Bessel's Equation and substitutions

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Homework Help Overview

The discussion revolves around solving a second-order differential equation related to Bessel's equation, specifically focusing on finding a general solution in terms of Bessel functions Jv and Yv. The original poster introduces a substitution of variables, indicating a transformation from x to z, where z = x².

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the substitution z = x² to compute derivatives but encounters difficulties with the second derivative. They express uncertainty about how to compute the derivative of dy/dz with respect to x. Another participant provides clarification on the chain rule, suggesting a method for computing the derivative.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the substitution and the necessary derivatives. Clarifications have been offered regarding the application of the chain rule, but no consensus or complete solution has been reached yet.

Contextual Notes

The original poster mentions needing to show details of their work, indicating a requirement for thoroughness in their approach. There is also a hint of confusion regarding the use of Bessel functions, particularly the interchangeability of Jv and Yv.

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Homework Statement


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)

Homework Equations



All given in part 1.

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.
 
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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]
 
Thank you!
 
i am also struggling with how I am going to compute
 

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