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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.

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    This is correct. You can post the details o$f$ your c$a$lculations as an answer and accept it in some time.2012-12-01

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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.