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Series Solution to ODE

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine a series solution to the following ODE about x0 = 0:

    [tex]xy'' + y' + xy = 0[/tex]

    3. The attempt at a solution

    I'll try to keep this concise.

    I first divided through by x and made the usual guesses for the form of the series. Subbing those in gave:

    [tex]\sum_{2}^{\infty}n(n-1)a_{n}x^{n-2}+ \sum_{1}^{\infty}na_{n}x^{n-2} + \sum_{0}^{\infty}a_{n}x^{n}[/tex]

    Then I shifted the first two series up 1 and the third one down 1 and multiplied through by x:

    [tex]\sum_{1}^{\infty}n(n+1)a_{n+1}x^{n}+ \sum_{0}^{\infty}(n+1)a_{n+1}x^{n} + \sum_{1}^{\infty}a_{n-1}x^{n}[/tex]

    Then to get the second series to start at 1, I moved the lower index up 1 and added the n = 0 term to make up for it:

    [tex]\sum_{1}^{\infty}n(n+1)a_{n+1}x^{n}+ \sum_{1}^{\infty}(n+1)a_{n+1}x^{n} + \sum_{1}^{\infty}a_{n-1}x^{n}+a_{1}[/tex]

    Now you can combine the series:

    [tex]\sum_{1}^{\infty}[n(n+1)a_{n+1}+(n+1)a_{n+1}+a_{n-1}]x^{n} + a_{1}[/tex]

    But then setting coefficients equal to zero gives a1 = 0, but don't you need to determine recurrence relations for a0 and a1?
  2. jcsd
  3. Jul 9, 2011 #2


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    No, your solution will have only even terms. The coefficient of the odd ones will be 0.
  4. Jul 9, 2011 #3


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    Because the coefficient of y'' is 0 at x= 0, you will not be able to find two independent solutions as power series. However, once you have found a single solution, even as a series, you can use the standard "order-reducing" method to reduce top a first order equation for the second independent solution- which should be something like ln(x) times the first solution.
  5. Jul 9, 2011 #4
    Thanks for the replies vela and Halls! Okay, so I'm just going to disregard the odd terms then:

    [tex]n(n + 1) a_{n+1} + (n + 1) a_{n+1} + a_{n-1} = 0[/tex]
    [tex]a_{n-1} = -a_{n+1}(n+1)^2[/tex]
    [tex]a_{0} = - 4a_{2} = - 16a_{4} = - 36a_{6} = - 64a_{8} = ...[/tex]
    [tex]a_{0} = - 2^2 a_{2} = - 4^2a_{4} = - 6^2a_{6} = - 8^2a_{8} = ...[/tex]
    [tex]a_{0} = - (2k)^2 a_{2k}:k=1,2,3...[/tex]
    [tex]a_{2k} = - \frac{a_{0}}{(2k)^2} :k=1,2,3...[/tex]

    [tex]y(x) = a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+a_{5}x^5+a_{6}x^6...[/tex]
    [tex]y(x) = a_{0}+a_{2}x^2+a_{4}x^4+a_{6}x^6...[/tex]
    [tex]y(x) = a_{0}- \frac{a_{0}}{(2(1))^2}x^2-\frac{a_{0}}{(2(2))^2}x^4-\frac{a_{0}}{(2(3))^2}x^6...[/tex]

    [tex] y(x) = a_{0}[1 + \sum_{n=1}^{\infty}\frac{x^{2n}}{(2n)^2}][/tex]
    [tex] y(x) = a_{0}[1 + \frac{1}{4}\sum_{n=1}^{\infty}(\frac{x^{n}}{n})^2][/tex]

    How's that? Also,

    Does that mean that I should guess a solution of the form y2(x) = v(x) y1(x) (for some v(x) and the y1 I just found) and then solve for v(x)?
  6. Jul 9, 2011 #5


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    That's not quite right. You found [itex]a_{n+2} = -a_n/(n+2)^2[/itex], so the first few terms are
    a_2 & = -a_0/2^2 = -a_0/4 \\
    a_4 & = -a_2/4^2 = -(-a_0/4)/16 = a_0/64 \\
    a_6 & = -a_4/6^2 = -(a_0/64)/36 = -a_0/2304
    which isn't what you got. The solution you should get is the Bessel function J0(x).
    Last edited: Jul 9, 2011
  7. Jul 9, 2011 #6
    I did? Oh yeah, I guess I did. So in this part:

    [tex]a_{n-1} = -a_{n+1}(n+1)^2[/tex]

    It's cool if I just sub in n + 1 for n (?) and then get:

    [tex]a_{n} = -a_{n+2}(n+2)^2[/tex]
    [tex]a_{n+2} = -\frac{a_{n}}{(n+2)^2}[/tex]


    [tex]y(x) = a_{0}- \frac{a_{0}}{4}x^2-\frac{a_{0}}{4\cdot16}x^4-\frac{a_{0}}{4\cdot16\cdot36}x^6...-\frac{a_{0}}{4\cdot16\cdot36\cdots(2n)^2}x^{2n}[/tex]

    I'm having a hard time determining a general formula for the series other than just

    [tex]y(x) = a_{0}[1-\sum_{n=1}^{\infty}\frac{x^{2n}}{4 \cdot16 \cdot36 \cdots(2n)^2}][/tex]
  8. Jul 9, 2011 #7

    Char. Limit

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    Well, let's see. Ignoring the x^(2n) for now, what I see is a product that can be written like this:

    [tex]\prod_{k=1}^n \frac{1}{(2k)^2} = \prod_{k=1}^n \frac{1}{4 k^2}[/tex]

    Now, I'm not certain if this is true, but I believe that you can separate the two, to get this:

    [tex]\left( \prod_{k=1}^n \frac{1}{4}\right) \left( \prod_{k=1}^n \frac{1}{k^2} \right)[/tex]

    The first product is rather easy to evaluate, and the second one can be manipulated like so to give another product that's easy to evaluate:

    [tex]\prod_{k=1}^n \frac{1}{k^2} = \left( \prod_{k=1}^n \frac{1}{k} \right)^2[/tex]

    Can you see what the product that's being squared evaluates to?

    Note: After a bit of testing, I now am relatively certain you can separate the one product into two.
  9. Jul 9, 2011 #8


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  10. Jul 9, 2011 #9
    Nice! I didn't know that trick.

    Regarding Char.Limit's suggestions, I think that:

    [tex] \prod_{k=1}^n \frac{1}{4} = \frac{1}{4^n} [/tex]


    [tex]\left( \prod_{k=1}^n \frac{1}{k} \right)^2 = \left[ (1/1)(1/2)(1/3)(1/4) \cdots (1/n) \right]^2 = \left[ \frac{1}{n!} \right]^2 = \frac{1}{(n!)^2}[/tex]

    So I've got:

    [tex]y(x) = a_{0}[1-\sum_{n=1}^{\infty}\frac{x^{2n}}{(4^n)(n!)^2}] [/tex]

    Thanks for the responses everyone by the way. How's that looking so far? Apparently it's a "modified Bessel function of the first kind..."

  11. Jul 9, 2011 #10


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    You forgot the factor of (-1)n.
  12. Jul 9, 2011 #11
    But wouldn't that make some terms positive?
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