# Thread: HELP!! differential equation and bessels function

1. ## 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.

thanks!!!!!!

2. 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.

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

just one more question.
how do you know which variable to use??
for instance, how do you choose the 'change of variable'?

4. Originally Posted by panda
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

5. 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.

6. 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 $\displaystyle u(x)=y(x)/x^{\alpha}$ 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