Integration of Logistic Model

[BraedenP]

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 $$\frac{dy}{dt}=ky\left ( 1-\frac{y}{C} \right)$$
and that $y(t)$ 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: $$\frac{1}{y}+\frac{1}{C-Y} dy = k dt$$
and then integrated to get: $$ln(y)-ln(C-Y)=kt+N$$, (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: $N=ln(\frac{1}{3})-1987k$ and then substituted this equation back into the equation using the other two values, "1999" and "6":

$$ln(\frac{3}{7})=1999k + ln(\frac{1}{3})-1987k$$

I then solved for k, getting: $$k=\frac{ln(\frac{9}{7})}{12}$$

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!

 

[lanedance]

at this point how about
$$ln(y)-ln(C-y) =ln(\frac{y}{C-y}) = kt + N$$

$$\implies \frac{y}{C-y} = e^{kt + N}$$

$$\implies y(1+e^{kt + N}) = Ce^{kt + N}$$

$$\implies y = \frac{Ce^{kt + N}}{1+e^{kt + N}} = \frac{C}{e^{-(kt+N)}+1} = \frac{C}{Be^{-kt}+1}$$

where B is a constant