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.
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 $(31,-22)$).