I would like to understand how to solve the equation
$$ \left(g(v)V'(v)+f(v)V(v)\right)'=-c_ig(v)^{1/2},\quad \eta_i<v<\zeta_i $$ (here $'$ means derivative with respect to $v$)
with boundary conditions $V'(\eta_i)=-a_i$ and $V'(\zeta_i)=-b_i$ as well as the condition $\int_{\eta_i}^{\zeta_i} V(v)\, dv=0$.
Moreover, we have that $g'(v)=-2f(v)$.
The general solution is claimed to be $$ V=g^{1/2}\left(\alpha_i+\beta_i\int_0^v g^{-3/2}-c_i\int_0^v\left(g^{-3/2}\int_0^s g^{1/2}\right)\right)\quad (*) $$ where $\alpha_i, \beta_i$ are integration constants that are to be determined by the boundary conditions and the integral condition.
I would like to confirm $(*)$, in particular because the solution is given in this ugly notation without using integration variables.
First of all, I think one can integrate from any $s$ to $v$ to get $$ g(v)V'(v)+f(v)V(v)=-c_i\int_s^v g^{1/2}(t)\, dt + C, $$ where $C$ is a constant (fundamental theorem of calculus).
Then, dividing by $g$, one gets $$ V'(v)=-\frac{f(v)}{g(v)}V(v)-c_ig(v)^{-1}\int_s^v g(v)^{1/2}\, dv+g^{-1}(v)C=:w(v)V(v)+q(v). $$ (The assumption seems to be that $g(v)\neq 0$.)
The homogeneous equation (i.e. for $q(v)=0)$, has the solution $$ V_{\text{hom}}(v)=D\exp\left(-\int_s^v w(t)\, dt\right)=D\exp\left(\frac{1}{2}\int_s^v\frac{g'(t)}{g(t)}\, dt\right)=D\left(\frac{g(v)}{g(s)}\right)^{1/2}, $$ where $D$ is a constant. Here, I have used that $f=-g'/2$.
Next, I choose $s=0$ (why not?) and make the variation of constants ansatz $D(v)\left(\frac{g(v)}{g(0)}\right)^{1/2}$ which gives the special solution
$$ V_{\text{sp}}(v)=\exp\left(-\int_0^vw(s)\, ds\right)\int_0^v q(t)\exp\left(\int_0^t q(x)\, dx\right)\, dt $$ $$ =\left(\frac{g(v)}{g(0)}\right)^{1/2}\left[\frac{C}{g^{-1/2}(0)}\int_0^v g^{-1}(t)g^{-1/2}(t)\, dt -\frac{c_i}{g(0)^{-1/2}}\int_0^v\left( g^{-1}(t)\int_0^t g(s)^{1/2}\, ds g^{-1/2}(t)\right)\, dt\right] $$ $$ =g^{1/2}(v)\left[C\int_0^v g^{-3/2}(t)\, dt - c_i\int_0^v \left(g^{-3/2}(t)\int_0^t g^{1/2}(s)\, ds\right)\, dt\right] $$
Hence, for the general solution, $V(v)=V_{\text{hom}}(v)+V_{\text{sp}}(v)$, I get (for $s=0$ as above),
$$ V(v)=g^{1/2}(v)\left[\frac{D}{g^{1/2}(0)}+C\int_0^v g^{-3/2}(t)\, dt-c_i\int_0^v \left(g^{-3/2}(t)\int_0^t g^{1/2}(s)\, ds\right)\, dt\right]. $$
If I now set $$ \alpha_i:=\frac{D}{g^{1/2}(0)},\quad\beta_i:=C, $$ I get $(*)$.
(1) Am I right? Do you agree?
(2) Just a little additional question besides: In the paper where solution $(*)$ is claimed, the solution $V_{\text{hom}}(v)=\left(\frac{g(v)}{g(s)}\right)^{1/2}$ is called the fundamental solution at s. For me this is very confusing because I would rather say that this is the fundamental solution at $v$ (and say that $s$, as the lower integration bound, can be chosen freely, say, as $s=0$ as done above). Do you have an explanation why they say "solution at $s$" instead of "solution at $v$"?