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Math Help - Help with antidifferentiation

  1. #1
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    Help with antidifferentiation

    Hello, I need help with the antidifferentiation of this question:

     \frac{dW}{dt} =  (m-n-kW) \cdot W

    \frac{dt}{dW} =  \frac{1}{(m-n-kW) \cdot W}

    What I need to know is how to antidifferentiate(or integrate) with respect to W.
    After that, I can solve the rest.
    Although if you must know,
    m=0.10, n=0.06, k = 0.00005
    Many thanks!!!
    Last edited by mr fantastic; August 17th 2009 at 04:31 AM. Reason: Fixed the formatting of the latex.
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  2. #2
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    Hi mrnacho

    What are m, n, and k?

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  3. #3
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    m, n and k are to be ignored.
    What I need to know is how to antidifferentiate(or integrate) with respect to W.
    After that, I can solve the rest.
    Although if you must know,
    m=0.10, n=0.06, k = 0.00005
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  4. #4
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    Hi mrnacho

    Ok, so m, n, and k are constants. It's really important information

    So :

    \frac{dW}{dt}=(m-n-kW)*W

    \frac{dt}{dW}=\frac{1}{(m-n-kW)*W}

    dt=\frac{1}{(m-n-kW)*W}dW

    You can use partial fraction to solve this :

    dt=\left(\frac{A}{m-n-kW}+\frac{B}{W}\right)dW
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  5. #5
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    Yeah, that makes sense. But I kinda forgot how to solve this. I did separate them and then, I got lost.
    1= AW + B(m-n-kW)
    And I havent even anti differentiate it yet :/
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  6. #6
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    Hi mrnacho

    Of course you haven't done the integration, because you have to find the value of A and B first.

    Hope this help : Partial-Fraction Decomposition: General Techniques
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  7. #7
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    This is a big more complex but finally, my value of A is
    <br /> <br />
\frac{k}{m-n}<br />

    and my value of B is:

    <br /> <br />
\frac{1}{m-n}<br />
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  8. #8
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    Now Im stumped after getting the values and subbing, this is what I get...

    <br /> <br />
\frac{dt}{dW}=\frac{k}{m^2+n^2+nkW+mkW-2mn}+\frac{1}{Wm-Wn}<br />

    How do I integrate that with respect to W?
    The numbers look so scary....
    Last edited by mr fantastic; August 17th 2009 at 02:05 PM. Reason: Merged posts
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  9. #9
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    Hi mrnacho

    dt=\left(\frac{A}{m-n-kW}+\frac{B}{W}\right)dW

    dt=\left(\frac{\frac{k}{m-n}}{m-n-kW}+\frac{\frac{1}{m-n}}{W}\right)\; dW

    \int dt=\int \left(\frac{\frac{k}{m-n}}{m-n-kW}+\frac{\frac{1}{m-n}}{W}\right)\; dW

    Because A and B are constants, it can be taken out from the integral.

    t=\frac{k}{m-n} \int \frac{dW}{m-n-kW}+\; \frac{1}{m-n}\int \frac{dW}{W}
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