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Linear dependence and Wronskian

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Okay so the question is to show that these 2 functions are linearly dependent.
    ie. they are not both solutions to the same 2nd order, linear, homogeneous differential equation for non zero choices of, say M, B and V

    2. Relevant equations
    f(x) = sin(Mx)
    g(x) = Bx + V

    3. The attempt at a solution

    So they would be dependent if their Wronskian is equal to 0.
    W = Bsin(Mx) - (Bx + V)Mcos(Mx)

    I'm stuck at this point.. I can't seem to find any identities to make this = 0. I've tried expanding too.. I don't think it helps much. Any help is appreciated!
     
  2. jcsd
  3. Feb 10, 2010 #2

    Dick

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    Those functions aren't linearly dependent, they are linearly independent. That makes your job easier. Just find a value of x where the Wronskian DOESN'T vanish.
     
  4. Feb 10, 2010 #3

    LCKurtz

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    From your i.e., part, I think you misunderstand the question. You aren't asked to show the functions are linearly dependent, which they aren't. You are asked to show those two functions are not solutions to the same second order linear DE. That is a different question.

    The point of this exercise is to use the theorem that says two solutions of such an equation will have a Wronskian that is either identically zero or never zero. These two functions can only be solutions of the same equation if they satisfy that Wronskian property. So all you have to do is show the Wronskian is neither identically zero nor never zero. Don't you just love double negatives?

    [Edited for technical accuracy]
     
    Last edited: Feb 11, 2010
  5. Feb 10, 2010 #4

    Dick

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    Thanks, LCKurtz. I guess I wasn't fully plumbing the depth of the i.e. and double negative here.
     
  6. Feb 11, 2010 #5

    Mark44

    Staff: Mentor

    From the context, I don't think i.e. stands for id est. What does it mean here?
     
  7. Feb 11, 2010 #6
    okay that makes much more sense. It's the way the question was phrased that had me confused. Thanks.
     
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