A question in my textbook asks me to write down the general solution to:
$\frac{d}{dx}(x^2\frac{dR(x)}{dx}) + [k^2x^2 - n(n+1)]R(x) = 0$ in terms of Bessel functions.
Now two similar questions before this one were:
$\frac{d}{dx}(x^2\frac{dR}{dx}) + [x^2 - n(n+1)]R = 0$ which I found the solution to be $R(x) = AJ_{n+\frac{1}{2}}(x) + BY_{n+\frac{1}{2}}(x)$
and
$x^2\frac{d^2R(x)}{dx^2}) + x\frac{dR(x)}{dx} + [k^2x^2 - v^2]R(x) = 0$ which I found the solution to be $R = AJ_v(kx) + BY_v(kx)$
so would the solution to the question be: $R = AJ_{n+\frac{1}{2}}(kx) + BY_{n+\frac{1}{2}}(kx)$ ? and if not, where did I go wrong?
The differential equation \begin{align} D(x^{2} \ Dy) + [(kx)^{2} - n(n+1)]y = 0 \end{align} can be placed in familiar form by making the substitution \begin{align} y(x) = \frac{1}{\sqrt{x}} f(x). \end{align} It is seen that \begin{align} Dy(x) &= \frac{1}{\sqrt{x}} \left( Df - \frac{x}{2} f \right) \\ D^{2} y(x) &= \frac{1}{\sqrt{x}} \left( D^{2} f - \frac{1}{x} Df + \frac{3}{2 x^{2}} f \right) \end{align} for which the differential equation becomes \begin{align} x^{2} D^{2} f + x D f + [ (kx)^{2} - \left( n + \frac{1}{2}\right)^{2} ] f = 0. \end{align} The solution to this differential equation is given by \begin{align} f(x) = A J_{n+1/2}(kx) + B Y_{n+1/2}(kx). \end{align} Using this in the transformation made the solution for $y$ becomes \begin{align} y(x) = \frac{A}{\sqrt{x}} J_{n+1/2}(kx) + \frac{B}{\sqrt{x}} Y_{n+1/2}(kx). \end{align} Without loss of generality, this may be seen in the form \begin{align} y(x) = A \sqrt{ \frac{\pi}{2 x} } J_{n+1/2}(kx) + B \sqrt{ \frac{\pi}{2 x} } Y_{n+1/2}(kx). \end{align} These solutions are known as the spherical Bessel functions and are represented as, in solution form, \begin{align} y(x) = A j_{n}(kx) + B y_{n}(kx). \end{align}