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Homework Help: Two-Point Boundary Value Problem

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    y''+[tex]\lambda[/tex]y=0
    y'(0)=0
    y'(pi)=0

    2. Relevant equations



    3. The attempt at a solution

    What's puzzling me is the case when we check if the eigenvalue is zero.
    y''=0
    y'=C1
    y=C1x+C2

    Now when I check the first boundary value I get C1=0
    now How do I check the second one ? with the pi...
    It doesn't make sense plugging into the first derivative again because I have no x value (only a constant).
    The answers show this:


    lambda=0 is an eigenvalue and the general solution is y0(x)=1

    How did they get this ?

    Thanks.
     
  2. jcsd
  3. May 8, 2010 #2

    lanedance

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    y'=C1
    y'(0)=0, y'(pi)=0 --> C1 = 0
    so
    y=C2, constant, which up to a scalar multiple is the same as 1
     
  4. May 8, 2010 #3
    Hello,


    I didn't really understand your last statement.
    the constant is the scalar multiple of 1 ?
    I think I'm missing something...
     
  5. May 8, 2010 #4

    lanedance

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    whenever you find it solution to a DE it is only determined up to a scalar multiple

    in this case any constant satisfies the DE & bc's
     
  6. May 8, 2010 #5
    Oh, right....
    but why did they choose
    Yo(x)=1
    an arbitrary choice ?

    thanks.
     
  7. May 8, 2010 #6

    lanedance

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    based on what you've said, yeah i think so
     
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