How are order and degree defined for this DE? (cos(y'') + xy' = 0)

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Discussion Overview

The discussion revolves around determining the order and degree of the differential equation \(\cos(y'') + xy' = 0\). Participants explore different methods of analysis, including the implications of expressing the equation in various forms and the challenges posed by non-polynomial characteristics.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that the equation is second order because \(y''\) is the highest derivative, but notes that it does not have a degree since it is not polynomial.
  • Another participant questions the claim that expressing \(y''\) as \(\cos^{-1}(-xy')\) results in a degree of one, suggesting that this interpretation reduces the equation to first degree.
  • A different viewpoint suggests linearizing the equation by approximating \(\cos(y'')\) as 1 for small \(y''\), leading to a simplified first order equation, while cautioning about the accuracy of this approach.
  • Some participants agree that the equation is second order and emphasize that its non-polynomial nature means it lacks a finite degree.
  • There is a suggestion that the equation can be treated as effectively first order by substituting \(y' = u(x)\), although this perspective is not universally accepted.

Areas of Agreement / Disagreement

Participants generally agree that the equation is second order and lacks a defined degree due to its non-polynomial nature. However, there are competing views on how to interpret the degree and the implications of linearization, indicating that the discussion remains unresolved.

Contextual Notes

The discussion highlights limitations in defining degree for non-polynomial differential equations and the potential inaccuracies introduced by linearization methods.

maverick280857
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Hi everyone

Consider the differential equation

[tex]\cos(y'') + xy' = 0[/tex]

How do you determine the order and degree of such a DE?

One way is to write

[tex]y'' = \cos^{-1}(-xy')[/tex]

and say that the order is 2 degree is 1.

But if I do not use the inverse cosine, and observe that the first member on the left hand side is a power series in [itex]y''[/itex], then the order is still 2, but the degree is not defined. What is the resolution to this problem?

PS--This is not homework.
 
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That is a second order equation because y" is the highest derivative.

It does NOT HAVE a degree because it is not polynomial.

I'm not sure why you say that [itex]y''= cos^{-1}(-xy')[/itex] has degree one. Is it because y'' is to first power? You can always solve a differential equation for the highest derivative so that way of looking at "degree" reduces any equation to first degree!
 
maverick280857 said:
Hi everyone

Consider the differential equation

[tex]\cos(y'') + xy' = 0[/tex]

How do you determine the order and degree of such a DE?
As mentioned about it does not have a degree.

One thing that physicists and engineers do in such a situation is linearize the differential equation. Since [tex]\cos (y'')\approx 1[/tex] for small [tex]y''[/tex]. And so we get,
[tex]1+xy'=0[/tex].

But we need to be careful about the accuray.
 
it looks second order and not being polynomial in the derivatives has no finite degree.
 
Yes, but it is effectively a first order equation, as all you have to do is let [tex]y^{\prime} = u(x)[/tex].
 

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