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Math Help - Non-existenant solution to differencial equation

  1. #1
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    Non-existenant solution to differencial equation

    Can, y=sin(x^2) be a solution to,
    y''+py'+qy=0
    With initial conditions, y(0)=A and y'(0)=B
    Where p and q are continous in an open interval containing 0?
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  2. #2
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    I solved it, thank you anyways.

    If you are interested how it was solved:

    The equation,
    y''+py'+qy=0
    Has basis of dimension 2 if p,q are continous in an neigborhood of x=0.

    Thus, there exists two linearly independent solutions that span the set. One is given as sin(x^2). Let the other linearly independent solution be y.
    Then we compute the Wronskian,
    [sin(x^2) y]
    [2xcos(x^2) y']
    Thus,
    y'sin(x^2)-2xycos(x^2)
    But at x=0 the Wronskian is zero.
    Thus, there does not exist a linearly independant solution with sin(x^2), a contradiction.
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