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Math Help - Differential Equations

  1. #1
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    Differential Equations

    Solve by separation of variables:

    dP/dt = P-P^2

    This is what I have so far:
    P/(P-1)=ce^t

    but I need to get that to:
    P=(ce^t)/(1+ce^t)

    Any help would be greatly appreciated.

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yellowrose View Post
    Solve by separation of variables:

    dP/dt = P-P^2

    This is what I have so far:
    P/(P-1)=ce^t

    but I need to get that to:
    P=(ce^t)/(1+ce^t)

    Any help would be greatly appreciated.

    Thanks!
    you have a problem here. i get (P - 1)/P = Ae^t

    this yields: P = 1/(1 + Ae^t)
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    you have a problem here. i get (P - 1)/P = Ae^t

    this yields: P = 1/(1 + Ae^t)
    Would you be able to tell me how you got (P-1)/P=Ae^t?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yellowrose View Post
    Would you be able to tell me how you got (P-1)/P=Ae^t?
    oh wait. i made an error. i wrote P(P - 1) when it should have been P(1 - P). that changes things a bit.

    \frac {dP}{dt} = P - P^2 = P(1 - P)

    \Rightarrow \frac {dP}{P(1 - P)} = dt

    \Rightarrow \left( \frac 1P + \frac 1{1 - P} \right)dP = dt ......through algerba, or partial fractions if you want to work hard

    \Rightarrow \ln |P| - \ln |1 - P| = t + C

    \Rightarrow \ln \left| \frac P{1 - P} \right| = t + C

    \Rightarrow \frac P{1 - P} = Ae^t

    \Rightarrow P - P^2 = Ae^t

    \Rightarrow P^2 - P + Ae^t = 0

    this is a quadratic in P, use the quadratic formula to solve for P, so:

    P = \frac {1 \pm \sqrt{1 - 4Ae^t}}{2}

    now simplify ... er, if you can, or want to
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