Another model for a growth function for a limited population is given by the Gompertz function, which is a solution to the differential equation
$\frac{dP}{dt} = c \ln\left(\frac{K}{P}\right)$ $P$
where $c$ is a constant and $K$ is the carrying capacity. Answer the following questions.
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Solve the differential equation with a constant $c = 0.05$, carrying capacity $K = 4000$, and initial population $P_0 = 750$.
Answer: $P(t) = $ -
With $c = 0.05$, $K = 4000$, and $P_0 = 750$, find $\displaystyle \lim_{t \to \infty} P(t)$.
Limit:
Best Answer
We are given:
$$\dfrac{dP}{dt} = c \ln\left(\frac{K}{P}\right)P$$
With a constant $c = 0.05 = \dfrac{1}{20}$, carrying capacity $K = 4000$, and initial population $P_0 = 750$.
This DEQ is separable as:
$$\dfrac{1}{c \ln\left(\frac{K}{P}\right)P}~dP = dt$$
Substituting the constants and integrating yields the following:
$$\int \dfrac{20}{\ln \left(\dfrac{4000}{p}\right) p}~dp = \int dt$$
This gives us:
$$\large P(t) = 4000 e^{e^{-(c-t)/20}}$$
We now solve for the constant, using $P(0)=750$, yielding:
$$P(t) = 125\ 2^{5-4 e^{-\frac{t}{20}}} 3^{e^{-\frac{t}{20}}}$$
Note: If this exact approach is to difficult to follow, use the numeric $c=0.05$ and you will get:
$$\large P(t) = 4000 e^{-1.67398^{e^{-0.05 t}}}$$
The limit is $4000$ since the two exponential terms go to zero, that is $125 \times 2^5$.