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Differential Eqns. (checking an answer + some questions)

  1. Dec 14, 2004 #1
    Any help is highly appreciated.

    Thank you

    Problem

    [tex] y^{\prime \prime} = xy [/tex]

    My Solution

    If

    [tex] y = \sum _{n=0} ^{\infty} c_n x^n \Longrightarrow xy = \sum _{n=0} ^{\infty} c_n x^{n+1} = \sum _{n=1} ^{\infty} c_{n-1} x^n [/tex]

    and

    [tex] y^{\prime \prime} = \sum _{n=2} ^{\infty} \left( n-1 \right) n c_n x^{n-2} = \sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n} [/tex]

    Then

    [tex] \sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n} = \sum _{n=1} ^{\infty} c_{n-1} x^n [/tex]

    Hence, we may find the recursion relation

    [tex] c_{n+2} = \frac{c_{n-1}}{\left( n+1 \right) \left( n+2 \right)} \qquad n=1,2,3,\ldots [/tex]

    and so we have:

    [tex] c_0 [/tex]

    [tex] c_1 [/tex]

    [tex] c_2 [/tex]

    [tex] c_3 = \frac{c_0}{2\cdot 3} [/tex]

    [tex] c_4 = \frac{c_1}{3\cdot 4} [/tex]

    [tex] c_5 = \frac{c_2}{4\cdot 5} [/tex]

    [tex] c_6 = \frac{c_3}{5\cdot 6} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6} [/tex]

    [tex] c_7 = \frac{c_4}{6\cdot 7} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7} [/tex]

    [tex] c_8 = \frac{c_5}{7\cdot 8} = \frac{c_2}{4\cdot 5\cdot 7\cdot 8} [/tex]

    [tex] \vdots [/tex]

    Well, since this is a second-degree differential equation, we must have [tex] c_2 = 0 [/tex]. It also follows that

    [tex] c_{3n} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \qquad n=1,2,3,\ldots [/tex]

    and

    [tex] c_{3n+1} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \qquad n=1,2,3,\ldots [/tex]

    Therefore

    [tex] y = c_0 + c_1 x + c_0 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n}}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \right] + c_1 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n+1}}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \right] [/tex]

    Questions

    1. Have I got it right? :smile:

    2. If so, can I write the denominators in a more elegant way? I mean:

    [tex] 2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 2\cdot 5\cdot 8\cdot \cdots \cdot \left( 3n-1 \right) \right] = n!3^n \times ( ? ) [/tex]

    and

    [tex] 3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n+1 \right) \right] = n!3^n \times ( ? ) [/tex]

    where [tex] ( ? ) [/tex] represents the other factor which I haven't been able to figure out so far.
     
  2. jcsd
  3. Dec 14, 2004 #2

    Tide

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    Homework Helper

    You essentially have it! With a little reworking you can put it into the standard form of the Airy functions (your equation is Airy's DE which describes waves travelling in a medium with a linear variation in density or water waves with linear variation of the depth, e.g.).

    It's a bit much to LaTeX so you can just look here for details: http://functions.wolfram.com/BesselAiryStruveFunctions/AiryAi/06/01/01/
     
  4. Dec 14, 2004 #3
    Thanks for your input. I've just checked out that website. The formulas there are a bit scary, but then I looked up a problem in my calc book about the Airy function. It looks simpler:

    [tex] A(x) = 1 + \frac{x^3}{2\cdot 3} + \frac{x^6}{2\cdot 3\cdot 5\cdot 6} + \frac{x^9}{2\cdot 3\cdot 5\cdot 6\cdot 8\cdot 9} +\cdots [/tex]

    Anyway, it seems to me that I get to [tex]A(x)[/tex] from

    [tex] y = c_0 + c_1 x + c_0 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n}}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \right] + c_1 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n+1}}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \right] [/tex]

    if I let [tex]c_0 = 1[/tex] and [tex]c_1 = 0[/tex].

    Regards
     
  5. Dec 14, 2004 #4

    Tide

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    That works! By the way, the functions are usually indicated as Ai(x) and Bi(x) [not A(x) and B(x)] with Ai(x) going to 0 as x -> infinity while Bi(x) diverges in that limit.
     
  6. Dec 14, 2004 #5
    Let me see how that works...

    We get [tex]A_i (x)[/tex] if [tex]c_0 = 1[/tex] and [tex]c_1 = 0[/tex]. Conversely, [tex]B_i (x)[/tex] is obtained if [tex]c_0 = 0[/tex] and [tex]c_1 = 1[/tex].
     
  7. Dec 14, 2004 #6

    Tide

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    Actually, no. The Ai and Bi are combinations of the two infinite series you wrote. One of those grows and the other one decays as x -> +infinity. I misspoke (miswrote) eariler when I compared your A(x) with Ai(x) - they are not the same.
     
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