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If you consider the classic differential equation used in textbooks to model animal population growth and forget adding in the carrying capacity terms:

$dP/dt$ = rP

It would seem that the above DE would benefit from some variable rate r instead of some fixed rate r in order to achieve a more accurate model.

I tried googling around for this and I'm guessing I simply wasn't using the best keywords. I'm wondering if someone could give an example of a DE where a variable rate r is used and how one usually solves such an equation.

Thanks

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  • $\begingroup$ I suppose that you want $r$ to be function of $P$ itself. Is this correct ? $\endgroup$ – Claude Leibovici Jul 17 '17 at 4:31
  • $\begingroup$ @ClaudeLeibovici oops, i interpreted as a function of $t$, good question $\endgroup$ – qbert Jul 17 '17 at 4:32
  • $\begingroup$ @ClaudeLeibovici yes that is correct in this instance. though a function of t in other scenarios would also work $\endgroup$ – H_1317 Jul 17 '17 at 4:32
  • $\begingroup$ @qbert. It could be any of $t$ or $P$. As you wrote, this can make the problem more difficult but doable (probably "easier" with $t$ (?)). $\endgroup$ – Claude Leibovici Jul 17 '17 at 4:36
  • $\begingroup$ @ClaudeLeibovici i would think so. It also looks cleaner in integral form if your goals is to find an explicit formula for p(t) $\endgroup$ – qbert Jul 17 '17 at 4:48
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If you try this out with $r(t)$ as a non constant function of time:

$$ \frac{dp}{dt}=r(t)p(t)\implies \ln(p)=\int r(t)\mathrm dt\implies p(t)=\exp\left ( \int r(t)\right ) $$ by the usual separation of variables method.

Whether this turns out to be something nice analytically depends on how easy $r(t)$ is to integrate. However, representing the solution in terms of an integral may be a totally valid solution for your purposes and the method is pretty much identical to the $r$ constant case.

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  • $\begingroup$ then i suppose the r(p) case would just end up being a more complicated integral to solve on the left with dp $\endgroup$ – H_1317 Jul 17 '17 at 4:41
  • $\begingroup$ @H_1317 correct. As long as you have a product of two single variate functions, you can separate and integrate just fine $\endgroup$ – qbert Jul 17 '17 at 4:47
  • $\begingroup$ What about initial conditions? I think $p(0)$ is missing here: as $\ln(p(0))$ in front of the 2nd expression, and $p(0)$ in front of the 3rd expression (from here) $\endgroup$ – Jesse Knight May 31 at 14:20
  • $\begingroup$ @JesseKnight yes certainly, seems like I was being sloppy with integration constants $\endgroup$ – qbert May 31 at 14:22

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