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Homework Help: Critical points of system of DE

  1. Dec 18, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Find all the critical points of the system of DE:
    [itex]x'=x-x^2-2xy[/itex] and [itex]y'=2y-2y^2-3xy[/itex].


    2. Relevant equations
    [itex]y'=f(x, \vec y)[/itex] has critical point if [itex]f(\vec x )=0[/itex].


    3. The attempt at a solution
    I'm self studying DE's and I'm not understanding well what critical points are.
    It looks like I must equate x' to 0 and isolate x (which I've done and it gave me [itex]x=0[/itex] or [itex]x=1-2y[/itex]) and equal y' to 0 and isolate y, which I've done and got [itex]y=0[/itex] or [itex]y=1-\frac{3x}{2}[/itex].
    Now I don't know what to do.
     
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  3. Dec 18, 2011 #2

    Dick

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    I think you've got a fine start. The critical points are just where (x',y')=(0,0). (x,y)=(0,0) is certainly one. Combine your two conditions for x and y in pairs and solve each pair. You'll get four critical points. The one I gave you combines x=0 and y=0. Now combine x=0 and y=1-3x/2. There are two more combinations to try.
     
  4. Dec 19, 2011 #3

    fluidistic

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    Ah ok thank you.
    So for example for [itex](x,y)=(0,1-\frac{3x}{2})[/itex].
    Gives [itex]y'=3x-\frac{9x^2}{2}[/itex] evaluated in x=0 gives 0. But on the other hand since [itex]y=1-\frac{3x}{2}[/itex], [itex]y'=3/2[/itex] which differ from the previous answer, 0. So that this isn't a critical point? Is the reasoning good? Or it's because 3/2 isn't equal to [itex]3x-\frac{-9x^2}{2}[/itex] that it isn't a critical point?

    Edit: I'm confused about the notation. Is x'=dx/dy and y'=dy/dx?

    Edit2: I tried the other combinations and none gave me critical points. Only the (0,0) is critical point.
     
    Last edited: Dec 19, 2011
  5. Dec 19, 2011 #4

    vela

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    Both x and y are functions of a parameter, t, and the derivative is with respect to this parameter. The differential equations mean
    \begin{align*}
    \frac{dx}{dt} &= x-x^2-2xy \\
    \frac{dy}{dt} &= 2y-2y^2-3xy
    \end{align*}
    For the case x=0 and y=1-(3/2)x, you have y = 1-(3/2)(0) = 1. If you plug (x,y)=(0,1) back into the original differential equations, you will indeed find that dx/dt=0 and dy/dt=0.
     
  6. Dec 20, 2011 #5

    fluidistic

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    Ah I see vela, thank you. In this case I obtain that these 4 points are indeed critical.
    Would it have been possible to obtain a point that wasn't critical?
    Edit: nevermind the answer is yes, in another problem I get that not all of the points are critical.
     
    Last edited: Dec 20, 2011
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