• Support PF! Buy your school textbooks, materials and every day products Here!

ODE Population problem

  • Thread starter cue928
  • Start date
  • #1
130
0
I know that the rate of change with time of a population is proportional to the square root of t. T=0 is y = 100. Population increases at rate of 20 per month.

I started out by trying to do dy/dt = p^.5. I am used to the population problems where I use y=Ce^(rt) but am having trouble making the jump to this kind. How do you account for the increase of 20/month?
 

Answers and Replies

  • #2
256
2
The population always increases at a rate of 20 per month, or is that an initial value?
 
  • #3
130
0
Yes, always increases at rate of 20/month and initial population is 100.
 
  • #4
256
2
Than the question makes no sense to me. How can the rate of increase in the population be of 20/month and proportional to the square root of t at the same time?
 
  • #5
130
0
You and me both then. Here is the exact question:
The time rate of change of a rabbit population P is proportional to the square root of P. At time t = 0 (months) the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. How many after one year? The answer I guess is 484.

How do you set up logistics equations?
 
  • #6
256
2
So 20/month is an initial value.

The rate of change of P, i.e. it's derivative with respect to P is proportional to the square root of P:

[tex] \frac{dP(t)}{dt} = \alpha P^{1/2} [/tex]

where alpha is a constant with respect to. You also know that

[tex] \frac{dP(t)}{dt} \bigg|_{t=0}} = 20 \ rabbits/month [/tex]

and that

[tex]P(0) = 100 \ rabbits[/tex]

Which is all the info you need to determine P(t)
 
Last edited:
  • #7
130
0
I appreciate what you wrote but I'm still unsure on where to start. I see that I was wrong to have a p^.5 on the right. Should I go ahead and start separating variables and then moving forward?
 
  • #8
256
2
I appreciate what you wrote but I'm still unsure on where to start. I see that I was wrong to have a p^.5 on the right
OOOPSSS, my mistake. I corrected it in the previous post.

Should I go ahead and start separating variables and then moving forward?

Of course you should. Even in the case where you shouldn't, equations don't bite you know.
 
  • #9
130
0
But how do you incorporate the dp/dt = 20 and P(0) = 100? I've done the sep of variables, no problem there. Sitting on 2p^.5 = (alpha)t + C.
 
  • #10
130
0
Made another attempt at this, can you tell me where I'm going wrong?
I came up with this equation after sep of variables: 2p^.5 = Bt + C.
1) With t = 0 and initial population of 100, I came up with a C value of 100.
2) With t = 1, population = 120, so I came up with a "B" value of 120.
3) But then when I do time = 12, I don't get anywhere near the 484 rabbits that I should have.
 
  • #11
Matterwave
Science Advisor
Gold Member
3,965
326
Er, did you solve for p in your equation so you have some p(t) on the left side instead of 2p^.5?
 
  • #12
256
2
2) With t = 1, population = 120, so I came up with a "B" value of 120.
This is wrong. What you have is P'(0) = 20. So use the derivative of P to find B. Do you realize that the rate of change in the population is the derivative of P?
 
Last edited:

Related Threads on ODE Population problem

  • Last Post
Replies
21
Views
4K
Replies
2
Views
2K
Replies
2
Views
902
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
12
Views
732
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
971
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
865
Top