Bessel function and its derivative in a boundary condition I have a function $u_{nj}=u_{nj}(r)$ of the radius variable $r$ (cylindrical coordinates) that satisfy the following Bessel equation
$$ r^{2} u''_{nj}+ r u'_{nj} +(k_{nj}^{2}-n^{2})u_{nj}=0, $$
where $n$ is integer, $j$ is the index denoting the $j$-th zero, and the solution must have finite (physical) amplitude everywhere within the domain inside a cylinder of radius $r=a$ (so typically this excludes second-kind Bessel functions and only allows the first kind, $J_{n}$).
This function ($u_{nj}$) also needs to satisfy the following boundary condition at $r=a$
$$ \left[ u_{nj} +(1+i)\beta_{0} a M u'_{nj} \right]_{r=a}=0,  $$
where $\beta_{0}, M, a$ are all constants, and we have $M<<1$ (small).
Somehow the result that satisfies both the equation and boundary condition above turns out to be (after taking first-order approximation in the small parameter $M$) given by
$$ u_{nj}=J_{n}(k_{nj} r), \ \ \ \ \text{where}  \ \ \ k_{nj}=\frac{V_{nj}}{a} [1-(1+i)\beta_{0} M],$$
and $V_{nj}$ denotes the $j$-th zero of the $J_{n}$ function.
I am having difficulty seeing how this result was reached. In particular, how can one attempt to solve an equation of the form
$$\left[ J_{nj} + (\text{constant}) J'_{nj} \right]_{r=a}=0,$$
to reach such result? Am I missing something trivial here?
Any help would be appreciated.
 A: From the condition
\begin{equation}
 u_{nj}(a) +(1+i)\beta_{0} a M u'_{nj}(a)=0
\end{equation}
with $u_{nj}=J_{n}(k_{nj} r)$ we must have
\begin{equation}
 J_{n}(k_{nj}a) +(1+i)\beta_{0} k_{nj}a M J'_{n}(k_{nj}a)=0 \tag{1}\label{1}
\end{equation}
For $M\ll1$, the first term dominates and thus $k_{nj}a\simeq V_{nj}$, where $V_{nj}$ is a zero of $J_n$. To go a step forward, we assume that
\begin{equation}
 k_{nj}a= V_{nj}+\kappa M+o(M)
\end{equation}
This expression and a first order Taylor expansion
\begin{align}
 J_{n}(k_{nj}a)&\sim J_n(V_{nj})+ \kappa M J'_n(V_{nj})\\
 &\sim\kappa M J'_n(V_{nj})
\end{align}
plugged into the condition (\ref{1}) gives,
\begin{equation}
 \kappa M J'_n(V_{nj})+(1+i)\beta_{0} \left(V_{nj}+\kappa M  \right) M J'_{n}(V_{nj}+\kappa M)=0
\end{equation}
or, by keeping the terms up to the first order in $M$,
\begin{equation}
 \kappa M J'_n(V_{nj})+(1+i)\beta_{0} V_{nj} M J'_{n}(V_{nj})=0
\end{equation}
which reads
\begin{equation}
  M J'_n(V_{nj})\left[\kappa+(1+i)\beta_{0}V_{nj}\right]=0
\end{equation}
This condition is fulfilled if
\begin{equation}
 \kappa=-(1+i)\beta_{0} V_{nj}
\end{equation}
We deduce
\begin{equation}
 k_{nj}a\simeq V_{nj}\left( 1- (1+i)\beta_{0}M\right)
\end{equation}
