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ODEs: Word problem involving (I think) phase lines

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    If the population y of rats on a farm at time t (in weeks) satisfies:
    dy/dt = -y(y-100)/50
    then how many rats per week should be killed to eradicate the population?


    2. Relevant equations
    None known.


    3. The attempt at a solution
    The ODE dy/dt is autonomous, so I can use a phase line.
    I found the equilibrium points to be at y=0 and y=100, and found that for the interval 0<y<100, solutions were increasing, and for both the intervals 100<y<%infinity, and -%infinity<y<0, solutions were decreasing, but the third interval is probably undefined, since a negative population is not feasible.

    And now I'm lost. I don't know what to do after I've worked this out! Thanks in advance for your hints.
     
  2. jcsd
  3. Sep 21, 2011 #2

    dynamicsolo

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    Homework Helper

    We want to modify the population system by "harvesting" rats (as population dynamics people so delicately put it) in each "time step" so that the population y will be driven to zero for any initial population. As things stand now, y = 0 is an unstable equilibrium ( dy/dt > 0 ), repelling the population toward the stable equilibrium of y = 100 ( dy/dt > 0 for y < 100 , dy/dt < 0 for y > 100 ). So, after some time has passed, the farm should expect to have to cohabit with 100 rats.

    We introduce a so-called "harvesting term" H which equals the number of rats per week we shall... okay, kill! The differential equation is now

    [tex] \frac{dy}{dt} = \frac{y \cdot (y-100)}{50} - H .[/tex]

    If we graph this and pick some positive value like, say, H = 10 (exterminate 10 rats a week), we'll see that the equilibria have moved. The important thing is that now dy/dt < 0 at y = 0 , so this population level is now in an "attractive" region, instead of a "repelling" one. Unfortunately, there is still a stable (attracting) equilibrium at some positive y value. Unacceptable! What is the smallest value we can make H (remember, we have to keep an integer number of rats) so that dy/dt < 0 for any positive value of y . This will guarantee that for any initial rat population the farm has, the number of rats will ultimately fall to zero.
     
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