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Math Help - Hard 1st order ODE Question

  1. #1
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    Hard 1st order ODE Question

    (dy/dx)^2 - 4x(dy/dx) + 6y = 0

    The dy/dx squared term is kind of nasty. I've tried with all the substitutes I could think of, none of them worked.

    Any hints would be appreciated. Thanks!!!
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  2. #2
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    Quote Originally Posted by chemenger View Post
    (dy/dx)^2 - 4x(dy/dx) + 6y = 0

    The dy/dx squared term is kind of nasty. I've tried with all the substitutes I could think of, none of them worked.

    Any hints would be appreciated. Thanks!!!
    Well:

    dy/dx = 2x +/- sqrt(4 x^2 -6)

    so with some assumptions about smoothness of dy/dx you should be able to
    proceed.

    RonL
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  3. #3
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    First off, thanks for the quick reply

    Shouldn't that be dy/dx = 2x +/- sqrt(4 x^2 -6y)
    I think you were missing a y.
    I've actually thought about using the quadratic eqn. But I wasn't sure how to deal with the stuff that's under the sqrt sign.
    Could you make it a bit clear? Thanks again.
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  4. #4
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    Quote Originally Posted by chemenger View Post
    First off, thanks for the quick reply

    Shouldn't that be dy/dx = 2x +/- sqrt(4 x^2 -6y)
    I think you were missing a y.
    Yes it should, unfortunatly with the y there its not so simple

    RonL
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  5. #5
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    Maybe your just have to approximate the solution.

    Because it have the special form,
    y'=f(x,y)
    Apply Newton's Method.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    Maybe your just have to approximate the solution.

    Because it have the special form,
    y'=f(x,y)
    Apply Newton's Method.
    I can't. The question says solve this ode, which means I'm gonna need an analytical solution.
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  7. #7
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    Quote Originally Posted by chemenger View Post
    I can't. The question says solve this ode, which means I'm gonna need an analytical solution.
    How about a series solution? (looks a bit tricky though)

    RonL
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  8. #8
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    I remember the lecturer said something about Laplace Transform, but I'm sure how I can relate that to this question.
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