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Series solution up to a term, power series

  1. Jul 2, 2011 #1
    1. The problem statement, all variables and given/known data
    consider the initial value problem (1-x)y,,+xy,-2y=0 find the series solution up to the term with x6


    2. Relevant equations
    (1-x)y,,+xy,-2y=0


    3. The attempt at a solution
    assuming the answer has the form [itex]\Sigma[/itex]anxn
    that gives y,,=[itex]\Sigma[/itex]nanxn-1 and y,,=[itex]\Sigma[/itex]n(n-1)anxn-2 then plugging these back in and getting rid of the xn-2 and xn-1 you get the equation [itex]\Sigma[/itex](n+2)(n+1)an+2xn-[itex]\Sigma[/itex]n(n+1)an+1xn+[itex]\Sigma[/itex]nanxn-2[itex]\Sigma[/itex]anxn so what I'm wondering is how do you replace the an+1 with an an so you can solve for an+2
     
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  3. Jul 2, 2011 #2

    tiny-tim

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    hi dp182! :smile:
    just subtract 1 from everything :wink:

    (eg ∑an+1xn = ∑anxn-1)
     
  4. Jul 2, 2011 #3

    vela

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    You don't. The recurrence relation involves an, an+1, and an+2. There's nothing wrong with that.
     
  5. Jul 2, 2011 #4
    It makes sense that the recurrence relation for a_n+2 involves a_n+1 and a_n. It would require you to define a_0 and a_1 and Second order differential equations require two boundary/initial conditions for a solution.
     
  6. Jul 2, 2011 #5
    So If I solve for an+2 I get an+2=n(n+1)an+1+an(2-n)/(n+2)(n+1) so to get my terms up to x6 I will just input values of n, so my series will be in terms of a0and a1?
     
  7. Jul 2, 2011 #6

    vela

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    Yup, but you made an algebra error solving for an+2 (or omitted very needed parentheses).
     
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