Translating a linear system to have critical points at the origin?

[Fractal20]

So suppose you have a nice simple linear system
dx/dt = a₁x + b₁y + c₁
dy/dt = a₂x + b₂y + c₂

and a critical point (x₀, y₀) ≠ (0,0). If you want to analyze the stability in the usual linear method, then I assume there should be some transformation to have the critical point at the origin. So is this system equivalent to the system that results from simply removing c₁, c₂? I see that in terms of taking the Jacobian, the constants don't matter at all. Furthermore, I also see that if there is no b's in either equation of a line, then of course they can only intersect at (0,0) -> that is a critical point as well. So I guess what I am asking is if the linear system for a given a1,a2,b1,b2 but and c1,c2 are all equivalent? Thanks!

*edit
Okay, I thought about it more and realized that substituting x' = x - x_0 and y' = y - y_0 where x_0, y_0 are the critical points. So can one just ignore the constants or is this intermediate step needed for rigor?

 

[jvicens]

I would like to clarify that the system resulting from removing the constants c1 and c2 is not necessarily equivalent to the original system. While it may be mathematically equivalent in terms of the Jacobian, it is not necessarily equivalent in terms of the behavior and stability of the system.

The constants c1 and c2 represent the influence of external factors on the system, and removing them may change the dynamics of the system. In some cases, the system may still behave similarly, but in others, the behavior may be drastically different.

Therefore, it is important to consider the constants in the analysis of the stability of the system. The intermediate step of substituting x' and y' is a valid approach, as it helps to transform the system to a more manageable form, but it is not necessary for rigor. The key is to understand the impact of the constants on the behavior of the system and to consider them in the analysis.