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Math Help - Solving differential equation

  1. #1
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    Question Solving differential equation

    Trying to solve this differential equation by separating variables. Where c is a constant and K is the carrying capacity. P is population

    \frac{{dP}}{{dt}} = c\ln \left( {\frac{K}{P}} \right)P

    I started with this

    <br />
\begin{array}{l}<br />
 \frac{{dP}}{{dt}} = c\ln \left( {\frac{K}{P}} \right)P \\ <br />
 dP = c\ln \left( {\frac{K}{P}} \right)P.dt \\ <br />
 \frac{{dP}}{P} = c\left( {\ln (K) - \ln (P)} \right)dt \\ <br />
 \end{array}<br />

    than I'm not sure if I can do this.

    <br />
\begin{array}{l}<br />
 \frac{{dP}}{P} - c\ln (P) = c\ln K.dt \\ <br />
 \frac{1}{P} - c\ln (P)dp = c\ln K.dt \\ <br />
 \end{array}<br />

    and if it allowed than I'm lost at this point anyhow. I'm really not liking these, I seem lost in the process.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The equation can be written as...

    \frac{dp}{dt}= c\cdot p\cdot (\ln k - \ln p) (1)

    Separing the variables we obtain...

    \frac {dp}{p\cdot (\ln k - \ln p)}= c\cdot dt (2)

    ... and integrating...

    \int \frac {dp}{p\cdot (\ln k - \ln p)} = -\ln (\ln k - \ln p) = c\cdot t + \ln a (3)

    ... where a>0 is a constant. From (3) with little more work...

     \ln \frac{k}{p}= a\cdot e^{- c\cdot t} \rightarrow p= k\cdot e^{-(a\cdot e^{-c\cdot t})} (4)

    Kind regards

    \chi \sigma
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  3. #3
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    Hi I would you mind going through how you integrated the left hand side of the equation? ie the part \int {\frac{1}{{P\ln \left( {\frac{K}{P}} \right)}}} dp I'm a little confused to what occurred there. Thanks in advance.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Setting \ln \frac {k}{p}= \ln k - \ln p the integral becomes...

    \int \frac {dp}{p\cdot \ln \frac{k}{p}} = \int \frac {dp}{p\cdot (\ln k - \ln p)} (1)

    Now if you take into account that...

    \frac {d}{dp} (\ln k -\ln p) = - \frac {1}{p} (2)

    ... the integral (1) is...

     \int \frac {dp}{p\cdot (\ln k - \ln p)} = -\ln (\ln k - \ln p) + a (3)

    ... where a is an 'arbitrary constant'...

    Kind regards

    \chi \sigma
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  5. #5
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    Sorry, still a little confused, at equation (2).
    \frac {d}{dp} (\ln k -\ln p) = - \frac {1}{p}<br />
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  6. #6
    MHF Contributor chisigma's Avatar
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    Since \ln k doesn't depend from p is...

    \frac {d}{dp} (\ln k - \ln p) = - \frac {d}{dp} \ln p= - \frac {1}{p}

    Kind regards

    \chi \sigma
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