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Homework Help: Airy equation and solution

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data

    So the Airy equation states that y''-xy'=0. My problem is to proof that the improper integral
    1/[tex]\pi[/tex] [tex]\int[/tex]cos(1/3*t^3+x*t) from 0 to [tex]\infty[/tex] satisfies this equation.

    I've tried differentiating under the integral sign, but all I've gotten is the integrand to be
    y''+ty = -1/pi [tex]\int[/tex] cos(1/3*t^3+x*t)(t^2+x) from zero to infinity. Naturally I do u substitution, but my final answer comes out to be -sin(infinity)/Pi

    What should I be doing?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 25, 2010 #2

    vela

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    After the substitution, you have

    [tex]y''-xy=\int_0^\infty \cos u\,du[/tex]

    That integral doesn't converge because the integrand doesn't vanish as [itex]u\rightarrow\infty[/itex]. The trick is to introduce an integrating factor [itex]e^{-\lambda u}[/itex] and then take the limit as [itex]\lambda\rightarrow 0^+[/itex].
     
  4. Apr 25, 2010 #3
    I'm sorry, but i don't quite understand what I do from there. If the integrand is cos u e^-k*u
    it'll converge, but do i find k and why you take the limit as k go to 0 from the right?

    this problem is supposed to be over the heads of students in our class b/c its extra credit and I'm a freshman in a sophomore class, so I don't know if I'm missing anything.
     
  5. Apr 25, 2010 #4

    vela

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    No, you don't find k; you're taking a limit as k goes to 0. The idea is that e-ku will go to 1 as k goes to 0, so the integrand is again just cos u. You have to approach from the right so that -ku<0, otherwise e-ku would go to positive infinity when you plug in the upper limit of the integral.
     
  6. Apr 25, 2010 #5
    ahhh, my answer works out to zero now. thank you.
     
  7. Apr 30, 2010 #6
    The only thing I dont get is, if I'm trying to do a proof of y''+xy=0 for a given equation, why am i allowed to put an integrating factor into the equation?? Wouldn't that show I was unable to prove what I wanted to prove?
     
  8. Apr 30, 2010 #7

    vela

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    I should have said "convergence factor," not "integrating factor."

    Honestly, I don't know the mathematical justification for when such factors are allowed and what their requirements are. It's just a technique that I occasionally ran across in physics, but the mathematical theory behind it was never explained.
     
  9. Apr 30, 2010 #8
    I'll make sure to read up on convergence factors once my tests are over. Thanks for the help!
     
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