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).