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Phase line problem, my given solution is wrong I believe

  1. Jul 13, 2012 #1
    I was given the solution to this problem:
    y'=-y^2+2y-1, plot phase line, solve it, in terms of y(0)=y0, what happens as t goes to infinity.
    So then the phase line shows x=1 is 'semistable'. Also the solution is y= 1+ (y0-1)/(1+(y0-1)t) but now he claims this, which I believe is wrong:
    if y0=1: y(t)=0 for all t>0
    if y0>1, y(t) goes to 0 as t goes to infin.
    if y0<1, y(t) goes to neg. infin. (also confused here because he now writes, "as t (arrow up) 1/(1-y0)" )
     
  2. jcsd
  3. Jul 13, 2012 #2

    tiny-tim

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    hi ericm1234! :smile:

    it's easier to analyse if you write it as y - 1 = 1/(t - C) :wink:

    (which is exactly what you get if you solve the equation directly)
     
  4. Jul 13, 2012 #3
    Agreed Tiny-Tim. But now, when t=0, c=-1/(y0-1). Ok, my issue now is with the analysis that I listed above, which I believe is wrong, as I wrote it (from the solutions to the problem given to me by the teacher)
     
  5. Jul 14, 2012 #4

    tiny-tim

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    yes, it's wrong :redface:
     
  6. Jul 14, 2012 #5

    HallsofIvy

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    In terms of the phase line, [itex]y'= -(y- 1)^2[/itex] so that y= 1 is an equilibrium point. For y< 1, y- 1< 0 but it is squared so that [itex]y'= -(y-1)^2[/itex] is positive and y is increasing toward y= 1 NOT toward negative infinity. For y> 1, y- 1> 0 but with the squaring we still have [itex]y'= -(y-1)^2[/itex] positive and y increases away from y= 1 to infinity.
     
  7. Jul 14, 2012 #6
    I think you disregarded the negative sign then. y' is negative for all y, except y=1.
     
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