Consider the linear system

$$\frac{dx}{dt}= -3x+2y, \frac{dy}{dt}= ax+6y, a \neq -9$$

classify the fixed point at the origin?

Is the correct approach to investigate the steady states and how these points will change according to the point a, so consider the range from -$\infty$ to $\infty$.

Many thanks in advance.

  • $\begingroup$ yes, it is correct. Your only steady state is the origin, and the matrix is non-singular (due to the assumption on $a$), therefore, calculating the eigenvalues of your matrix will do. $\endgroup$ – Artem May 14 '12 at 9:00
  • $\begingroup$ @Artem i get the following for the eigenvalues though, $\lambda^{2}-3\lambda-2a=0$ ? $\endgroup$ – user24930 May 14 '12 at 9:07
  • $\begingroup$ Nope, you missed something :) $\endgroup$ – Artem May 14 '12 at 9:11
  • $\begingroup$ @Artem can't see it :/ $\endgroup$ – user24930 May 14 '12 at 9:22
  • $\begingroup$ it should be $\lambda^2-3\lambda-2a-18=0$. Now you need to calculate the eigenvalues, and, depending on the found values, classify the fixed point at the origin. $\endgroup$ – Artem May 14 '12 at 9:24

It's not necessary to find eigenvalues explicitly: for $2\times 2$ matrices, the trace and determinant give enough information. Here $$\mathrm{tr} = -3,\quad \det = -18-2a$$ As $a$ moves from $-\infty$ to $\infty$, the matris moves down on the trace-determinant plane, from the region of stable spirals ($\mathrm{tr}^2-4\det<0$) to stable nodes ($\mathrm{tr}^2-4\det\ge 0$, $\det>0$) and finally to saddles ($\det<0$). We skip the inconvenient value $a=-9$, which would yield a non-isolated equilibrium ($\det=0$).

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