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Finding a power series solution to a differential equation?

  1. Apr 23, 2017 #1
    1. The problem statement, all variables and given/known data
    "Find the recurrence relation in the power series solution for ##y''-xy'-y=0## centered about ##x_0=1##."

    2. Relevant equations
    ##y=\sum_{n=0}^\infty a_nx^n##
    Answer as given in book: ##(n+2)a_{n+2}-a_{n+1}-a_n=0##

    3. The attempt at a solution
    ##y=\sum_{n=0}^\infty a_n(x-1)^n##
    ##xy'=x\sum_{n=0}^\infty na_n(x-1)^{n-1}=x\sum_{n=1}^\infty na_n(x-1)^{n-1}=x\sum_{n=0}^\infty (n+1)a_{n+1}n(x-1)^{n}##
    ##y''=\sum_{n=0}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=0}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=2}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=0}^\infty a_{n+2}(n+2)(n+1)(x-1)^{n}##

    ##y''-xy'-y=0=\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}-x(n+1)a_{n+1}-a_n](x-1)^n##

    ##(n+2)(n+1)a_{n+2}=x(n+1)a_{n+1}+a_n##

    I don't know what I'm doing wrong. It's not matching the answer given in the back of the book.
     
  2. jcsd
  3. Apr 23, 2017 #2
    Never mind, I found what I did wrong.

    ##xy'=x\sum_{n=0}^\infty na_n(x-1)^{n-1}=[1+(x-1)]⋅\sum_{n=0}^\infty na_n(x-1)^{n-1}=\sum_{n=1}^\infty na_n(x-1)^{n-1}+\sum_{n=0}^\infty na_n(x-1)^{n}=\sum_{n=0}^\infty (n+1)a_{n+1}(x-1)^{n}+\sum_{n=1}^\infty na_n(x-1)^{n}##
     
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