Linearizing \dot{x}+√x = 0 when the derivative is undefined at equilibrium

FaroukSchw · Apr 9, 2012

Hello ,
I am trying to linearize [itex]\dot{x}[/itex]+√x = 0. The only equilibrium point is at x=0; but the derivative is not defined at this point. Does anybody have a suggestion?
Regards.

Office_Shredder · Apr 9, 2012

Why are you trying to linearize it?

FaroukSchw · Apr 9, 2012

hello Office_Shredder
I just got interested in it when talking to a friend about problems that we can encounter

HallsofIvy · Apr 9, 2012

[itex]\sqrt{x}[/itex] cannot be linearized at x= 0 precisely because its derivative does not exist.

Linearizing \dot{x}+√x = 0 when the derivative is undefined at equilibrium

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Linearizing \dot{x}+√x = 0 when the derivative is undefined at equilibrium

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