# Thread:

1. ## 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.

2. 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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8841039538383484, "perplexity_flag": "middle"}