# Growth model word problem w/differential equations

• Apr 29th 2008, 07:35 PM
emttim84
Growth model word problem w/differential equations
Instructions:

Biology. A population of eight beavers has been introduced into a new wetlands area. Biologists estimate that the maximum population the wetlands can sustain is 60 beavers. After 3 years, the population is 15 beavers. If the population follows a Gompertz growth model, how many beavers will be in the wetlands after 10 years?

Gompertz growth model:

$\displaystyle \frac{dy}{dt}=kyln\frac{60}{y}$

Here's what I did.

$\displaystyle dy=kyln\frac{60}{y}(dt)$

$\displaystyle \frac{dy}{yln\frac{60}{y}}=kydt$

$\displaystyle \int \frac{1}{y(ln60-lny)}dy=\int kdt$

$\displaystyle u=ln60-lny$
$\displaystyle du=-\frac{1}{y}dy$
$\displaystyle -du=\frac{1}{y}dy$

$\displaystyle - \int \frac {1}{u}du = \int kdt$

$\displaystyle -ln|u| = kt + C$

$\displaystyle -|ln60-lny| = e^{kt+c}$

$\displaystyle -ln\frac{60}{y} = Ce^{kt}$

$\displaystyle \frac{60}{y}=e^{-Ce^{kt}}$

$\displaystyle 60=e^{-Ce^{kt}}y$

My values for the variables are:

$\displaystyle y=0, t=0 y=15, t=3 y=?, t=10$

However, the solutions manual states the answer for the general soln is

$\displaystyle y=60e^{-Ce^{kt}}$ which is a difference of signs for the constant in the exponential. I'm not sure where I went wrong but I don't want to start plugging in values until I have the right general solution. Help please? :)
• Apr 29th 2008, 07:40 PM
Mathstud28
Quote:

Originally Posted by emttim84
Instructions:

Biology. A population of eight beavers has been introduced into a new wetlands area. Biologists estimate that the maximum population the wetlands can sustain is 60 beavers. After 3 years, the population is 15 beavers. If the population follows a Gompertz growth model, how many beavers will be in the wetlands after 10 years?

Gompertz growth model:

$\displaystyle \frac{dy}{dt}=kyln\frac{60}{y}$

Here's what I did.

$\displaystyle dy=kyln\frac{60}{y}(dt)$

$\displaystyle \frac{dy}{yln\frac{60}{y}}=kydt$

$\displaystyle \int \frac{1}{y(ln60-lny)}dy=\int kdt$

$\displaystyle u=ln60-lny$
$\displaystyle du=-\frac{1}{y}dy$
$\displaystyle -du=\frac{1}{y}dy$

$\displaystyle - \int \frac {1}{u}du = \int kdt$

$\displaystyle -ln|u| = kt + C$

$\displaystyle -|ln60-lny| = e^{kt+c}$

$\displaystyle -ln\frac{60}{y} = Ce^{kt}$

$\displaystyle \frac{60}{y}=e^{-Ce^{kt}}$

$\displaystyle 60=e^{-Ce^{kt}}y$

My values for the variables are:

$\displaystyle y=0, t=0 y=15, t=3 y=?, t=10$

However, the solutions manual states the answer for the general soln is

$\displaystyle y=60e^{-Ce^{kt}}$ which is a difference of signs for the constant in the exponential. I'm not sure where I went wrong but I don't want to start plugging in values until I have the right general solution. Help please? :)

• Apr 29th 2008, 07:46 PM
emttim84
Isn't $\displaystyle -|ln\frac{60}{y}| =$ whatever the same as -$\displaystyle ln\frac{60}{y}=$whatever since the negative sign is on the outside of the absolute value anyway?
• Apr 29th 2008, 07:47 PM
Mathstud28
Quote:

Originally Posted by emttim84
Isn't $\displaystyle -|ln\frac{60}{y}| =$ whatever the same as -$\displaystyle ln\frac{60}{y}=$whatever since the negative sign is on the outside of the absolute value anyway?

no...no it doesnt...because say you have $\displaystyle \pm{x}\cdot{-1}$

all this will do is change "when" the negative occurs..not the existence of it...it is still +/-
• Apr 29th 2008, 08:08 PM
emttim84
Quote:

Originally Posted by Mathstud28
no...no it doesnt...because say you have $\displaystyle \pm{x}\cdot{-1}$

all this will do is change "when" the negative occurs..not the existence of it...it is still +/-

Ahh ok...so because of the +/-, since we're talking about population growth, and you can't have a - if it's growth, then it's assumed that the + value of the absolute value should be used?