Register to reply

Solve Bessel's equation through certain substitutions

by ILikeMath
Tags: bessel, function, substitution
Share this thread:
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?
Phys.Org News Partner Science news on
Suddenly, the sun is eerily quiet: Where did the sunspots go?
'Moral victories' might spare you from losing again
Mammoth and mastodon behavior was less roam, more stay at home
Oct5-09, 07:45 PM
P: 110
try [tex]z^{2}=4x^{4}[/tex]
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]


[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....

Oct6-09, 06:14 PM
P: 5
Solve Bessel's equation through certain substitutions

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

Register to reply

Related Discussions
Bessel's Equation Solution Proof Differential Equations 16
Differential equation reducible to Bessel's Equation Differential Equations 9
Bessel's equation Differential Equations 1
Integration - Trig Substitutions (solve for Y) Calculus & Beyond Homework 5
S.E in cylinder - Bessel's equation Introductory Physics Homework 5