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Solving an ODE with power series method

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve ##(1-x)y''+y=0## at the point ##x_0=0##. Use this solution to find a solution to ##xy''+y=0## around the point ##x_0=1##.

    2. Relevant equations

    3. The attempt at a solution

    ##(1-x)y''+y=0##
    ##(x-1)y''=y##
    ##\displaystyle\sum_{k=2}^\infty a_k k (k-1)\left(x^{k+3}-x^{k+2}\right)=\displaystyle\sum_{k=0}^\infty a_k x^k##

    I tried to find a recurrence relation for ##a_k##, but I keep getting 0 for all k. I realize that y = 0 is a solution, but I don't know how that would help me in the second part.
     
  2. jcsd
  3. Mar 10, 2013 #2

    ehild

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    Compare the coefficients of xk: they have to be the same on both sides.

    What do yo get for the constant term? What is the linear term?

    ehild
     
  4. Mar 10, 2013 #3
    Right, that's where my problem occurs:

    ##y''=\sum_{k=2}^\infty a_k k(k-1)\left(x^{k+3}-x^{k+2}\right)=a_2 (2\cdot1)\left(x^5-x^4\right)+a_3 (3\cdot2)\left(x^6-x^5\right)+a_4 (4\cdot3)\left(x^7-x^6\right)+\cdots##

    ##y=\sum_{k=0}^\infty a_k x^k=a_0 + a_1x + a_2x^2+ a_3x^3+a_4x^4+\cdots##

    Matching up coefficients, I have
    ##a_0=0\\ a_1=0\\ a_2=0 \\ a_3=0\\
    a_4=-(2\cdot1)a_2=0 \\ a_5=(2\cdot1)a_2-(3\cdot2)a_3=0\\ \text{and so on...}##
     
  5. Mar 10, 2013 #4

    ehild

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    The second derivative of y is wrong. The correct equation is

    [tex]\displaystyle\sum_{k=2}^\infty a_k k (k-1)\left(x^{k-2}-x^{k-1}\right)=\displaystyle\sum_{k=0}^\infty a_k x^k[/tex]

    ehild
     
  6. Mar 11, 2013 #5
    Oh, right. I had (k + 2) on my mind for some reason. Thanks for pointing that out!
     
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