Is my charastic equation for y''-4y' +5y =0 correct?

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Discussion Overview

The discussion revolves around the correct formulation of the characteristic equation for the differential equation y'' - 4y' + 5y = 0. Participants explore the definition and derivation of the characteristic equation, as well as the correct notation and terminology associated with it.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant attempts to present a solution involving e^{2t}(C_1\cos{t} + C_2\sin{t}), which is challenged by others as incorrect.
  • Another participant asserts that the initial response is not even an equation and suggests consulting a textbook for clarification on the characteristic equation.
  • A participant clarifies that the characteristic equation is also referred to as the auxiliary equation.
  • One participant proposes the characteristic equation as x^2 - 4x + 5, explaining the process of substituting y = e^(rt) and performing derivatives to derive the characteristic equation.

Areas of Agreement / Disagreement

Participants do not agree on the correctness of the initial formulation of the characteristic equation, with multiple competing views on what constitutes the correct expression and terminology.

Contextual Notes

There are unresolved issues regarding the definitions used and the steps taken to derive the characteristic equation, which may depend on the interpretation of terms like "characteristic equation" and "auxiliary equation."

RadiationX
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I'm trying to rewrite my notes from my differential equations class last week. my proffesor asked us to write the charastic equation of this:[tex]y''-4y' +5y =0[/tex] is this correct?
[tex]e^{2t}(C_1\cos{t} + C_2\sin{t})[/tex]
 
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No, it's not- it's not even an equation! I would suggest you look up "characteristic equation" in your textbook.
 
HallsofIvy said:
No, it's not- it's not even an equation! I would suggest you look up "characteristic equation" in your textbook.


i meant this.[tex]y=e^{2t}(C_1\cos{t} + C_2\sin{t})[/tex]
 
The characteristic equation is what you elsewhere call the auxiliary equation, if I'm not mistaken.
 
x^2 - 4x + 5, i.e. make every derivative a power, so you get x^2 -4x + 5x^0

or just substitute y = e^(rt) then perform the derivatives in the given differential equation and you get the characteristic equation out front in terms of the variable r this time.
 

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