Why is this eigenvalue problem solved by $\phi(r) = J_v(\alpha r)$

Question

I can't seem to see how the bessel function $J_v(ar)$ solves the problem. The eigenvalue problem has an $\alpha^ 2\phi(r)$ term

Ive tried writing $x = \alpha r$ in the expression for $J$ but cant seem to get it to work out

Answer 1

By substituting $\alpha r$ for $r$ in the first display we obtain $$ J''_{\nu}(\alpha r) + \frac{1}{\alpha r} J'_{\nu}(\alpha r) + \Bigr(1 - \frac{\nu^2}{\alpha^2 r^2}\Bigr) J_{\nu}(\alpha r) = 0, $$ so, after multiplying both sides by $\alpha^2$, $$ \alpha^2 J''_{\nu}(\alpha r) + \frac{\alpha}{r} J'_{\nu}(\alpha r) + \Bigr(\alpha^2 - \frac{\nu^2}{r^2}\Bigr) J_{\nu}(\alpha r) = 0. $$ But $\phi'(r) = \alpha J'_{\nu}(\alpha r)$ and $\phi''(r) = \alpha^2 J'_{\nu}(\alpha r)$.