So suppose you have a nice simple linear system(adsbygoogle = window.adsbygoogle || []).push({});

dx/dt = a_{1}x + b_{1}y + c_{1}

dy/dt = a_{2}x + b_{2}y + c_{2}

and a critical point (x_{0}, y_{0}) ≠ (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 c_{1}, c_{2}? 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?

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

Can you offer guidance or do you also need help?

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