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Second derivative of an autonomous function

  1. Sep 22, 2014 #1
    For the derivative: dy/dt = ry ln(K/y)

    I am trying to solve the second derivative. It seems like an easy solution, and I did:

    d^2y/dt^2 = rln(K/y)y' + ry(y/K)

    which simplifies to:

    d^2y/dt^2 = (ry')[ln(K/y) + 1/Kln(K/y)

    Unfortunately, the answer is d^2y/dt^2 (ry')[ln(K/y) - 1] and I don't quite see where I went wrong. Any help would be greatly appreciated!
  2. jcsd
  3. Sep 23, 2014 #2


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    Homework Helper

    You're missing a [itex]y'[/itex] from the last term and you haven't differentiated [itex]\ln(K/y)[/itex] correctly. Since [tex]\ln(K/y) = \ln K - \ln y[/tex] we have [tex]\frac{d}{dy} (\ln(K/y)) = \frac{d}{dy} (\ln K - \ln y) = -\frac 1y.[/tex]
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