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

  1. Aug 9, 2007 #1
    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.
     
  2. jcsd
  3. Aug 9, 2007 #2

    HallsofIvy

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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!
     
  4. Aug 9, 2007 #3
    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.
     
  5. Aug 9, 2007 #4

    mathwonk

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    it looks second order and not being polynomial in the derivatives has no finite degree.
     
  6. Aug 18, 2007 #5
    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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