Solving a Re-arranging Problem: Making r the Subject

[lostidentity]

I'm trying integrate the following equation and make r the subject
$$\frac{dr}{dt} = \Phi - \Psi \frac{2}{r}\frac{dr}{dt}$$

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

$$r + 2\Psi\ln{r} = \Phi{t} + r_0$$

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

Many thanks.

 

[tiny-tim]

hi lostidentity!

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) )

 

[lostidentity]

I'm wondering if there is another way to solve the ODE I gave in the previous post, i.e.

$$\left(1+2\frac{\Psi}{r}\right)\frac{dr}{dt} = \Phi$$

 

[Bill Simpson]

I'm wondering if there is another way to solve the ODE I gave in the previous post, i.e.

$$\left(1+2\frac{\Psi}{r}\right)\frac{dr}{dt} = \Phi$$

Check this very very carefully

$$r = 2 \Psi lambert\left(\frac{e^{\frac{t\Phi+c}{2 \Psi}}}{2 \Psi}\right)$$

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.