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Bessel's Equation and substitutions

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a general solution in terms Jv of and Yv . Indicate
    whether you could also J-v use instead of Yv. Use the
    indicated substitution. Show the details of your work.

    9x2y''+9xy'+(36x4-16)y=0

    Substitution (z=x2)

    2. Relevant equations

    All given in part 1.

    3. The attempt at a solution

    Given z=x2, [itex]\frac{dz}{dx}[/itex]=2x
    Therefore [itex]\frac{dy}{dx}[/itex]=[itex]\frac{dy}{dz}[/itex]*[itex]\frac{dz}{dx}[/itex]=2x*[itex]\frac{dy}{dz}[/itex]

    But I need the second derivative of y with respect to x to make the substitution, this is where I run into trouble. Using the chain rule, I get this:
    [itex]\frac{d^{2}y}{dx^{2}}[/itex]=2*[itex]\frac{dy}{dz}[/itex]+[itex]\frac{d}{dx}[/itex]([itex]\frac{dy}{dz}[/itex])*2x

    I have no clue how to compute [itex]\frac{d}{dx}[/itex]([itex]\frac{dy}{dz}[/itex])
    Any help would be appreciated.
     
  2. jcsd
  3. Nov 9, 2013 #2

    HallsofIvy

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    For any function [itex]\phi[/itex], [itex]\frac{d\phi}{dx}= \frac{d\phi}{dz}\frac{dz}{dx}[/itex]. In particular, if [itex]\phi= \frac{dy}{dz}[/itex] then [itex]\frac{d}{dx}\left(\frac{dy}{dz}\right)= \frac{d^2y}{dz^2}\left(\frac{dz}{dx}\right)[/itex]
     
  4. Nov 9, 2013 #3
    Thank you!
     
  5. Feb 17, 2016 #4
    i am also struggling with how im going to compute
     
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