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Autonomous Systems and Stability

  1. Nov 26, 2006 #1
    I have a problem regarding the equations dx/dt=x-xy and dy/dt=y+2xy. I need to find the critical points of this system and denote if they are stable or asymptotic or whatever. I flipped through the section on this and you can find the critical points by setting dx/dt and dy/dt to zero and solving for x and y. This gives (0,0) and (1,-1/2). But you still don't know what their nature is, so I tried to solve the system. I did this by dividing one by the other and getting dy/dx=y+2xy/x-xy. However I am stuck trying to solve this problem. I can't find a integrating factor for this in terms or either x or y (we haven't learnt anything beyond that) in order to make it exact. How else can I solve this? I gather I can turn this into a matrix and solve it somehow by I am not sure.
  2. jcsd
  3. Nov 26, 2006 #2
    Did you try looking at the eigenvalues of the linearized system at the critical points?
  4. Nov 26, 2006 #3


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    Did you notice it's separable? (Oh, and you really should be using parentheses where appropriate. 1+2/2+3 is 5, not 3/5)

    But as LeBrad suggests, you should be able to figure out the answer without solving the equation. I suggest you do that first, and then solve to check your answer!
  5. Nov 27, 2006 #4


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    Recheck your equilibrium points! x= 1, y= -1/2 do NOT make dx/dt and dy/dt 0.
    As Hurkyl and LeBrad suggested, you don't need to solve the system to determine if it is stable. In fact, that's the whole point! Most non-linear systems can't be solved that easily. The stability can be determined by looking at the "linearized" system at each equilibrium point. (That's what LeBrad was talking about: non-linear systems don't have eigen-values.)
    Last edited by a moderator: Nov 27, 2006
  6. Nov 29, 2006 #5
    I been busy the last few days and haven't had a chance to take a look at these forums. How do I find the linearized system? I took another look at the equation and realized it is indeed separable!
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