Using the Jacobean to Linearize at nonlinear system at an equilibrium point

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[@evanmiller4471]

Coming in clutch the night before a final. Thank you sir!

[@crazymonkeyVII]

this is exactly what I was looking for. Easy to understand and very enlightening. Thank you!

[@krewjaime4482]

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[@kartermarcus3565]

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[@matthewjames7513]

Thanks for the video! Can you please tell me what the key assumptions are behind the linearization theorem? When does it apply and when does it not apply?

[@sandraalvarenga359]

So simple! Thank you!

[@ThePritt12]

Hey:) I think thats not the way to linearize a system. You have to use the n-dim- taylor-Formula and not only the jacobian. You're loosing the constants. This way it works for determining stability but not for getting the linearized system in the right coordinate systems. Get my point?

[@brendan12882]

The first term vanishes to zero at a critical point. dy/dt=0 and dx/dt=0 are what defines a critical point. The H.O.T. also vanish if the system is nice. What is left is the Jacobian matrix multiplied by a small vector deviance from the critical point aka an appropriate linearization.

[@digges7516]

really nice video!

[@SouthernMagnoliaWheelCo]

This was extremely helpful! Thank you! I'm taking an online class at UM and it's more difficult to ask questions virtually. On example 3 the Jacobian evaluated at (0,0). I cannot seem to follow where the upper right term in the matrix comes from. I think it should be [0,1;-1,-1] instead of [0,-1;-1,-1]. Did I miss a step? Thanks again great video!!!

[@omkarsh5435]

Yes, its a fault, it should have been [0,1;-1,-1]

[@tahashaker1621]

Thank you so much

[@mungadze9102]

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[@dewanmohammedabdulahad5766]

How we consider the equilibrium point ? Is it random? one of your example you choose 4 equilibrium point, how it helps?

[@joantriadoaymerich]

The way is to fix derivatives at 0