1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook