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

[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

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

I believe the method I should be using is to use y(x)=J_p(κx)=J_p($\widetilde{x}$) 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.

 

[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, $J(kx)$.