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Critical Points

  1. Aug 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Find all the critical points:

    [tex]\frac{dx}{dt} = 1 - xy[/tex]

    [tex]\frac{dy}{dt} = x - y^3[/tex]

    2. Relevant equations
    [tex]\frac{dx}{dt} = 0[/tex]

    [tex]\frac{dy}{dt} = 0[/tex]

    3. The attempt at a solution

    Here's what I did:

    [tex]1 - xy= 0[/tex] .......... [tex]x - y^3 = 0[/tex]

    [tex]1 - xy= 0[/tex] .......... [tex]x = y^3[/tex]

    [tex]1 - y^4= 0[/tex] .......... [tex]x = y^3[/tex]

    [tex]y= 1, -1[/tex] .......... [tex]x = y^3[/tex]

    [tex]y= 1, -1[/tex] .......... [tex]x = 1, -1[/tex]

    So the critical points are:

    [tex](1, 1) (1, -1) (-1, 1) (-1, -1)[/tex]

    But in the book the answer is only:

    [tex](1, 1) (-1, -1)[/tex]

    Can someone explain to me why?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 11, 2007 #2


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    Staff Emeritus
    Science Advisor

    It doesn't really matter, but are those caculated from a given function or were you give the "direction field"?

    I find your way of writing this a bit peculiar! You know that dx/dt= 1- xy= 0 and dy/dt= x- y3= 0. That's your first line. The second line repeats the first equation and solves the second for x: x= y3.
    In the third line you replace the x in the first equation by y3, from the second equation, and have 1- y4= 0. That has the two (real) solutions y= 1 and y= -1. If y= 1, then the first equation becomes 1- x= 0 so x= 1: a critical point is (1,1). If y= -1 then the first equation become 1+ x= 0 so x= -1: a critical point is (-1, -1).

    Because you get x= 1 only when y= 1, not when y=-1. Similarly, x= -1 is correct only when y= -1. You are looking for critical points, not just values of x and y that you then combine willy-nilly.
  4. Aug 11, 2007 #3
    Oh now I understand. Thanks a lot!
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