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Bessel Differential Equation Problem

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Use the substitution [itex]x = e^t[/itex] to solve the following differential equation in terms
    of Bessel functions:

    [itex]\frac{d^{2}y}{dt^2} + (e^{2t} - \frac{1}{4})y = 0[/itex]

    2. Relevant equations



    3. The attempt at a solution


    So, using the Chain Rule, [itex]\frac{d^{2}y}{dt^2} = e^{2t}\frac{d^{2}y}{dx^2} = x^2\frac{d^{2}y}{dx^2}[/itex], so our differential equation becomes:[tex]x^2\frac{d^{2}y}{dx^2} + (x^{2} - \frac{1}{4})y = 0[/tex].
    The general solution is [itex]y = c_1J_{1/2}(x) + c_2J_{-1/2}(x)[/itex]. After that we need to replace x with [itex]e^t[/itex]. Is this correct?

    The second question asks to express our answer in terms of the elementary functions. What is exactly meant by this?
     
  2. jcsd
  3. Jun 1, 2012 #2
    Actually, using the chain rule would give (y(x))'=y'(x)*x'(t)
    a second application (y(x))''=(y'(x)*x'(t))=y''(x)*(x'(t))^2)+y'(x)*x''(t)
     
  4. Jun 1, 2012 #3
    elementary functions means answer it in terms of regular functions, hint sine and cosine (if this were a linear homogeneous equation, what would it look like.)
     
  5. Jun 1, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In general, the Bessel functions cannot be written in terms of elementary functions- which is why Bessel functions have whole books devoted to them! However, the Bessel functions of order 1/2 can be:
    [tex]J_{1/2}(x)= \sqrt{\frac{2}{\pi x}}sin(x)[/tex]
    [tex]J_{-1/2}(x)= \sqrt{\frac{2}{\pi x}}cos(x)[/tex]
     
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