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Determining what value y will approach eventually

  1. Sep 15, 2011 #1
    Determining what value "y" will approach eventually

    I'm having a bit of trouble with an equilibrium problem. The autonomous equation given is:

    y' = (2y+8)(y-3)(y-8)^3

    The first task is to find and classify the stability of each equation, which I've done.

    Points: y=-4, y=3, y=8


    The bit that I am unsure of how to do comes next.

    a.) If y(8) =-10, to what value will y approach eventually?

    b.) If y(-4) = 8, to what value will y approach eventually?

    I've seen it done with direction fields, but I've never really understood those.. I've also heard that there is a way to determine what value y will approach without actually solving the differential equation.

    Can anyone help?
     
  2. jcsd
  3. Sep 15, 2011 #2

    HallsofIvy

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    Re: Determining what value "y" will approach eventually

    The crucial point is where y'=0, where it is positive, and where it is negative.
    Remember, from very basic calculus, that y' tells how y changes. In particular, if y' is negative, y is decreasing and if y' is positive, y is increasing.

    Obviously, [itex]y'= (2y+8)(y- 3)(y- 8)^3[/itex] will be 0 where 2y+ 8= 0, y- 3= 0, or y- 8= 0. It should be easy to see where y' is positive or negative. For example, if y< -4, each of 2y+ 8, y- 3, and y- 8 are negative. The product of 5 negative numbers is negative so y' is negative and y decreases. Since y' will still be negative, there is nothing to stop y form going to negative infinity. If -4< x< 3, 2y+ 8 is positive while y- 3 and y- 8 are still negative. The product of four negative numbers and one positive is positive so y is increasing. But it can't just keep increasing- beyond y= 3, that changes. So if y starts with a value between -4 and 3, it increase toward y= 3 but stops their (because y'(3)= 0).
     
    Last edited: Sep 15, 2011
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