Solve Bessel's equation through certain substitutions


by ILikeMath
Tags: bessel, function, substitution
ILikeMath
ILikeMath is offline
#1
Oct5-09, 06:55 PM
P: 5
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?
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206PiruBlood
206PiruBlood is offline
#2
Oct5-09, 07:45 PM
P: 109
try [tex]z^{2}=4x^{4}[/tex]
ILikeMath
ILikeMath is offline
#3
Oct5-09, 08:49 PM
P: 5
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....

ILikeMath
ILikeMath is offline
#4
Oct6-09, 06:14 PM
P: 5

Solve Bessel's equation through certain substitutions


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


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