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DE solve?

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve the euqation [tex] dP/dt = k(M - P)P[/tex] to show that it equals: [tex] P(t) = (MP_{0}) /( P_{0} + (M - P_{0})e^{-kMt})[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex] \[ \int \frac {dP} {P(M - P)}\] = k \int dt \ [/tex]

    [tex] \frac {1} {M}\ \ln( \frac {P} {M - P}) + C = Kt + C [/tex]

    I combine the constants of integration... i can do this right?
    Then i get rid of the log by taking the exponent of both sides.

    [tex] P/(M - P) = e^{mkt + c} [/tex]

    I then turn [tex] e^{c}[/tex] into [tex] P_{0} [/tex]

    Next i divide through by [tex] (M - P) [/tex]

    [tex] P = (M - P)P_{0}e^{mkt} [/tex]

    next i combine the [tex] P [/tex] then divide through by the remainder.

    [tex] P = (mP_{0}e^{mkt})/(1 + P_{0}e^{mkt}) [/tex]

    and i end up with:
    [tex] P = (mP_{0})/(e^{-mkt} + P_{0}) [/tex]

    In the equation i am suppose to get i dont see how there could possibly be three terms in the denominator.
    Last edited: Nov 11, 2007
  2. jcsd
  3. Nov 10, 2007 #2


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    You could just differentiate P and substitute it into the differential equation.
  4. Nov 10, 2007 #3
    thanks for the reply cristo.

    I was in the process of editing my post (i am still getting use to LaTex) when you replied. i could do that but my teacher want us to just solve it. Also this has been bugging me for a while, i have been working on it for the past few hours and cannot get it. My teacher... well a grad student TA for my DE lab couldn't solve it either.

    I also would like to know what is wrong with my method... it seems like it should work but... its not.

    one thing think might be wrong with my method is the [tex] P_{0} [/tex] initial condition.
  5. Nov 11, 2007 #4


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    Staff Emeritus
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    HOW did you "turn [itex]e^{c}[/tex] into [itex]P_0[/itex]"?

    When t= 0, You have [itex]P_0/(M-P_0)= e^c[/itex]

  6. Nov 11, 2007 #5
    Thanks a million. works out fine after that.
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