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I've just been learning the definition of an isolated equilibrium point. From my understanding of this definition, I would expect (as an example) the point $x=1$ to be an isolated fixed point for the differential equation $\dfrac{\mathrm dx}{\mathrm dt} = x-1$. Is this correct?

BAI
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M Smith
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1 Answers1

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Yes, that is correct.

$1$ is an isolated equilibrium point as we can find a neighbourhood around it which does not contain any other equilibrium points (there's no other equilibrium points for the equation).

Skyhit2
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