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]