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Homework Help: Finding critical points for phase planes

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    X'=x-y^2 Y'=x^2 -xy -2x

    2. Relevant equations
    Find the critical point of this system.

    3. The attempt at a solution
    I know one point is (0,0).
    When I tried it I got sqrt(x)= 2 and sqrt(x)= 1 but the point (1,1) doesn't fit later parts of the question. I'm also aware that since a sqrt can have positive and negative values then (4, -2) can also equal zero. Am I missing something obvious?
  2. jcsd
  3. Dec 3, 2015 #2


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    Science Advisor

    Frankly, I don't know what you mean by "When I tried it I got sqrt(x)= 2..." When you tried what? And why write "sqrt(x)= 2" rather than "x= 4"? In any case, a critical point of the system will be where x'= x- y^2= 0 and y'= x^2- xy- 2x= 0. From the first equation, x= y^2. Replacing x with y^2 in the second equation gives y^4- y^3- 2y^2= y^2(y^2- y- 1)= y^2(y+ 1)(y- 2)= 0.
  4. Dec 3, 2015 #3
    Sorry by 'try it' I meant solved for zero and I used sqrt(x) because I was trying to solve it in x instead of y. Your way makes much more sense and solved the problem! Thanks a million
  5. Dec 3, 2015 #4


    Staff: Mentor

    There are two things wrong with this sentence.
    1) By definition, the square root of a nonnegative number represents a single value. For example, ##\sqrt{4} = + 2##. Period. You are mistaken to think that ##\sqrt{4} = \pm 2##.
    2) "then (4, -2) can also equal zero." -- No. (4, -2) is a point in the plane. It cannot equal a single number such as zero.

    BTW, (4, -2) is also NOT a solution of the equations below.
    Set both X' and Y' to 0 and solve the resulting equations simultaneously.
    IOW, solve
    ##x - y^2 = 0##
    ##x^2 - xy - 2x = 0##
    It helps to use one of the equations to substitute in the other.
    There are three solutions, one of which is (0, 0), which you already found. (1, 1) is NOT a solution, nor is (4, -2)..
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