Consider the following $1$st order eigen ODE system of 2 components $(\alpha,\beta)(x)$ defined for $x\in[0,L]$ $$ F_{l+1}\beta+m\alpha=\lambda\alpha\\ F_{-l}\alpha-m\beta=\lambda\beta $$ where $\lambda$ is the eigenvalue, $l$ is integer, $m(0)$ is finite and $m(L)\rightarrow\infty$, and $F_k=-i(\frac{d}{dx}+\frac{k}{x})$.
Since it is just $1$st order, I thought only 2 boundary conditions (b.c.) are necessary, which can be most easily taken as $(\alpha,\beta)(L)=0$ since $m(L)$ diverges. So one doesn't need or just shouldn't involve b.c. at $x=0$? Is this true?
However, on the other hand, if I do the very simple asymptotic analysis at $x\rightarrow0$ for these equations, I find that $$ l\neq0,-1\qquad (\alpha,\beta)(0)=0\\ l=-1\qquad (\alpha,\beta')(0)=0 $$ with $l=0$ similar to $l=-1$. What I did is basically just noting that, when $k\neq0$, any component acted by $F_k$ is $0$ at $x=0$. And using this, we notice the other component's derivative vanishes. It seems that b.c. at $x=0$ is natural as well.
How to reconcile these two considerations? Where am I wrong?