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A D.E. question

  • Thread starter gordda
  • Start date
20
0
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:)
 
88
0
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.
 
20
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ok i get 1 but in 2 how do i show that dp/dt is >0 or <0.
 
88
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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:
20
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shouldn't i use values more closer to the range
 
88
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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.
 
88
0
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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