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.

Respuesta :

Answer:

The solution is

[tex]P(t) = 1000e^{-1}e^{e^{-0.05t}}[/tex].

Step-by-step explanation:

We have the Gompertz's differential equation

[tex]\frac{dP}{dt} = -cP\ln(KP)[/tex].

Notice that this a separable equation. So,

[tex]\frac{dP}{P\ln(KP)} = -cdt[/tex].

Now, integrating in both sides:

[tex]\int\frac{dP}{P\ln(KP)} = -\int cdt[/tex].

Recall that

[tex]\int\frac{dP}{P\ln(KP)} = \ln(|\ln(KP)|)[/tex].

Thus,

[tex]\ln(|\ln(KP)|) = -ct+A[/tex],

and taking exponential:

[tex]|\ln(KP)| = A'e^{-ct}[/tex].

Taking another exponential

[tex]P(t) = \frac{A''}{K} e^{e^{-ct}}[/tex].

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

[tex]P(t) = \frac{A''}{3000} e^{e^{-0.05t}}[/tex].

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

[tex]P(0) = \frac{A''}{3000} e^{e^{0}} = \frac{A''}{3000}e=1000[/tex].

Then,

[tex]A'' = \frac{3\cdot 10^6}{e}[/tex].

Hence, the solution is

[tex]P(t) = 1000e^{-1}e^{e^{-0.05t}}[/tex]

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