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For the following system find the fixed points \begin{cases} x'(t) &= x^2 - y,\\ y'(t) &= x-y. \end{cases}

I got $y=x^2$ and $y=x$.

These are non linear systems and so we need to compute the fixed points at its Jacobian matrix.

However, I am not sure on how to do this since I don't know the stability at the fixed points. Hence, I will not be able to draw a phase portrait for it.

  • 3
    Can you solve the equation $x^2-x=0$?2012-12-01
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    i beleive i understand what you are trying to say. Are the fixed points(0,0) and (1,1) since i got x=0,1 after solving the equation2012-12-01
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    Right. Do you know what the Jacobi matrix is?2012-12-01
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    yes i have solved the jacobian matrix and have obtained at (0,0) it is stable, spiral at(1,1) it is a saddle point thank you for the help2012-12-01
  • 0
    This is correct. You can post the details of your calculations as an answer and accept it in some time.2012-12-01

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