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Homework Help: Integrate ln [f'(x)] + ln [f(x)] = 0

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Integrate z'' + (1/z)(z'^2) = 0 to find an equation of motion

    This is a GR problem, using geodesic equations of motion, so I have derivatives of coordinate z wrt some parameter λ, where z is a function of λ, and z'(λ) is the first derivative wrt λ

    2. Relevant equations

    ∫(z'/z) = ln z

    3. The attempt at a solution

    I have gone from

    z'' + (1/z)(z'^2) = 0


    z''/z' + z'/z = 0

    Integrate once to get

    ln (z') + ln z = 0

    But I'm not sure how to integrate again. Looking at a previous problem (which I have the solution for) it should be something like

    1/z = Cλ + D

    But I don't know how to get my problem into this form!

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 19, 2012 #2


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    Staff Emeritus
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    Homework Helper
    Gold Member

    Use the following two facts.

    ln(z') + ln(z) = in(z'z) .

    (z2)' = 2z'z .
  4. Apr 19, 2012 #3
    Thanks for the reply Sammy.

    Can I do the following?

    Use your tip to say ln (z'z) = 0

    Exponentiate both sides to get z'z = e^0 = 1

    Integrate both sides to get 1/2 z^2 = λ + C

    I don't think this is right because the answer is not in the format I was expecting, eg f(z) = λC + D . . . am I allowed to do the exponential step or do I have to physically integrate ln(z'z) or ln [(1/2z^2)'] ?
  5. Apr 19, 2012 #4


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    Science Advisor

    What you have in your last post is a first order equation. It's general solution can't be "Cλ+ D" because that involves two arbitrary constants and the general solution to a first order equation will involve only one.

    Going back to your first post you have
    where you have forgotten the constant of integration. You should have
    ln(z')+ ln(z)= C where C can be any number. Now that becomes
    ln(zz')= C so zz'= e^C.

    Integrate that to get (1/2)z^2= e^Cλ+ D which becomes (1/2)z^2=C'λ+ D with C'= e^C.
  6. Apr 19, 2012 #5
    That's great, thankyou! I knew it wasn't quite right .. .

    Cheers for all the help.
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