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Thread: dP/dt = kP(m - P)

  1. #1
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    dP/dt = kP(m - P)

    dP/dt=kP(M-P) find the function for P
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  2. #2
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    $\displaystyle \frac{dP}{dt} = kP(M-P)$

    $\displaystyle \frac{dt}{dP} = \frac{1}{kP(M-P)}$

    $\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$.


    Now applying Partial Fractions we have

    $\displaystyle \frac{A}{P} + \frac{B}{M-P} = \frac{1}{P(M-P)}$

    $\displaystyle \frac{A(M-P) + BP}{P(M-P)} = \frac{1}{P(M-P)}$

    $\displaystyle A(M-P)+BP=1$

    $\displaystyle AM - AP + BP = 1$

    $\displaystyle (B-A)P + AM = 0P + 1$


    So $\displaystyle B - A = 0, AM = 1$.

    This means $\displaystyle A = \frac{1}{M}$ and $\displaystyle B = \frac{1}{M}$.


    Therefore:

    $\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$

    $\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{MP} + \frac{1}{M(M-P)}\right]$

    $\displaystyle \frac{dt}{dP} = \frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)$

    $\displaystyle t = \int{\frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

    $\displaystyle t = \frac{1}{kM}\int{\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

    $\displaystyle t = \frac{1}{kM}\left(\ln{|P|} - \ln{|M-P|}\right) + C$

    $\displaystyle t = \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} + C$

    $\displaystyle \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} = t-C$

    $\displaystyle \ln{\left|\frac{P}{M-P}\right|} = kMt - kMC$

    $\displaystyle \left|\frac{P}{M-P}\right| = e^{kMt - kMC}$

    $\displaystyle \left|\frac{P}{M-P}\right| = e^{-kMC}e^{kMt}$

    $\displaystyle \frac{P}{M-P} = \pm e^{-kMC}e^{kMt}$

    $\displaystyle \frac{P}{M-P} = Ae^{kMt}$ where $\displaystyle A= \pm e^{-kMC}$

    $\displaystyle \frac{P-M+M}{M-P} = Ae^{kMt}$

    $\displaystyle \frac{P-M}{M-P} + \frac{M}{M-P} = Ae^{kMt}$

    $\displaystyle -1 + \frac{M}{M-P} = Ae^{kMt}$

    $\displaystyle \frac{M}{M-P} = Ae^{kMt} + 1$

    $\displaystyle \frac{M - P}{M} = \frac{1}{Ae^{kMt} + 1}$

    $\displaystyle M - P = \frac{M}{Ae^{kMt} + 1}$

    $\displaystyle P = M - \frac{M}{Ae^{kMt} + 1}$

    $\displaystyle P = \frac{AMe^{kMt} + M - M}{Ae^{kMt} + 1}$

    $\displaystyle P= \frac{AMe^{kMt}}{Ae^{kMt} + 1}$.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Prove It View Post
    $\displaystyle \frac{dP}{dt} = kP(M-P)$

    $\displaystyle \frac{dt}{dP} = \frac{1}{kP(M-P)}$

    $\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$
    Why do you invert dp/dt? Do you not integrate P wrt t?

    Just wondering
    Thanks
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  4. #4
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    Because your function is given in terms of $\displaystyle P$. That means you will need to integrate with respect to $\displaystyle P$.
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Prove It View Post
    Because your function is given in terms of $\displaystyle P$. That means you will need to integrate with respect to $\displaystyle P$.
    ok, fair enough, but can you not just integrate p wrt t and then group the p terms at the end and make it the subject of the expression?
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  6. #6
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    Because you don't know what function $\displaystyle P$ is in terms of $\displaystyle t$.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    $\displaystyle \frac{dP}{dt} = kP(M-P)$

    $\displaystyle \frac{dt}{dP} = \frac{1}{kP(M-P)}$

    $\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$.


    Now applying Partial Fractions we have

    $\displaystyle \frac{A}{P} + \frac{B}{M-P} = \frac{1}{P(M-P)}$

    $\displaystyle \frac{A(M-P) + BP}{P(M-P)} = \frac{1}{P(M-P)}$

    $\displaystyle A(M-P)+BP=1$

    $\displaystyle AM - AP + BP = 1$

    $\displaystyle (B-A)P + AM = 0P + 1$


    So $\displaystyle B - A = 0, AM = 1$.

    This means $\displaystyle A = \frac{1}{M}$ and $\displaystyle B = \frac{1}{M}$.


    Therefore:

    $\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$

    $\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{MP} + \frac{1}{M(M-P)}\right]$

    $\displaystyle \frac{dt}{dP} = \frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)$

    $\displaystyle t = \int{\frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

    $\displaystyle t = \frac{1}{kM}\int{\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

    $\displaystyle t = \frac{1}{kM}\left(\ln{|P|} - \ln{|M-P|}\right) + C$

    $\displaystyle t = \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} + C$

    $\displaystyle \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} = t-C$

    $\displaystyle \ln{\left|\frac{P}{M-P}\right|} = kMt - kMC$

    $\displaystyle \left|\frac{P}{M-P}\right| = e^{kMt - kMC}$

    $\displaystyle \left|\frac{P}{M-P}\right| = e^{-kMC}e^{kMt}$

    $\displaystyle \frac{P}{M-P} = \pm e^{-kMC}e^{kMt}$

    $\displaystyle \frac{P}{M-P} = Ae^{kMt}$ where $\displaystyle A= \pm e^{-kMC}$

    $\displaystyle \frac{P-M+M}{M-P} = Ae^{kMt}$

    $\displaystyle \frac{P-M}{M-P} + \frac{M}{M-P} = Ae^{kMt}$

    $\displaystyle -1 + \frac{M}{M-P} = Ae^{kMt}$

    $\displaystyle \frac{M}{M-P} = Ae^{kMt} + 1$

    $\displaystyle \frac{M - P}{M} = \frac{1}{Ae^{kMt} + 1}$

    $\displaystyle M - P = \frac{M}{Ae^{kMt} + 1}$

    $\displaystyle P = M - \frac{M}{Ae^{kMt} + 1}$

    $\displaystyle P = \frac{AMe^{kMt} + M - M}{Ae^{kMt} + 1}$

    $\displaystyle P= \frac{AMe^{kMt}}{Ae^{kMt} + 1}$.
    Thanks if M= 197.3 how would i go about to find K?
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