Integration of Logistic Model

  • Thread starter BraedenP
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Homework Statement



The following is a population logistics model where t is the year, C is the carrying capacity, k is some unknown constant, and y(t) is the population at time t.

I am given [tex]\frac{dy}{dt}=ky\left ( 1-\frac{y}{C} \right)[/tex]
and that [itex]y(t)[/itex] is the population size for the year t.

I am also given that when t=1987, y=5 AND when t=1999, y=6. C is always 20.

I am supposed to figure out (a) the population in 2050 where C=20, and (b) the population in 2100 where C=10.

Homework Equations



The Attempt at a Solution



I rearranged the equation to form: [tex]\frac{1}{y}+\frac{1}{C-Y} dy = k dt[/tex]
and then integrated to get: [tex]ln(y)-ln(C-Y)=kt+N[/tex], (where N is the constant of integration to avoid confusion with C)

This is where things start going wrong. I substituted the "1987" and "5" values into the equation and solved for N, getting: [itex]N=ln(\frac{1}{3})-1987k[/itex] and then substituted this equation back into the equation using the other two values, "1999" and "6":

[tex]ln(\frac{3}{7})=1999k + ln(\frac{1}{3})-1987k[/tex]

I then solved for k, getting: [tex]k=\frac{ln(\frac{9}{7})}{12}[/tex]

All of these logarithms prevent me from calculating a nice clean answer for (a), and I have no clue where to start with regard to (b). Can anyone point me in the right direction?

Thanks!
 

Answers and Replies

  • #2
lanedance
Homework Helper
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at this point how about
[tex]
ln(y)-ln(C-y)
=ln(\frac{y}{C-y}) = kt + N
[/tex]

[tex]
\implies \frac{y}{C-y} = e^{kt + N}
[/tex]

[tex]
\implies y(1+e^{kt + N}) = Ce^{kt + N}
[/tex]

[tex]
\implies y
= \frac{Ce^{kt + N}}{1+e^{kt + N}}
= \frac{C}{e^{-(kt+N)}+1}
= \frac{C}{Be^{-kt}+1}
[/tex]

where B is a constant
 
Last edited:

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