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A D.E. question

  1. Jul 31, 2005 #1
    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

  2. jcsd
  3. Jul 31, 2005 #2
    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.
  4. Jul 31, 2005 #3
    ok i get 1 but in 2 how do i show that dp/dt is >0 or <0.
  5. Jul 31, 2005 #4
    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: Jul 31, 2005
  6. Jul 31, 2005 #5
    shouldn't i use values more closer to the range
  7. Jul 31, 2005 #6
    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.
  8. Jul 31, 2005 #7
    if you need review of solving inequality equations, try here:
    proceed page by page thru the various review topics to the extent needed.
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