Solving Diff. Equations: Population Growth & Changes

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Homework Help Overview

The discussion revolves around solving differential equations related to population growth. The original poster presents two specific equations: one for population growth modeled by a logistic function and another differential equation that describes population changes over time. Participants are exploring methods to analyze the behavior of these equations, particularly in terms of population growth rates and conditions for increasing or decreasing populations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the approach of differentiating the population function to find growth rates and question the method of evaluating the second derivative. There is also a focus on determining when the population is increasing or decreasing based on specific ranges of population values.

Discussion Status

Some participants have offered guidance on evaluating the differential equation by suggesting specific population values to test. There is an ongoing exploration of inequalities related to the population model, with participants questioning the best approach to demonstrate the conditions for population growth and decline.

Contextual Notes

Participants are working within the constraints of the problem as presented, including the initial conditions and the specific ranges of population values that need to be analyzed. There is an emphasis on understanding the implications of the inequalities derived from the differential equations.

gordda
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Hello
I am having major difficulties with two Differential equation.
1. If I have the equation P=2000/(1+9e^.06t). where p=population and t=time. How do I work out when the population is growing the fastest? To work this question out do I just make t the subject and diff twice and let it equal to zero?

2. The diff equation dp/dt= .08p(1-p/1000)(1-200/p) has an inital population of Po how do I show that the population is increasing if 200<P<1000 and decreasing if 0<P<200.

Please help

Thanx:)
 
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gordda said:
Hello
I am having major difficulties with two Differential equation.
1. If I have the equation P=2000/(1+9e^.06t). where p=population and t=time. How do I work out when the population is growing the fastest? To work this question out do I just make t the subject and diff twice and let it equal to zero?

2. The diff equation dp/dt= .08p(1-p/1000)(1-200/p) has an inital population of Po how do I show that the population is increasing if 200<P<1000 and decreasing if 0<P<200.
1) set d2P/dt2 = 0 and solve for t is a good beginning.
2) show that (dp/dt)>0 when 200<P<1000, and (dp/dt)<0 when 0<P<200. it's not difficult.
 
ok i get 1 but in 2 how do i show that dp/dt is >0 or <0.
 
gordda said:
ok i get 1 but in 2 how do i show that dp/dt is >0 or <0.
you're thinking it's more difficult than it really is.
begin by evaluating (dp/dt) at p=100 and p=500.
which term changes sign? do any other terms change sign?
now generalize these observations over the 2 given ranges of p.
 
Last edited:
shouldn't i use values more closer to the range
 
gordda said:
shouldn't i use values more closer to the range
you can.
however, what are the solutions to the 2 inequalities below (for p > 0)?
(1-200/p) < 0
(1-200/p) > 0
and under what condition does the following inequality hold:
(1-p/1000) > 0
the above inequality solutions give the 2 ranges of p.
 
geosonel said:
you can.
however, what are the solutions to the 2 inequalities below (for p > 0)?
(1-200/p) < 0
(1-200/p) > 0
and under what condition does the following inequality hold:
(1-p/1000) > 0
the above inequality solutions give the 2 ranges of p.
if you need review of solving inequality equations, try here:
http://www.sosmath.com/algebra/inequalities/ineq01/ineq01.html
proceed page by page thru the various review topics to the extent needed.
 

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