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?