Population Model

  • Thread starter KillerZ
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Homework Statement



The differential equation [tex]\frac{dP}{dt} = P(a - bP)[/tex] is a well-known population model. Suppose the DE is changed to:

[tex]\frac{dP}{dt} = P(aP - b)[/tex]

Where a and b are positive constants. Discuss what happens to the population P as time t increases.

Homework Equations



[tex]\frac{dP}{dt} = P(aP - b)[/tex]

The Attempt at a Solution



Well I thought the population would increase because in the original equation on the intervals:

critical points
[tex] P(t) = 0[/tex]
[tex] P(t) = a/b[/tex]

[tex]-\infty < P < 0[/tex] it is decreasing
[tex]0 < P < a/b[/tex] it is increasing
[tex]a/b < P < \infty[/tex] it is decreasing

so in the changed equation:

critical points
[tex] P(t) = 0[/tex]
[tex] P(t) = b/a[/tex]

[tex]-\infty < P < 0[/tex] it is increasing
[tex]0 < P < b/a[/tex] it is increasing
[tex]b/a < P < \infty[/tex] it is increasing
 

Answers and Replies

  • #2
HallsofIvy
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The derivative is positive (and so the function is increasing) when P(aP- b)> 0. Now, if P< 0, both P and aP-b are negative so P(aP- b)> 0. For 0< P < b/a, P is positive but aP- b is still negative so P(aP-b)< 0 and P is decreasing. Was that a typo?

From that, it seems to me that what happens to the population "as time t increases" depends upon the initial value. What happens if P(0) is less than b/a? What happens if P(0) is larger than b/a? What happens if P(0)= b/a?
 
  • #3
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Ok I got it thanks.
 

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