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Math Help - Integration by separation of variables

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Kiwi_Dave View Post
    \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
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