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?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

Loading...

Similar Threads - Translating linear system | Date |
---|---|

B Need some help with understanding linear approximations | Feb 17, 2018 |

I Integrating scaled and translated indicator function | Nov 20, 2017 |

I Taylor expansion of f(x+a) | Nov 1, 2017 |

I Generalizing the translation operator | Jan 25, 2017 |

**Physics Forums - The Fusion of Science and Community**