# [SOLVED] Diff Eq's population model **MIDTERM TOMORROW**

• Feb 24th 2009, 11:22 PM
ibanez270dx
[SOLVED] Diff Eq's population model **MIDTERM TOMORROW**
Hello,
Thanks for reading this... I've got my midterm tomorrow and I just want to make sure I know every nook and cranny of the review sheet... Here's the problem:

A population of goldfish in a certain pond is modelled by:

dG/dt = G*(1-6/30)*(1-200/G)*(1-300/G)

where G is the number of goldfish in the pond. Identify equilibrium points and their stability. If there are initially 17 goldfish in the pond, what is the ultimate fate of their population? Note: you DO NOT have to actually solve this differential equation.

I found the equilibrium points to be G=0, 30, 200, and 300. After drawing a phase diagram and plugging in numbers to the equation, I found that G=0 and G=200 are unstable, and G=30 and G=300 are stable. However, I cannot figure out how I am supposed to find the fate of the population without solving the DE... Can somebody point me in the right direction?

• Feb 25th 2009, 04:59 AM
Jester
Quote:

Originally Posted by ibanez270dx
Hello,
Thanks for reading this... I've got my midterm tomorrow and I just want to make sure I know every nook and cranny of the review sheet... Here's the problem:

A population of goldfish in a certain pond is modelled by:

dG/dt = G*(1-6/30)*(1-200/G)*(1-300/G)

where G is the number of goldfish in the pond. Identify equilibrium points and their stability. If there are initially 17 goldfish in the pond, what is the ultimate fate of their population? Note: you DO NOT have to actually solve this differential equation.

I found the equilibrium points to be G=0, 30, 200, and 300. After drawing a phase diagram and plugging in numbers to the equation, I found that G=0 and G=200 are unstable, and G=30 and G=300 are stable. However, I cannot figure out how I am supposed to find the fate of the population without solving the DE... Can somebody point me in the right direction?

First off, is your DE

$\displaystyle \frac{dG}{dt} = G \left(1 - \frac{G}{30} \right) \left(1 - \frac{G}{200} \right)\left(1 - \frac{G}{300} \right)$

this actually coincides with the stability of your critical points. Since you determined that $\displaystyle G=0$ was unstable and $\displaystyle G=50$ stable, then the initial population of $\displaystyle G=17$ would tend to $\displaystyle G=50$ in time.
• Feb 25th 2009, 06:23 AM
arpitagarwal82
Danny is right.

There are for critical points G= 0, 30,200, 300.
Now initially G = 17.

You can see that dG/dt is > 0 at G = 7.
so function G is increasing in nature around G = 17.
at G = 30. dG/dt = 0.

so when G = 30 is reached, there will be no growth in funtion. So finally G wil tend to take the value of 30.
Hope this is helpful.
• Feb 25th 2009, 06:59 AM
HallsofIvy
Jeff, note that these responses are assuming that the equation is really
dG/dt = G*(1-G/30)*(1-G/200)*(1-G/300)

NOT dG/dt = G*(1-6/30)*(1-200/G)*(1-300/G) as you initially wrote.
• Feb 25th 2009, 05:42 PM
ibanez270dx
Thank you all for your replies! That actually does make a lot of sense now that I look at my phase diagrams. I have a hard time relating these kinds of concepts, so I really appreciate all the help!

HallsofIvy, yes I saw that the equation they responded with was incorrect, but I think the concept is still the same. If it's not, please let me know or I'll be perpetually confused!

Just to let you know, I think I did really well on my midterm! (Rock)

Thanks again,
- Jeff
• Feb 25th 2009, 06:06 PM
Jester
Quote:

Originally Posted by ibanez270dx
Thank you all for your replies! That actually does make a lot of sense now that I look at my phase diagrams. I have a hard time relating these kinds of concepts, so I really appreciate all the help!

HallsofIvy, yes I saw that the equation they responded with was incorrect, but I think the concept is still the same. If it's not, please let me know or I'll be perpetually confused!

Just to let you know, I think I did really well on my midterm! (Rock)

Thanks again,
- Jeff

Yeah (Clapping)
• Feb 25th 2009, 09:42 PM
arpitagarwal82
Quote:

Originally Posted by ibanez270dx

Just to let you know, I think I did really well on my midterm! (Rock)

Congrats buddy. Keep solving the problems and you will do ever better. :)