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

[ck99]

Homework Statement

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

Homework Equations

∫(z'/z) = ln z

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!

Homework Equations

The Attempt at a Solution

 

[SammyS]

Homework Statement

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

Homework Equations

∫(z'/z) = ln z

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!

Use the following two facts.

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

(z²)' = 2z'z .

 

[ck99]

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)'] ?

 

[HallsofIvy]

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

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

Integrate once to get

ln (z') + ln z = 0

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.

 

[ck99]

That's great, thankyou! I knew it wasn't quite right .. .

Cheers for all the help.