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Math Help - Laguerre's Differential Equation

  1. #1
    Super Member 11rdc11's Avatar
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    Laguerre's Differential Equation

    I need just a little help with this question. I almost solved it except for a minor detail at the end.

    Here is the question.

    The equation

    xy''(x) + (1-x)y'(x) +my(x) = 0,

    where m is a nonnegative interger, is called Laguerre's differential equation. Show that for each m, this equation has a polynomial solution of degree m. These polynomials are denoted by L_{n}(x) and are called Laguerre polynomials. The first few Laguerre polynomials are

    L_{0}(x) = 1
    L_{1}(x) = -x +1
    L_{2} = x^2-4x+2

    Ok so I did a power series solution for the problem and arrived at

    a_{0} = constant

    a_{1} = -ma_{0}

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

    so

    a_{2} = a_{0}\frac{(-m)(1-m)}{4}

    a_{3}=a_{0}\frac{(-m)(1-m)(2-m)}{(4)(9}

    a_{4} = a_{0} \frac{(-m)(1-m)(2-m)(3-m)}{(4)(9)(16)}

    and this is where I get confused.

    I see that there is a factorial pattern so I'm trying to use factorial notation to condense.

    so would it be

    \sum_{n=2}^{\infty} \frac{(-m)_{n}}{(1)^2_{n}}x^n

    or

    \sum_{n=2}^{\infty} \frac{(-m)_{n}}{(1^2)_{n}}x^n

    So would the final solution be

    m \geq 0

    y(x) = a_{0} -ma_{1}x + whatever series is correct?

    This solution would show depending what m is chosen that the answer would cancel out certain terms.
    Last edited by 11rdc11; September 18th 2010 at 09:58 AM.
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  2. #2
    Super Member 11rdc11's Avatar
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    The second part of the question is this

    Use the results to obtain the 1st few terms in a series expansion about x=0 for a genreal solution for x > 0 to Laguerre's differential equation for n = 0 and 1. Does this just mean plug in 0 and 1 into the general solution I found earlier?
    Last edited by 11rdc11; September 18th 2010 at 09:57 AM.
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