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Second Order Linear ODE - Power Series Solution to IVP

  1. Nov 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Let y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞) be a power series solution of

    (x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0

    Note that x=0 is an ordinary point.

    2. Relevant equations
    y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞)
    y'(x)=[itex]\sum[/itex](kckxk-1) (k=1 to ∞)
    y''(x)=[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞)

    3. The attempt at a solution

    (x2-1)[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞) +[itex]\sum[/itex](kckxk) (k=1 to ∞)+[itex]\sum[/itex]ckxk (k=0 to ∞) -2x=0 ??

    I'm not having an issue with the power series themselves, I'm just not sure how to incorporate in the "2x" term when I'm setting up the series equation. We didn't cover this scenario in class and I couldn't find anything like it in my textbook.

    I've been trying to incorporate it as a series itself
    ie. 2[itex]\sum[/itex]ckxk (k=0 to ∞) where C0=0, or 2[itex]\sum[/itex]ckxk (k=1 to ∞)
    but I'm not sure I can even do that mathematically?


    Thank you!
     
  2. jcsd
  3. Nov 27, 2013 #2

    Dick

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    Your are going to equate powers of x to get a relation between the c_k values, right? The -2x will only contribute the x^1 term, yes?
     
  4. Nov 27, 2013 #3
    So I can leave it in as a constant coefficient term, say 2C1x? and then use that in combination with the other constants I've pulled out?
     
  5. Nov 27, 2013 #4

    Dick

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    Mmm. Sort of, but the coefficient your x^1 term is only going to have a -2 in it. Without any C_k term in front of it. It's just a constant.
     
    Last edited: Nov 27, 2013
  6. Nov 27, 2013 #5
    Oh! so I can just leave it be and put it with the other x1 coefficient terms?
     
  7. Nov 28, 2013 #6

    Dick

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    Sure.
     
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