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How is this an ordinary point? (Frobenius method for DE's)

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Find two linearly independent power series solutions for the differential equation y′′ − xy = 0 about the ordinary point x0 = 0. Your answer should include a general formula for the coefficients.

    3. The attempt at a solution

    Im having trouble seeing how x0 = 0 is an ordinary point (i assume ordinary point means regular singular point?).

    For it to be an ordinary point (x-x0)p(x) and (x-x0)q(x) have to be analytic at 0 right?

    (x-x0)p(x) = x(0/x) = 0

    (x-x0)q(x) = x(-1/x2) = -1/x which is not analytic at 0?

    So how can x0=0 be an ordinary point?
    Please help!!
     
  2. jcsd
  3. Nov 17, 2009 #2

    lanedance

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    Homework Helper

    so after a quick google to refresh my memory

    if the DE is in the form
    [tex] y'' + p(x) y' + q(x) y = 0 [/tex]

    if p(x) & q(x) are analytic at x0, its an ordinary point, if either diverges, its a singular point

    if its a singular point, but both (x-x0)p(x) & (x-x0)2q(x) are analytic at x0 its a regular singular point

    http://mathworld.wolfram.com/RegularSingularPoint.html
     
  4. Nov 17, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, that's where you are wrong- an "ordinary point" is not a singular point at all! And that means you don't use Frobenius' method, just a regular series solution.

    No. For an ordinary point p(x) and q(x) must be analytic.

     
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