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Math Help - I need help solving a simple differential equation?

  1. #1
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    I need help solving a simple differential equation?

    hello!

    I have never learned how to solve differential equations using partial fractions (i think that is what it is called?). I was hoping someone could post the solution to the following differential equation - I do not need to know the derivation, I just need the correct solution:

    dP/dt=K*P(1-P/L)

    I suppose it is not simple, but I think it is quite a well known equation for population growth! I appreciate any solutions!
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  2. #2
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    Quote Originally Posted by Machete View Post
    hello!

    I have never learned how to solve differential equations using partial fractions (i think that is what it is called?). I was hoping someone could post the solution to the following differential equation - I do not need to know the derivation, I just need the correct solution:

    dP/dt=K*P(1-P/L)

    I suppose it is not simple, but I think it is quite a well known equation for population growth! I appreciate any solutions!
    It is the Logistic Equation see >>here<<

    CB
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  3. #3
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    sweet thanks
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  4. #4
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    Quote Originally Posted by Machete View Post
    hello!

    I have never learned how to solve differential equations using partial fractions (i think that is what it is called?). I was hoping someone could post the solution to the following differential equation - I do not need to know the derivation, I just need the correct solution:

    dP/dt=K*P(1-P/L)

    I suppose it is not simple, but I think it is quite a well known equation for population growth! I appreciate any solutions!
    \frac{dP}{dt} =K\cdot P\cdot \left(1-\frac{P}{L}\right)
    \displaystyle\frac{dP}{P\left(1-\frac{P}{L}\right)} =K dt
    \displaystyle\frac{dP}{P\cdot \frac{1}{L}\left(L-P\right)} =K dt
    L\cdot\displaystyle\frac{dP}{P\left(L-P\right)} =K dt
    \displaystyle\frac{dP}{P\left(L-P\right)} =\frac{K}{L} dt
    \displaystyle\frac{1}{P\left(L-P\right)} =\frac{1}{LP}+\frac{1}{L(L-P)}
    \displaystyle\int\frac{dP}{P\left(L-P\right)} =\int\left(\frac{1}{LP}+\frac{1}{L(L-P)}\right) dP=\frac{1}{L}\ln P-\frac{1}{L}\ln(L-P)= \displaystyle=\frac{1}{L}\left(\ln P -\ln (L-P)\right)=\frac{1}{L}\ln\frac{P}{L-P}
    \displaystyle \int \frac{dP}{P(L-P)}=\int\frac{K}{L}dt
    \displaystyle \frac{1}{L}\ln\frac{P}{L-P}=\frac{K}{L} t + C
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