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Differential Equations through Power Series Expansion

  1. Mar 28, 2009 #1
    When solving diff-eq's given initial values, e.g.

    y'' - 2y ' + y = 0

    y (0) = 0

    y ' (0) = 1

    Can one assume immediately that

    y(0) = c0

    and y ' (0) = c1


    Since these are the first 2 terms in the series?

  2. jcsd
  3. Mar 28, 2009 #2


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    One can assume that
    [tex]y(x) = \sum_{n = 0}^\infty c_n x^n.[/tex]

    Plugging in x = 0 gives y(0) = c0.
    [tex]y'(x) = \sum_{n = 0}^\infty n c_n x^{n - 1}[/tex]
    and taking x = 0 gives y'(0) = c1.

    So you can show it, if you assume the series expansion to hold.
  4. Mar 28, 2009 #3


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    You can't "immediately assume" that y(0)= c0 because you haven't yet said what "c0" means! You can immediately assume that y(0)= 0 because you are told that. You are also told that y'(0)= 1 and you can calculate that y"(0)= 2y'(0)- y(0)= 2. Further, since y"= 2y'- y for all x, y"'= 2y"- y' and y"'(0)= 2y"(0)- y'(0)= 2- 1= 1.

    Then y""= 2y"'- y" so y""(0)= 2y"'(0)- y"= 2- 2= 0. So far, y(x)= 0+ 1x+ (2/2)x2+ 0x3= x+ x2.

    You can continue differentiating y"= 2y'- y to find the higher degree terms.
  5. Mar 28, 2009 #4
    So if we do assume that a solution will take the form of:

    [tex]y(x) = \sum_{n = 0}^\infty c_n x^n.[/tex]

    Then as CompuChip says, it follows that whatever this series looks like, c0 will be the only term without an x variable, and is thus equal to y(0).

    Likewise, when the series is differentiated, c1 will be the only term with out an x, and thus will equal y '(0).

    Is this true?
  6. Mar 29, 2009 #5


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    Yes, of course. That's just the MacLaurin series:
    [tex]\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n[/tex]
    The coefficient of xn for every n is the nth derivative divided by n!.
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