Linearizing a differential equation

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SUMMARY

The discussion centers on the linearization of the differential equation (1 + 4k²ρ²)ẋ = ρω² - 4k²ρẋ² - 2gkρ, specifically in the context of problem 7.41 from Taylor's Classical Mechanics. The goal is to determine the stability of the equilibrium position ρ = 0. The linearized form of the equation is derived as ẋ = (ω² - 2gk)ρ, with the justification for neglecting the term 4k²ρẋ² based on the assumption that at equilibrium, both ρ and ẋ are approximately zero, rendering the term negligible.

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  • Understanding of differential equations and their linearization
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  • Knowledge of Taylor series expansions for approximating functions
  • Basic calculus, including derivatives and second derivatives
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eep
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This question popped up while trying to solve problem 7.41 from Taylor's Classical Mechanics book.

Basically we have the differential equation

[tex](1+4k^2\rho^2)\ddot{\rho} = \rho \omega^2 - 4k^2\rho{\dot{\rho}}^2-2gk\rho[/tex]

and we are looking if the equilibrium position [tex]\rho = 0[/tex] is stable.

After "linearizing" this you are supposed to end up with

[tex]\ddot{\rho} = (\omega^2 - 2gk)\rho[/tex]

but I really don't understand the correct way to arrive at this. Particularly, how are you supposed to justify dropping the [tex]4k^2\rho{\dot{\rho}}^2[/tex] term?
 
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eep said:
This question popped up while trying to solve problem 7.41 from Taylor's Classical Mechanics book.

Basically we have the differential equation

[tex](1+4k^2\rho^2)\ddot{\rho} = \rho \omega^2 - 4k^2\rho{\dot{\rho}}^2-2gk\rho[/tex]

and we are looking if the equilibrium position [tex]\rho = 0[/tex] is stable.

After "linearizing" this you are supposed to end up with

[tex]\ddot{\rho} = (\omega^2 - 2gk)\rho[/tex]

but I really don't understand the correct way to arrive at this. Particularly, how are you supposed to justify dropping the [tex]4k^2\rho{\dot{\rho}}^2[/tex] term?
My guess is that at the equilibrium position, ρ ≈ 0, and [itex]\dot{\rho}[/itex] is either zero or close to it, which would make [itex]\dot{\rho}^2[/itex] negligible.
 

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