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Find equilibrium points given 2 differential equation

  1. Mar 22, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\dot{x}[/itex] = -pxy + qx, [itex]\dot{y}[/itex] = rxy - sy

    where p,q,r and s are positive constants (p does not equal r)

    Question is : Determine all the equilibrium points for the system of differential equations given above, expressing your answers in terms of p,q,r and s


    3. The attempt at a solution

    I do know one point is (0,0) however am stuck in finding out the others. I factorise them to get

    x(-py + q) = 0
    y(xr - s) = 0

    Any hints will be much appreciated
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 22, 2013 #2

    HallsofIvy

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    Okay, and then what? If you are taking a differential equations course, we should be able to assume that you can do basic algebra. You should know "if ab= 0 then either a= 0 or b= 0."

     
  4. Mar 22, 2013 #3
    -py+q=0 so do I say y = p/q how do I find the x coordinate from that? Is that when x =0 or do I substitute it into my second equation?
     
  5. Mar 23, 2013 #4
    I mean y = q/p sorry. What's confusing is the fact that it is variables as the constants instead of numbers so any more hints would be grateful.
     
  6. Mar 23, 2013 #5

    HallsofIvy

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    ?? Constants are numbers.

    Your two equations are
    x(-py + q) = 0
    y(xr - s) = 0

    clearly, (0, 0) is a root. In fact if y= 0, -py+q is not 0 (unless q happens to be 0 which I am assuming is not true) so we must have x= 0 also.

    If y is NOT 0 then we must have xr- s= 0 so that x= s/r. Putting that into the first equation gives s/r(-py+ q)= 0 so that y= q/p. (s/r, q/p) is also an equilibrium point.

    If x is not 0 we can do the same thing but get the same point again.
     
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