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Math Help - Solving the second order equation?

  1. #1
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    Solving the second order equation?

    How do I solve an equation that is missing the independent variable?

    2y^2* y'' + 2y(y')^2 = 1
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Solving the second order equation?

    I have an idea but i'm not sure if it will work

    If y = ax^2 + bx + c then
    2y^2*y'' = 4th degree polynomial
    2y*(y')^2 = 4th degree polynomial

    so you would have to solve for the coefficients so that they cancel out.
    Last edited by jakncoke; February 25th 2013 at 08:00 PM.
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  3. #3
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    Re: Solving the second order equation?

    Quote Originally Posted by jakncoke View Post
    I have an idea but i'm not sure if it will work

    If y = ax^2 + bx + c then
    2y^2*y'' = 4th degree polynomial
    2y*(y')^2 = 4th degree polynomial

    so you would have to solve for the coefficients so that they cancel out.
    Hi !

    I don't think that the method proposed by jakncoke is convenient. OK., the order of the polynomials are the same. But this doesn't mean that they can be equal. In some case yes, but not in the present case.
    The solving is more difficult (see in attachment)
    Attached Thumbnails Attached Thumbnails Solving the second order equation?-ode.jpg  
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  4. #4
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    Re: Solving the second order equation?

    I should have read JJacqueline's post first- she did exactly the same thing.
    Last edited by HallsofIvy; February 26th 2013 at 03:28 AM.
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  5. #5
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    Re: Solving the second order equation?

    Quote Originally Posted by HallsofIvy View Post
    I should have read JJacqueline's post first- she did exactly the same thing.
    "Clothes do not make the man" : No "e" at the end of the name of the man !
    Thanks for the thanks
    JJ.
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