A Bessel's functions of the second kind (Neumann' functions) deduction

[Rulonegger]

Homework Statement

I need to obtain the Bessel functions of the second kind, from the expressions of the Bessel functions of the first kind.

Homework Equations

Laplace equation in circular cylindrical coordinates reads
\nabla^2\phi(\rho,\varphi,z)=0 with \nabla^2=\frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho \frac{\partial}{\partial \rho} \right)+\frac{1}{\rho^2}\frac{\partial^2}{\partial \varphi^2}+\frac{\partial^2}{\partial z^2}

The Attempt at a Solution

Supposing that \phi(\rho,\varphi,z)=R(\rho)\Phi(\varphi)Z(z) i get \begin{eqnarray} Z_{m}(z)=A_{1_m}e^{m z}+A_{2_m}e^{-m z} && && \Phi_{k}(\varphi)=A_{3_k}\cos{k \varphi}+A_{4_k}\sin{k\varphi} \end{eqnarray} with m and k being the eigenvalues from the ODE's associated with Z and \Phi, respectively.
Then the associated ODE for the radial component is \frac{d^2R}{d(k\rho)^2}+\frac{1}{k\rho}\frac{dR}{d(k\rho)}+\left(1-\frac{m^2}{(k\rho)^2}\right)R=0 and defining x=k\rho i get \frac{d^2R}{d x^2}+\frac{1}{x}\frac{dR}{d x}+\left(1-\frac{m^2}{x^2}\right)R=0 which is Bessel's differential equation. Using Frobenius method to remove the singularity, I'm able to compute Bessel functions of the first kind J_{m}(x)=\sum_{s=0}^{\infty}{\frac{(-1)^s}{s!(m+s)!}\left(\frac{x}{2}\right)^{m+2s}} for m integer. Changing n by -n and removing the zero terms in the series for J_{-m}(x), we see that
J_{-m}(x)=(-1)^{m}J_{m}(x) so both functions are linearly dependent, and we need to specify another function of x linearly independent of J_{m}(x) because the differential equation is a second order one.
Then, a common way to find a second function of a differential equation of the form y''(x)+p(x)y'(x)+q(x)y(x)=0 with a known solution y_{1}(x), we can suppose that y_{2}(x)=y_{1}(x)g(x). After deriving y_{2} and substituting in the differential equation, we get
\frac{g''}{g'}=-\frac{y_{1}'}{y_{1}}-p
but if i try to use this method to find the Neumann's functions Y_{m}(x), i cannot find the expression
Y_{m}(x)=\frac{\cos{mx}J_{m}(x)-J_{-m}(x)}{\sin{mx}} which commonly one finds that it is a "definition" for the Neumann's functions. Any ideas?

 

[Rulonegger]

Correction

I'm sorry, but the last expression was Y_{m}(x)=\frac{\cos{m\pi}J_{m}(x)-J_{-m}(x)}{\sin{m\pi}} Any kind of help would be greatly appreciated!