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Differential equation reducible to Bessel's Equation

  1. Jan 18, 2008 #1
    How do I reduce an equation to Bessel's equation and find a general solution to it:

    For example how do I solve this:

    x^2y" + xy' + (4x^4 - 1/4)y = 0 (set x^2 = z)
     
  2. jcsd
  3. Jan 18, 2008 #2

    HallsofIvy

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    Okay, apparently you have been told to "set z= x2". Have you tried that at all? Do you know how to use the chain rule to convert d2y/dx2 to d2y/dz2?

    To start you off, dy/dx= (dy/dz)(dz/dx)= (2x)(dy/dz).

    Then d^2y/dx^2= d(2x dy/dz)/dx= 2 dy/dz+ d(dy/dx)/dx= 2 dy/dz +(d^2y/dz^2)(2x).
     
  4. Nov 2, 2008 #3
    Hi all,

    How do i reduce the eqn. below to Bessel's eqn. (how can i use the transformation when i have y, x, and u)

    4x^2y'' - 20xy' + (4x^2 + 35)y = 0 (y = (x^3).u)
     
  5. Nov 2, 2008 #4

    HallsofIvy

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    Do you mean to set y= x3u(x) where u is some unknown function? Then just DO it!

    If y= x3u, then y'= 3x2u+ x3u' and y"= 6xu+ 6x2u'+ x3u". Replace y, y' and y" in the equation with those and you get another differential equation for u rather than y.
     
  6. Nov 2, 2008 #5
    I was not sure about u is a function of x, should that be so? I am asking this is because i do not understand (unless u is a func. of x) why we use a third variable for reducing.

    I will try what you suggested, and thank you for your help.
     
  7. Nov 2, 2008 #6

    HallsofIvy

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    If u is NOT a function of x, what is it? A constant? If that were so, you would be asking how to write an equation in x and y, a function of x, in terms of x only, and that cannot be done.
     
  8. Nov 2, 2008 #7
    Yes, you are right. I just could not think simple, (may be it can be written as u(x) instead of u, for being more clear, but u is commonly used as a function of x, i should have remembered that),anyway, i reduced the ode to Bessel's eqn. Thanks for your help again.
     
  9. Jul 27, 2009 #8
    hey pips can you help me solve this one:

    xy'' - y' + y = 0....in terms of bessel functions
     
  10. Jul 30, 2009 #9
    Try this substitution:

    [tex]X=2\sqrt{x}[/tex]

    [tex]Y=\frac{y}{x}[/tex]
     
  11. Sep 23, 2010 #10
    Hey could you help me solve
    x^2*y''+2x*y'+x^2*y=0
     
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