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Another model for a growth function for a limited population is given by the Gompertz function, which is a solution to the differential equation dPdt=cln(KP)P where c is a constant and K is the carrying capacity. Answer the following questions. 1. Solve the differential equation with a constant c=0.05, carrying capacity K=3000, and initial population P0=1000.

User Emillie
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Answer:

The solution is


P(t) = 1000e^(-1)e^{e^(-0.05t)}.

Explanation:

We have the Gompertz's differential equation


(dP)/(dt) = -cP\ln(KP).

Notice that this a separable equation. So,


(dP)/(P\ln(KP)) = -cdt.

Now, integrating in both sides:


\int(dP)/(P\ln(KP)) = -\int cdt.

Recall that


\int(dP)/(P\ln(KP)) = \ln(|\ln(KP)|).

Thus,


\ln(|\ln(KP)|) = -ct+A,

and taking exponential:


|\ln(KP)| = A'e^(-ct).

Taking another exponential


P(t) = (A'')/(K) e^{e^(-ct)}.

Now, let us substitute the given values for c and K:


P(t) = (A'')/(3000) e^{e^(-0.05t)}.

To find the value of the constant A'' we use the initial condition:


P(0) = (A'')/(3000) e^{e^(0)} = (A'')/(3000)e=1000.

Then,


A'' = (3\cdot 10^6)/(e).

Hence, the solution is


P(t) = 1000e^(-1)e^{e^(-0.05t)}

User Rifka
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