Re-arranging problem

  • #1
I'm trying integrate the following equation and make r the subject
[tex]\frac{dr}{dt} = \Phi - \Psi \frac{2}{r}\frac{dr}{dt}[/tex]

I first collect the derivative terms together and integrate the equation with respect to r and t to obtain

[tex]r + 2\Psi\ln{r} = \Phi{t} + r_0 [/tex]

where r0 is the constant of integration. My question is how would I make r the subject of the above equation?

Many thanks.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi lostidentity! :smile:
My question is how would I make r the subject of the above equation?

not possible!

(unless you define the LHS to be f(r), in which case it's r = f-1(RHS) :wink:)
 
  • #3
I'm wondering if there is another way to solve the ODE I gave in the previous post, i.e.

[tex] \left(1+2\frac{\Psi}{r}\right)\frac{dr}{dt} = \Phi[/tex]
 
Last edited:
  • #4
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I'm wondering if there is another way to solve the ODE I gave in the previous post, i.e.

[tex] \left(1+2\frac{\Psi}{r}\right)\frac{dr}{dt} = \Phi[/tex]

Check this very very carefully

[tex] r = 2 \Psi lambert\left(\frac{e^{\frac{t\Phi+c}{2 \Psi}}}{2 \Psi}\right)[/tex]

where c is some constant and where lambert gives the principle solution for w in z=w e^w.

When I substitute this back into the original ODE it seems to check, but do not trust this until you have triple checked it.
 
Last edited:

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