Exercise :

Find the Stationary Points and then their kind and behavior, of the system :

$$x' = 7x + 10y + 3, y' = -5x -7y + 1$$

Attempt :

By solving the linear system :

$$\begin{cases} 7x+10y+3=0 \\ -5x-7y+1 =0 \end{cases}$$

we get that $A=(31,-22)$ is our stationary point, which is also unique.

Letting new variables be :

$$\begin{cases} y_1 = x-31 \\ y_2 = y + 22 \end{cases}$$

we get the system of equations :

$$\begin{cases} y_1'=7y_1 +10y_2 \\ y_2' =-5y_1 -7y_2 \end{cases}$$

Obviously, $O(0,0)$ is a stationary point for this system of equations.

The system's matrix has complex eigenvalues $λ=\pm i$, which leads us to the conclusion that $O(0,0)$ and thus the initial stationary point $A=(31,-22)$ and because $ay_1 = 7y_1$ which means $a>0$, the stationary point is a focus or spiral point which is unstable.

Is my solution correct ? Is there something that I am missing or something that I've done wrong ? I've had a bit of confusion making the new system through the stationary point but I think it's correct, since the initial constants get eliminated.

Please clarify me if my process and my conclusions have been carried out correctly.

  • 1
    $\begingroup$ Let me strongly suggest to review your notes. Eigenvalues $\pm i$ certainly do not signal a focus, stable or unstable, but a $______$... $\endgroup$
    – Did
    Commented Nov 13, 2017 at 14:27
  • $\begingroup$ @Did Considering the value of the coefficient of $y_1$ and $y_2$ though which is $7>0$ in the form of $ay_1$ and $-ay_2$ isn't there a conclusion to be made ? Otherwise, if I have misunderstood something, proceed to helping me if you can, I'd appreciate it. $\endgroup$
    – Rebellos
    Commented Nov 13, 2017 at 14:29
  • $\begingroup$ As I said... what do your notes say the solutions are when the eigenvalues are both purely imaginary? (This is helping you, so you can skip the naughtiness and turn to doing mathematics instead.) $\endgroup$
    – Did
    Commented Nov 13, 2017 at 14:35
  • $\begingroup$ @Did "When the eigenvalues of a matrix A are purely complex, as they are in this case, the trajectories of the solutions will be circles or ellipses that are centered at the origin. The only thing that we really need to concern ourselves with here are whether they are rotating in a clockwise or counterclockwise direction." I apologise, I was mixed up with another note correlation complex eigenvalues to coefficients of a differential system. By that, the eigenvalues I found suggest that the trajectiories of the solutions will be centered around the stationary points $O(0,0)$. $\endgroup$
    – Rebellos
    Commented Nov 13, 2017 at 14:40
  • $\begingroup$ @Did So, furthermore, the stationary point of the initial differential system $A=(31,-22)$ should be a center and the trajectories circles/ellipses ? $\endgroup$
    – Rebellos
    Commented Nov 13, 2017 at 14:42

1 Answer 1


Your stationary point and corresponding eigenvalues are correct. The eigenvalues $\lambda = \pm i$ suggest center point not focus/spiral. That means any orbit will be periodic around origin (or point $(31,-22)$).


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