Integration by separation of variables

• May 23rd 2009, 04:32 PM
Kiwi_Dave
Integration by separation of variables
$\rho \frac {du}{d \rho}=1-e^u$

Use separation of variables to obtain:

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

When I separate the variables I get:

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

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

I have tried:
$(1+e^u+e^{2u}+e^{3u}+e^{4u}+...)du=\frac{d \rho} \rho$
$\therefore u-1+1+e^u+\frac12e^{2u} +\frac13e^{3u}+...=ln \rho+c$

But that doesn't seem to be getting me far.
• May 23rd 2009, 04:40 PM
skeeter
Quote:

Originally Posted by Kiwi_Dave
$\rho \frac {du}{d \rho}=1-e^u$

Use separation of variables to obtain:

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

When I separate the variables I get:

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

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

I have tried:
$(1+e^u+e^{2u}+e^{3u}+e^{4u}+...)du=\frac{d \rho} \rho$
$\therefore u-1+1+e^u+\frac12e^{2u} +\frac13e^{3u}+...=ln \rho+c$

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

$\frac{1}{1 - e^u} = \frac{1 - e^u + e^u}{1 - e^u} = \frac{1 - e^u}{1 - e^u} - \frac{-e^u}{1 - e^u} = 1 - \frac{-e^u}{1 - e^u}$

integrate the last expression