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Power series method for DE

  1. Aug 12, 2006 #1

    siddharth

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    Question: Find a power series solution in powers of x for the following differential equation

    [tex] xy' - 3y = k [/tex]

    My attempt:
    Assume
    [tex] y = \sum_{m=0}^{\infty} a_m x^m [/tex]

    So,
    [tex] xy' = \sum_{m=0}^{\infty}m a_m x^m [/tex]

    [tex] xy'-3y-k=0 [/tex]

    implies

    [tex] \sum_{m=0}^{\infty}m a_m x^m - 3\sum_{m=0}^{\infty} a_m x^m - k = 0 [/tex]

    and

    [tex] \left(a_1x+2a_2x^2 +3a_3x^3 +.... \right) - \left(k + 3a_0 + 3a_1x+3a_2x^2+3a_3x^3+... \right) = 0[/tex]

    Which means

    [tex] a_0=-k/3 [/tex]
    [tex]a_1-3a_1=0, a_1=0 [/tex]
    [tex]2a_2-3a_2=0, a_2=0 [/tex]
    [tex]3a_3 - 3a_3=0, a_3=? [/tex]
    [tex]... a_n=0, n>3 [/tex]

    The Question: Now, how do I find [tex] a_3 [/tex]?
     
    Last edited: Aug 12, 2006
  2. jcsd
  3. Aug 12, 2006 #2

    0rthodontist

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    It could be anything-substitute back.
     
  4. Aug 12, 2006 #3

    siddharth

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    Of course.What was I thinking? I think my brain degenerated over the summer hols :frown:
    Thanks for the help.
     
  5. Aug 12, 2006 #4
    Just to add to this, for a n'th order linear DE you should expect to find n arbitrary constants. So it shouldn't be too surprising that one of the coeffiecients is arbitrary given that this is a first order linear DE.
     
  6. Aug 12, 2006 #5
    Perfect!

    In fact the general solution for the given equation has the form

    [tex]y = C x^3 - \frac{k}{3}[/tex].
     
  7. Apr 3, 2007 #6
    i got the same question; but i am not sure that the general solution is a power series representation
     
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