1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    to

    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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integrate ln [f'(x)] + ln [f(x)] = 0
Loading...