I'm looking for a specific solution of:
$$z^2 w'' + z w' + (z^2 - \nu^2) w = a^2$$
where a is a constant (with a = 0 this would be the standard form for the integer order Bessel equation). The a = 0 case has solution $w(z) = \alpha J_\nu(z) + \beta Y_\nu(z)$.
Does anybody know of a specific solution to this non-homogeneous equation, or a transformation or technique that can be used to find such a solution?
It's not pretty, but I tried a series solution
$f(z) = \sum_{k=0}^\infty c_k z^k$
Presuming I didn't make any mistakes, that simplified to:
$f(z) = -\frac{a^2}{2 \nu} \sum_{k = 0}^\infty \frac{ \Gamma(\nu/2 - 2 k)}{16^k \Gamma(\nu/2 + 2 k + 1)} z^{2k}$