Solve Bessel's equation through certain substitutions

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Homework Help Overview

The discussion revolves around solving Bessel's equation represented by the differential equation x²y'' + xy' + (4x⁴ - 1/4)y = 0. Participants are exploring substitutions to transform this equation into a recognizable form.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts a substitution of z = x², leading to a transformed equation. They express uncertainty about whether this substitution is appropriate for obtaining a Bessel equation form. Other participants suggest an alternative substitution of z² = 4x⁴, but encounter complications with the resulting terms.

Discussion Status

Participants are actively engaging with the problem, trying different substitutions and discussing their implications. There is a sense of exploration as they work through the transformations, although the original poster indicates they have resolved their confusion by the end of the discussion.

Contextual Notes

There is an indication of difficulty in achieving the desired form of the equation, with specific attention to the complications introduced by the substitutions attempted. The original poster's final comment suggests a resolution, but the details of that resolution are not provided.

ILikeMath
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1. Homework Statement , relevant equation
x^{2}y'' + xy' + (4x^{4}-\frac{1}{4})y = 0

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

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

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

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?
 
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try z^{2}=4x^{4}
 
If I use z2 = 4x4 I get:
z = 2x2 and dz/dx = z' = 4x

(\frac{dy}{dx}) = \frac{dy}{dz} \frac{dz}{dx} = 4x \frac{dy}{dz}

Then

\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}}

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:
Never mind, I got it. Thanks for your help!
 

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