Hi may i like to ask how to solve this differential equation

$$\frac{dP}{dt}=c P(\frac{ln(k)}{p})$$

Where c is a constant and K is a carrying capacity.

(a) Solve this differential equation

(b) Compute $lim_{t\rightarrow \infty} P(t)$

(c) Graph the Function above for $K=1000$, $P_o=100$ and $c=0.005$, and compare it with the logistic function. What are the similarities and differences.

(d) Use the above differential equation to show that this function grows fastest when $P=\frac{k}{e}$

  • $\begingroup$ is the equation $\frac{dP}{dt} = c\ln{\left(\frac{K}{p}\right)}P$? $\endgroup$ – Chinny84 Mar 14 '14 at 11:40
  • $\begingroup$ What did you try ? Please show us your work. $\endgroup$ – Claude Leibovici Mar 14 '14 at 12:21
  • $\begingroup$ Oh i tried to use the separable equation method by bringing over the ln(k/p)p over to the left hand side but i do not know how to integrate P times a ln(P). Do i need to sepreate the P and ln(p) by partial fractions? 1/(pln(p))dP=cdt. I dont know how to integreats the 1/(pln(p)dp) $\endgroup$ – ys wong Mar 14 '14 at 12:29
  • $\begingroup$ Is part d. also mistyped? $\endgroup$ – Amzoti Mar 14 '14 at 12:51
  • $\begingroup$ Oh for part d it should be P=k/e $\endgroup$ – ys wong Mar 14 '14 at 12:55

This is the Gompertz Model and I believe it is typed wrong in the problem. Also, for part $d.$, I think it is supposed to be $p = \dfrac{k}{e}$.

We have:

$$\dfrac{dp}{dt} =c ~\ln \left(\dfrac{K}{p}\right)~p$$

This can be solved as a separable or exact equation.

After separating and setting up the integral, we have:

$$\int \dfrac{1}{\ln \left( \dfrac{k}{p} \right) p}~dp = \int c ~ dt$$

For the LHS integral, let $u = \ln \left( \dfrac{k}{p} \right) \implies ~du = -\dfrac{1}{p}~dp$.

Solving this gives us:

$$\large p(t) = k e^{-e^{-ct - w}}$$

Substituting the constants and IC, we have:

$$\large p(t) = 10^{-e^{-t/200} +~ 3}$$

You should now be able to continue answering the other parts.

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