I am having difficulties solving this differential equation for $P$, \begin{align*} \frac{\text{d}P}{\text{d}t}=kP\left(\frac{A-P}{A}\right)\left(\frac{P-m}{P}\right) \end{align*} The context is a population model, where $A$ is the maximum sustainable population, $m$ is the minimum sustainable population and $k$ are constants.

The final answer I get is: \begin{align*} P=\frac{m-Abe^{\frac{(A-m)}{A}kt}}{1-be^{\frac{(A-m)}{A}kt}} \end{align*} where $b=e^c$.

This was achieved after the following:

My final integral after using partial fractions is: \begin{align*} \int\frac{1}{A-m}\left(\frac{1}{A-P}+\frac{1}{P-m}\right)\ \text{d}P&=\int\frac{k}{A}\ \text{d}t \end{align*} My final simplification process after the integration is: \begin{align*} \frac{\ln|P-m|-\ln|P-A|}{A-m}&=\frac{k}{A}t +c\\ \therefore\ln\left|\frac{P-m}{P-A}\right|&=\frac{(A-m)}{A}kt+c\\ \therefore\frac{P-m}{P-A}&=be^{\frac{(A-m)}{A}kt}\;,\;b=e^c\\ \therefore P-m&=be^{\frac{(A-m)}{A}kt}(P-A)\\ \therefore P-m&=Pbe^{\frac{(A-m)}{A}kt}-Abe^{\frac{(A-m)}{A}kt}\\ \therefore P-Pbe^{\frac{(A-m)}{A}kt}&=-Abe^{\frac{(A-m)}{A}kt}+m\\ \therefore P&=\frac{m-Abe^{\frac{(A-m)}{A}kt}}{1-be^{\frac{(A-m)}{A}kt}} \end{align*} Is this correct, or have I made a mistake somewhere?

Thank you very much in advance.

  • 1
    $\begingroup$ Should it be $\frac{P-m}{m}$ instead of $\frac{P-m}{P}$ in the problem? $\endgroup$ – Empiricist Sep 24 '15 at 3:09
  • $\begingroup$ No the question does say $\frac{P-m}{P}$ $\endgroup$ – T.Walker Sep 24 '15 at 3:11
  • $\begingroup$ Then why not eliminate the $P$ in the numerator and the denominator? $\endgroup$ – Empiricist Sep 24 '15 at 3:12
  • $\begingroup$ Yes, my first step was $\frac{\text{d}P}{\text{d}t}=\frac{k}{A}(A-P)(P-m)$ $\endgroup$ – T.Walker Sep 24 '15 at 3:22

Good description of your process. I would like to know the initial value $P_0$. This is because it affects how one deals with the absolute value stuff. You just plain removed the absolute value sign, which for some initial values of $P$ can lead to error.

At a certain stage, you had $P-m=b\dots$. It should have been $P-m=\pm b\dots$.

  • $\begingroup$ $P$ is always positive because it is a population model, but there is no specific value for $P_0$ given. $\endgroup$ – T.Walker Sep 24 '15 at 3:07
  • 1
    $\begingroup$ Yes, $P$ is positive. But how is $P_0$ related to $A$ and to $m$? The sign depends on that. If $P_0$ is unspecified, then we get slightly different possible forms for $P$. $\endgroup$ – André Nicolas Sep 24 '15 at 3:11
  • $\begingroup$ I think $m<P<A$ would be true. $\endgroup$ – T.Walker Sep 24 '15 at 3:23
  • 1
    $\begingroup$ In principle not necessarily, since population can start below the minimum sustainable, or above the maximum. But if we assume that $m\lt P_0\lt A$, then $P$ will stay within those bounds. And if that is the case, then the equation you wrote is incorrect. It should have been $P-m$ is equal to the messy stuff in front times $A-P$, not $P-A$. Or else you can keep the $P-A$ but then you need a minus sign in front of the $b$. For note that $P-A$ is negative. Or else you can choose to make $b$ negative, but that is in general not done. $\endgroup$ – André Nicolas Sep 24 '15 at 3:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.