Gelfand Yaglom Theorem (boundary conditions)

Question

I have one question about the Gelfand Yaglom Theorem. Typically I see people who want to calculate the determinant of a Schrödinger Operator $\hat{\cal{M}}_{{\rm S}} = -\frac{{\rm d}^2}{{\rm d}t^2} + U(t)$ and then solve \begin{align} \hat{\cal{M}}_{\rm S} \, \phi = 0 \end{align} with the boundary conditions $\phi(0)=0$ and $\phi'(0)=1$. From a paper from Kirsten I know that in this case $\phi'(0)=1$ is for convenience in order for a contribution at $\infty$ in the $\lambda$-plane to vanish. I'm however wondering what happens if I consider a more general Sturm-Liouville type equation with the Operator $\hat{\cal{M}}_{{\rm SL}} = -\frac{{\rm d}}{{\rm d}t} \, g(t) \, \frac{{\rm d}}{{\rm d}t} + U(t)$. Now here it seems to be that the "convenient" boundary conditions are $\phi'(0)=\frac{1}{g(0)}$ (see Kleinert Pathintegrals p. 134). I haven't been able to figure out why this is so though...