Respuesta :
Answer:
A) [tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]
B) P(t→∞)=2000
C) [tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]
Step-by-step explanation:
Given differential eq is
[tex]\frac{dP}{dt}=c ln (\frac{K}{P})P[/tex] --- (1)
Eq is separable
[tex]\frac{1}{ln (\frac{K}{P})P}dP=cdt[/tex] --- (2)
[tex]let \\u = ln\frac{K}{P}\\du= \frac{1}{\frac{K}{P}}(\frac{-K}{P^{2}}).dP\\du=\frac{-1}{P}.dP\\dP=-P.du[/tex]
substituting in (2)
[tex]-\frac{du}{u}=dt[/tex]
Integrating both sides
[tex]\int {-\frac{1}{u}} \, du=\int{c}\,dt\\-ln|u|=ct +B\\ln|u|=-ct -B\\[/tex]
Back substituting value of u
[tex]ln |ln\frac{K}{P}|=-ct-B\\|ln\frac{K}{P}|=e^{-ct-B}\\ln|\frac{K}{P}|=be^{-ct}\\[/tex]---(3)
at t =0
[tex]ln|\frac{K}{P}|=be^{-ct}\\b=ln|\frac{K}{P}|\\b=ln\frac{2000}{500}\\b=ln|4|[/tex]
from (3)
[tex]ln|\frac{K}{P}|=be^{-ct}\\\frac{K}{P}=e^{ln4e^{-ct}}\\P(t)=\frac{K}{e^{ln4e^{-ct}}}[/tex]
[tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]
B) [tex]\lim{t \to \infty}[/tex]
[tex]P( {t \to \infty} )=\frac{2000}{e^{ln4e^{-0.1\infty}}}\\e^{-0.1\infty}=0\\\implies P( {t \to \infty} )=\frac{2000}{e^{0}}}\\\\P(\infty)=2000\\[/tex]
which is the carrying capacity.
C) To find the fastest growth rate we have to maximize [tex]\frac{dP}{dt}[/tex]
From given differential eq
[tex]\frac{dP}{dt}=cln|\frac{K}{P}|P[/tex]
so function to maximize is
[tex]f(P)=cln|\frac{K}{P}|P[/tex]
[tex]f'(P)=cln|\frac{K}{P}|+c\frac{1}{\frac{K}{P}}\frac{-K}{P^{2}}.P[/tex]
[tex]f'(P)=c[ln|\frac{K}{P}|-1][/tex]
To maximize find f'(P)=0
[tex]c[ln|\frac{K}{P}|-1]=0[/tex]
[tex]ln|\frac{K}{P}|=1[/tex]
[tex]\frac{K}{P}=e[/tex]
[tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]
