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.

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    $\begingroup$ Can you solve the equation $x^2-x=0$? $\endgroup$ – Artem Dec 1 '12 at 19:07
  • $\begingroup$ 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 equation $\endgroup$ – Saba Di Dec 1 '12 at 19:22
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    $\begingroup$ Right. Do you know what the Jacobi matrix is? $\endgroup$ – Artem Dec 1 '12 at 19:23
  • $\begingroup$ 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 help $\endgroup$ – Saba Di Dec 1 '12 at 19:29
  • $\begingroup$ This is correct. You can post the details of your calculations as an answer and accept it in some time. $\endgroup$ – Artem Dec 1 '12 at 19:36

The Jacobian is $$\begin{pmatrix} 2x & -1 \\ 1 & -1\end{pmatrix}$$ which has trace $2x-1$ and determinant $1-2x$.

  • At $(0,0)$, with trace $-1$ and determinant $1$, there is a stable spiral.
  • At $(1,1)$, with trace $1$ and determinant $-1$, there is a saddle point.

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