# HELP!! differential equation and bessels function

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• Jun 11th 2008, 07:54 AM
panda
HELP!! differential equation and bessels function
Hi,
I'm studying for my final exam at the moment and i found this question from the past exam paper. but i have no idea how to solve it.
can anyone please please help me!!!

thanks!!!!!!
• Jun 11th 2008, 08:54 AM
CaptainBlack
Quote:

Originally Posted by panda
Hi,
I'm studying for my final exam at the moment and i found this question from the past exam paper. but i have no idea how to solve it.
can anyone please please help me!!!

thanks!!!!!!

Make the change of variable $\displaystyle y'=y/x^{\alpha}$ and rewrite the de in terms of $\displaystyle y'$ and $\displaystyle x$. Then put $\displaystyle x'=\beta x^{\gamma}$ and you should find that $\displaystyle x'$ and $\displaystyle y'$ satisfy the Bessel equation of order $\displaystyle n$.

RonL
• Jun 11th 2008, 09:49 AM
panda
thanks for the reply
just one more question.
how do you know which variable to use??
for instance, how do you choose the 'change of variable'?
• Jun 11th 2008, 10:08 AM
CaptainBlack
Quote:

Originally Posted by panda
thanks for the reply
just one more question.
how do you know which variable to use??
for instance, how do you choose the 'change of variable'?

By comparing the Bessel equation with that given, and more directly looking at the form of the solution that you are to demonstrate. (in fact the suggested change of variables comes directly from the suggested solution)

RonL
• Jun 14th 2008, 06:58 AM
panda
using http://www.mathhelpforum.com/math-he...1ef959be-1.gif
i got
(y'x^2 - yxalpha)x^-alpha + yxx^-alpha - (2alphaxy)x^-alpha + beta^2.gemma^2.x^2gamma + alpha^2 -n^2.gemma^2)y = 0

is this right???? and i wasnt sure how to use x'=beta.x^gemma from here.
• Jun 14th 2008, 09:35 AM
CaptainBlack
Quote:

Originally Posted by panda
using http://www.mathhelpforum.com/math-he...1ef959be-1.gif
i got
(y'x^2 - yxalpha)x^-alpha + yxx^-alpha - (2alphaxy)x^-alpha + beta^2.gemma^2.x^2gamma + alpha^2 -n^2.gemma^2)y = 0

is this right???? and i wasnt sure how to use x'=beta.x^gemma from here.

You do know that the ' do not denote differentiation here don't you?

Try instead $\displaystyle u(x)=y(x)/x^{\alpha}$ first, from this compute $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ in terms of $\displaystyle x$, $\displaystyle u$, $\displaystyle \frac{du}{dx}$ and $\displaystyle \frac{d^2u}{dx^2}$ and substitute into the DE.

RonL