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Math Help - Bessel Equation

  1. #1
    Junior Member
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    Sep 2008
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    Question Bessel Equation

    I don't know how to solve the following problem!!

    Prove that the general solution of,

    y'' + \frac{1-2a}{x}y' + \left[ (bcx^{c-1})^{2} = \frac{a^{2}-n^{2}c^{2}}{x^{2}}\right]y =0

    is given by y=x^{a}[c_{1}J_{n}(bx^{c}) + c_{2}N_{n}(bx^{c}) ]

    Use this result to solve y''+xy=0

    Where do I start?? what substitution do I make to make it look like a bessel equation???? Any suggestion?
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  2. #2
    Member
    Joined
    May 2009
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    Firstly I'm assuming you meant a - inplace of the equals sign in the brackets. Also, naturally, I'll assume that a, b and c are all constants.
    To do this you simply use the substitution  y = x^a v(u) where u=b x^c to transform the equation in terms of u, the function v(u) and its derivatives wrt u. So to start you off you should find that

    y^{\prime} = ax^{a-1} \, v + x^a \, \frac{dv}{du} \cdot \frac{du}{dx}

    and, using Leibniz's differentiation of products rule),

    y^{\prime \prime} = a(a-1)x^{a-2} \, v + 2 a x^{a-1}\, \frac{dv}{du} \cdot \frac{du}{dx}  + x^a \left( \left(\frac{du}{dx} \right)^2 \cdot \frac{d^2 v}{du^2} + \frac{dv}{du} \cdot \frac{d^2 u}{dx^2} \right) .



    Further to this you only need to substitute for the derivatives of u wrt x using:

    \frac{du}{dx} = bc x^{c-1} and \frac{d^2 u}{dx^2} = bc(c-1) x^{c-2} .


    Assuming your original equation was correct this should do the trick transforming it into the Bessel equation. Hope that helped!
    Last edited by the_doc; May 10th 2009 at 07:15 AM.
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