Consider the following differential equation \begin{equation} \frac{d}{dx}\left[N(x)\frac{dw}{dx}\right]+\sigma^2\rho(x)w=f(x,\sigma),~~ 0<x<l, \end{equation} $0<N\in C^1(0,l)$, $0<\rho\in C(0,l)$, with boundary conditions $$ w(0,\sigma)=u(\sigma),~ w(l,\sigma)=0,~~ \sigma\in\mathbb{R}, $$ Transformation $$ w(x,\sigma)=\exp\left[\int\frac{\nu(\xi,\sigma)}{N(\xi)}d\xi\right], $$ reduces the homogeneous equation into Riccati differential equation $$ \frac{\partial \nu}{\partial x}+\frac{1}{N}\nu^2+\sigma^2\rho=0. $$ It has order one, therefore $\nu$ contains one integration constant, say, $c$.
When I pick particular $N$ and $\rho$, the fundamental solutions of initial equation are nothing else, but $$ w_1(x,\sigma)=\exp\left[\int\frac{\nu(\xi,\sigma;c=0)}{N(\xi)}d\xi\right],~ w_2(x,\sigma)=\exp\left[\int\frac{\nu(\xi,\sigma;c=l)}{N(\xi)}d\xi\right]. $$ Is there a way to prove this rigorously in general?
I thought to prove that the Wronskian is not trivial for those functions: $$ W\left[w_1,w_2\right]=\frac{\nu(x,\sigma;c=l)-\nu(x,\sigma;c=0)}{N(x)}\exp\left[\lambda+\mu\right]\neq0, $$ since $\nu(x,\sigma;c=l)-\nu(x,\sigma;c=0)\neq0$ a.e. in $0<x<l$, in which $$ \lambda=\int\frac{\nu(\xi,\sigma;c=0)}{N(\xi)}d\xi,~~ \mu=\int\frac{\nu(\xi,\sigma;c=l)}{N(\xi)}d\xi. $$ But this is the case also for $c=0$ and any $0<c<l$. So, I got stack.
Edit: Particular $N$ and $\rho$ are such that $N(x)=N_0g(x)$, $\rho(x)=\rho_0g(x)$, $N_0>0$, $\rho_0>0$.