Originally Posted by

**Kiwi_Dave** $\displaystyle \rho \frac {du}{d \rho}=1-e^u$

Use separation of variables to obtain:

$\displaystyle e^u=\frac 1 {1-\frac{c_2} \rho}$

When I separate the variables I get:

$\displaystyle \frac{du}{1-e^u}=\frac{d \rho}\rho$

But I can't see how to integrate the LHS or manipulate it to give the required form.

I have tried:

$\displaystyle (1+e^u+e^{2u}+e^{3u}+e^{4u}+...)du=\frac{d \rho} \rho$

$\displaystyle \therefore u-1+1+e^u+\frac12e^{2u} +\frac13e^{3u}+...=ln \rho+c$

But that doesn't seem to be getting me far.