304}x^6 + ...)


    However, Wolfram Alpha says the solution is:

    y = C_1 J_0 + C_2 Y_0

    ...where J_n and Y_n are Bessel Functions


    So, did I make a mistake? And if so, where?
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  2. #2
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    Re: Did I make a mistake with this ODE via Power Series?

    For solutions on the real line, it looks correct. I only checked the first two terms, but so far your solution is consistent with Wolfram Alpha for C_1=a_0 and C_2=0. However, keep in mind that your ODE is equivalent (for x\neq 0) to

    \left[\begin{array}{c}y_1\\y_2\end{array}\right]'=\left[\begin{array}{c}y_2\\-(y_2/x)-y_1\end{array}\right], which may have solutions which tend to \pm\infty at zero. So you don't get to assume that every solution can be expressed in a power series about zero.
    Last edited by hatsoff; July 10th 2011 at 03:49 PM.
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    Re: Did I make a mistake with this ODE via Power Series?

    Quote Originally Posted by hatsoff View Post
    For solutions on the real line, it looks correct. I only checked the first two terms, but so far your solution is consistent with Wolfram Alpha for C_1=a_0 and C_2=0. .
    Forgive me, but I'm not sure I understand. The answer from WolframAlpha had a bessel function paired with C_2. How does my answer of C_2 = 0 correspond with that?
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  4. #4
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    Re: Did I make a mistake with this ODE via Power Series?

    Well C_1J_0+C_2Y_0 is just a linear combination of Bessel functions J_0 and Y_0. Your solution is apparently the function a_0J_0+0Y_0.
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    Re: Did I make a mistake with this ODE via Power Series?

    Quote Originally Posted by hatsoff View Post
    Well C_1J_0+C_2Y_0 is just a linear combination of Bessel functions J_0 and Y_0. Your solution is apparently the function a_0J_0+0Y_0.
    Okay, thanks. I guess I thought I should be getting a result that was more universal to a_1, rather than a specific value of a_1.
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  6. #6
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    Re: Did I make a mistake with this ODE via Power Series?

    What are you assuming a form y = \sum_{n=0}^{\infty} x^n and not y = \sum_{n=0}^{\infty} x^{n+r} (method of Frobenius')?
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    Re: Did I make a mistake with this ODE via Power Series?

    Quote Originally Posted by Lancet View Post
    I need to solve the following ODE using Power Series:


    xy'' + y' + xy = 0


    Now, when I do it, this is what happens:


    y = \sum_{n=0}^{\infty} a_n x^n

    y' = \sum_{n=1}^{\infty} na_n x^{n - 1}

    y'' = \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2}

    x \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 2} + \sum_{n=1}^{\infty} na_n x^{n - 1} + x \sum_{n=0}^{\infty} a_n x^n = 0

    \sum_{n=2}^{\infty} n(n - 1) a_n x^{n - 1} + \sum_{n=1}^{\infty} na_n x^{n - 1} + \sum_{n=0}^{\infty} a_n x^{n + 1} = 0


    Respectively:

    k = n - 1
    k = n - 1
    k = n + 1

    \sum_{k=1}^{\infty} k(k + 1) a_{k + 1} x^k + \sum_{k=0}^{\infty} (k + 1)a_{k + 1} x^k + \sum_{k=1}^{\infty} a_{k - 1} x^k = 0

    \sum_{k=1}^{\infty} k(k + 1) a_{k + 1} x^k + \sum_{k=1}^{\infty} (k + 1)a_{k + 1} x^k + \sum_{k=1}^{\infty} a_{k - 1} x^k + a_1 = 0

    \sum_{k=1}^{\infty} [x^k[(k + 1)ka_{k + 1} + (k + 1)a_{k + 1} + a_{k - 1}]] + a_1 = 0

    a_1 = 0

    (k + 1)ka_{k + 1} + (k + 1)a_{k + 1} + a_{k - 1} = 0

    a_{k + 1} = \frac{-a_{k - 1}}{(k + 1)k + (k + 1)}


    y = a_0(1 - \frac{1}{4}x^2 + \frac{1}{64}x^4 - \frac{1}{2304}x^6 + ...)


    However, Wolfram Alpha says the solution is:

    y = C_1 J_0 + C_2 Y_0

    ...where J_n and Y_n are Bessel Functions


    So, did I make a mistake? And if so, where?
    You didn't any mistake at all!... You have hipotized a solution analytic in x=0 and what You have found is the series expansion of the zero order first kind Bessel function...

     y= c\ J_{0}(x)= c\ (1 - \frac{x^{2}}{2^{2}} + \frac{x^{4}}{2^{2}\ 4^{2}} - \frac{x^{6}} {2^{2}\ 4^{2}\ 6^{2}} + ...) (1)

    The other possible solution is the zero order second kind Bessel function Y_{0} (x) which in x=0 has a logarithmic singularity...

    Kind regards

    \chi \sigma
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  8. #8
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    Re: Did I make a mistake with this ODE via Power Series?

    Quote Originally Posted by Danny View Post
    What are you assuming a form y = \sum_{n=0}^{\infty} x^n and not y = \sum_{n=0}^{\infty} x^{n+r} (method of Frobenius')?

    I'm assuming a form:

    y = \sum_{n=0}^{\infty} a_n x^n
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  9. #9
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    Re: Did I make a mistake with this ODE via Power Series?

    Quote Originally Posted by chisigma View Post
    You didn't any mistake at all!... You have hipotized a solution analytic in x=0 and what You have found is the series expansion of the zero order first kind Bessel function...

    Thanks for the confirmation and explanation!
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