HELP!! differential equation and bessels function

**panda** · June 11th 2008, 07:54 AM

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!!!!!!

**CaptainBlack** · June 11th 2008, 08:54 AM

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

RonL

**panda** · June 11th 2008, 09:49 AM

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'?

**CaptainBlack** · June 11th 2008, 10:08 AM

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

**panda** · June 14th 2008, 06:58 AM

using

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.

**CaptainBlack** · June 14th 2008, 09:35 AM

Originally Posted by panda

using

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

RonL

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