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Asymptotic mathcing for a first order differential equation

  1. Jul 31, 2006 #1
    The first-order differential equation
    [tex]y' +(ex^2+1+1/x^2)y=0[/tex], with boudary value y(1) =1

    Using, asymptotic mathcing to study the behaviour of the sltion as e tends to +0, when x is not too large, the term [tex]ex^2 [/tex] is negligible so an approximate equation for y is
    [tex] y'_L +(1+1/x^2)y_L=0 [/tex].

    When x is large, [ tex ]ex^2 [ /tex ] is not negligible but [tex]1/x^2 [/tex] is. Therefore, an approximate equation valid as x tends to infinity is [tex] y'_R +(ex^2+1)y_R=0 [/tex].

    I think that te upper edge of the left region would be the largest value of x for which ex^2 is still small compared with 1. This would suggest that the left region consists of those x for which x<<e^-(1/2) as e tends to +0. But actually the region of validity of the left solution is e^(-1/3) (e tends to +0)....Can anyone explain this to me??

    Sams as the right region...Please kindly help
     
    Last edited: Aug 1, 2006
  2. jcsd
  3. Jul 31, 2006 #2

    jim mcnamara

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    I'm confused - [tex]e[/tex] is usually a constant
    [tex]e~=2.71828182845904523536028747135266249775724709369995...[/tex]
     
  4. Jul 31, 2006 #3
    sorry, I am referring e to be a small pertubation introduced...
    It would better to use a Greek word for it...
    but I have forgotten the word...
     
  5. Jul 31, 2006 #4

    HallsofIvy

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    e= epsilon: [itex]\epsilon[/itex]
     
  6. Jul 31, 2006 #5
    yes, HallsofIvy! Thank you.

    Do you mind helping me for this question?
     
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