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Solve Bessel's equation through certain substitutions

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data, relevant equation
    [tex] x^{2}y'' + xy' + (4x^{4}-\frac{1}{4})y = 0[/tex]

    2. The attempt at a solution
    I tried substituting z = x2

    From this I have [tex]\frac{dy}{dx} = 2x \frac{dy}{dz}[/tex]
    and [tex]\frac{d}{dx}(2x\frac{dy}{dz}) = 2xy'' + 2y'[/tex]

    Then the original equation becomes:
    [tex]2z^{3/2}y'' + 4zy' + (4z^{2}-\frac{1}{4})y = 0[/tex]

    where derivatives of y are now with respect to the new variable z.
    This does not look like a Bessel equation and I'm not sure how to make it look like one. Did I use the wrong substitution?

    I know how to solve once it's in the correct form, but could someone help me get it there please?
     
  2. jcsd
  3. Oct 5, 2009 #2
    try [tex]z^{2}=4x^{4}[/tex]
     
  4. Oct 5, 2009 #3
    If I use z2 = 4x4 I get:
    z = 2x2 and dz/dx = z' = 4x

    [tex] (\frac{dy}{dx}) = \frac{dy}{dz} \frac{dz}{dx} = 4x \frac{dy}{dz}[/tex]

    Then

    [tex]\frac{d}{dx}(4x\frac{dy}{dz})=4\frac{dy}{dz}+\frac{d}{dz}(\frac{dy}{dz})\frac{dz}{dx}=4\frac{dy}{dz}+4x\frac{d^{2}y}{dz^{2}}[/tex]

    Plugging in:
    x2(4y' + 4xy'') + x(4xy') + (4x4 - 1/4)y = 0
    if z2 = 4x4
    (z/2)3/2 y'' + zy' + (z2/4 - 1/16)y = 0
    so it still doesn't work. The x3 term messes the whole thing up....
     
    Last edited: Oct 5, 2009
  5. Oct 6, 2009 #4
    Never mind, I got it. Thanks for your help!
     
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