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Power Series Diff. Eq.

  1. Apr 11, 2006 #1
    Given:
    y'+2xy=0
    Find:
    Write sereis as an elementary function

    My solution so far:
    y=[Sum n=0, to infinity]C(sub-n)*x^n
    y'=[Sum n=1, to infinity]n*C(sub-n)*x^(n-1)


    y' can be transformed into:
    =[Sum n=0, to infinity](n+1)*C(sub-n+1)*x^n

    ([Sum n=0, to infinity](n+1)*C(sub-n+1)*x^n) + 2x([Sum n=0, to infinity]C(sub-n)*x^n)=0

    ([Sum n=0, to infinity](n+1)*C(sub-n+1)*x^n) +2([Sum n=0, to infinity]C(sub-n)*x^(n+1))=0


    My question:
    Can I transform [Sum n=0, to infinity]C(sub-n)*x^(n+1) into ([Sum n=1, to infinity]C(sub-n-1)*x^(n)?

    If so, then y(x)=([Sum n=,0. to infinity]((-1)^(n+1)*C(sub-o)*x^(n+1)*2^(n-1))/n!

    How do I transform that into an elementary function?

    Sorry about the ugly typing...
     
  2. jcsd
  3. Apr 12, 2006 #2

    dextercioby

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    Science Advisor
    Homework Helper

    Why don't you simply integrate the ODE and then write the series expansion for the solution...?

    Daniel.
     
  4. Apr 12, 2006 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In other words:
    Can you change
    [tex]\Sigma_{n=0}^\infty C_n x^{n+1}[/tex]
    to
    [tex]\Sigma_{n=1}^\infty C_{n-1}x^n[/tex]

    I think you will find it easier to use a different letter for the index:
    Let j= n+1. Then n= j-1 so Cn becomes Cj and xn+1 becomes xj[/sub]. Of course when n= 0, j= 1 so
    [tex]\Sigma_{n=0}^\infty C_n x^{n+1}[/tex]
    is the same as
    [tex]\Sigma_{j=1}^\infty C_{j-1} x^j[/tex]
    and since j is just a dummy index, you can certainly replace it with n.
     
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