# Bessels Equation J(κx) instead of J(x)

1. Oct 19, 2011

### gmanmtb

Hello all, I seem to be stuck trying to prove this one, I'm not sure if I'm headed down the right path or just missing something obvious

1. The problem statement, all variables and given/known data
The equation x2y''+xy'+(κ2x2-p2)y=0, where κ constant, appears frequently in applications. Prove that J(κx) is a solution by changing the form of Bessel's equation of order p. Do this by letting $\widetilde{x}$=κx in the above equation and apply the chain rule.

2. Relevant equations
Well the first solution of Bessel's equation is
Jp(x)=$\sum^{\infty}_{k=0}\frac{(-1)^{k}}{k!\Gamma(k+p+1)}(\frac{x}{2})^{2k+p}$

3. The attempt at a solution
I believe the method I should be using is to use y(x)=Jp(κx)=Jp($\widetilde{x}$) but from there I can derive y'=$\sum^{\infty}_{k=1}\frac{(-1)^{k}(2k+p)}{k!\Gamma(k+p+1)}(\frac{\widetilde{x}}{2})^{2k+p-1}$ and y''=$\sum^{\infty}_{k=2}\frac{(-1)^{k}(2k+p)(2k+p-1)}{k!\Gamma(k+p+1)}(\frac{\widetilde{x}}{2})^{2k+p-2}$ but from there I get fuzzy, I can't replace $\widetilde{x}$=κx in a way that works out.

Last edited: Oct 19, 2011
2. Oct 20, 2011

### jackmell

No. Make a change of indepedent variable and then solve for y in terms of $\overline{x}$

For example, if $\overline{x}=kx$, then:

$$\frac{dy}{d\overline{x}}=\frac{dy}{dx}\frac{1}{k}$$

solve for dy/dx, then compute the second derivative in terms of $\overline{x}$, solve for d^2y/dx^2, then substitute those into the original equation to obtain an equation in $y(\overline{x})$ for which $J(\overline{x})$ is a solution, that is, $J(kx)$.