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Math Help - nonlinear ODE

  1. #1
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    Smile nonlinear ODE

    solve the following ODE:
    d^2(x)
    ------ = -ln(x)-1
    dt^2

    I've tried a couple of things and got this:

    d^2(x)
    ------ = -ln(x*e)
    dt^2

    but couldn't solve it, your help is appriciated.
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  2. #2
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    Quote Originally Posted by zokomoko View Post
    solve the following ODE:
    d^2(x)
    ------ = -ln(x)-1
    dt^2

    I've tried a couple of things and got this:

    d^2(x)
    ------ = -ln(x*e)
    dt^2

    but couldn't solve it, your help is appriciated.
    The best I can do is to quadrature. Multiply your ODE by x'

    so

    x' x'' = -\ln x x' - x'

    and integrate giving

    \frac{1}{2} x'^2 = - x \ln x + c_1 so x' = \pm\, \sqrt{c_1 - 2 x \ln x}

    so (absorbing a 2 into c1)

    \int \frac{dx}{\sqrt{c_1 - 2 x \ln x}} = \pm \,t + c_2
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