I have the equation
$$(gV'+fV)'=-\frac{\overline{c}_i}{\varepsilon}g^{1/2}$$ where $(gV'+fV)'=gV''-fV'+f'V$
There are also some boundary conditions given, $V'(v_i^0)=-a_i, V'(v_i^*)=-b_i$, but their context is very long and I dont think I can write it down here.
Moreoever $g(u)=-2\int_{-1}^u f$ and $g'=-2f$.
Now, it is said that from this equation and the fact that the fundamental solution at s of the equation $gy'+fy=0$ is given by $(g(v)/g(s))^{1/2}$ one can obtain
$$ V=g^{-1/2}\left(\alpha_i+\beta_i\int_0^v g^{-3/2}-\frac{\overline{c}_i}{\varepsilon}\int_0^v\left(g^{-3/2}\int_0^s g^{1/2}\right)\right) $$ where $\alpha_i,\beta_i$ are some integration constants which are to be determined by the boundary conditions.
I dont understand what is done here. I am not asking for the exact computation of this formula but I would like to understand what is done here, In particular:
(1) Why considering $gy'+fy=0$ and why is its fundamental solution given by $(g(v)/g(s))^{1/2}$?
(2) How to get the formula for V then after understanding (1)?
Hint:
Integrate once
$$(gV'+fV)'=-\frac{\overline{c}_i}{\varepsilon}g^{1/2}$$
to get
$$gV'+fV=-\frac{\overline{c}_i}{\varepsilon}\int g^{1/2}\,dv+C.$$
This is a linear ODE of the first order, and you start by finding the solution of the homogeneous part,
$$gV'+fV=0$$ or
$$\frac{V'}{V}=-\frac fg,$$ giving
$$V=\frac Dg.$$
Now a particular solution can be obtained by variation of the constant, i.e. setting $V=\dfrac{D(v)}g$ and plugging in the equation.