Do you know any example of a positive on ${\mathbb R}$ and a bounded function $p(x)>0$ for which all solutions of Sturm–Liouville equation are known (ie some formula)?
$-(p(x)f'(x))'=\lambda f(x)$
I am specifically interested in the case $p(x)=e^{-x^2}$, but any example is of interest.
On
You can assume a power series in $\lambda$: $$ f_{\lambda}(x)=\sum_{n=0}^{\infty}\lambda^n g_n(x). $$ This will yield a solution of $Lf_{\lambda}=\lambda f_{\lambda}$, where $$ Lf = -(pf')'. $$ Substituting: $$ \sum_{n=0}^{\infty}\lambda^nLg_n = \sum_{n=0}^{\infty}\lambda^{n+1}g_n, $$ which implies that $Lg_n=g_{n-1}$ for $n \ge 1$ and $Lg_0=0$. To satisfy particular endpoint conditions, you may choose to specify conditions for $g_0$, and then choose $0$ endpoint conditions for $g_n$, $n=1,2,3,\cdots$. That is, $$ g_0(0)=A, \;\;\; g_0'(0)=B,\\ g_n(0)=0,\;\; g_n'(0)=0,\;\;\; n \ge 1. $$ In your case, $Lf = -(e^{-x^2}f')'$. So $g_0(x)$ must satisfy $$ (e^{-x^2}g'_0)'=0, \\ g_0(0)=A,\;\; g_0'(0)=B. $$ Therefore, $$ e^{-x^2}g_0'(x) = B \\ g_0'(x) = Be^{x^2} \\ g_0(x) = A+B\int_{0}^{x}e^{x^2}dx $$ Next, $Lg_1 = g_0$ is solved subject to $g_1(0)=g_1'(0)=0$. That is, $$ -(e^{-x^2}g_1')' = g_0,\\ g_1(0)=0,\; g_1'(0)=0. $$ This is solved by direct integration: $$ -e^{-x^2}g_1'(x) = \int_0^x g_0(x_0)dx_0 \\ g_1(x) = -\int_0^{x}e^{x_1^2}\int_0^{x_1} g_0(x_0)dx_0 dx_1 $$ The above establishes the recursion that you need in order to solve for the series $f_{\lambda}(x)$ given at the top. Your integrals are going to involve Gaussians, which means you're not going to be able to explicitly perform the integrations.
$$-(p(x)f'(x))'=k f$$ See for example: $p(x)=x^2$, the the ODE is $$x^2f''+2xf'+kf=0$$ This is Eular's homogeneous ODE. Take $f=x^m$, then you get $m(m-1)+2m-k=0 \implies k=\frac{-1\pm\sqrt{1-4k}}{2}=m_1,m_2$ So the solution is $$f(x) =C_1 x^{m_1}+ C_2 x^{m_2}$$