I'm trying to reproduce the calculations of a paper. The goal is to solve the following ODE: $\left( 1-x^2 \right) y'' - 4xy' + (c-2)y = 0$, inside the interval $[-1,1]$. The solution must be normalizable, i.e., $\int y(x)dx < \infty$.
For $c = 0$, they say that the solution can be considered as a sum of two delta functions $\delta(x-1)$ and $\delta(x+1)$. But as far as I know, the delta functions make sense when integrated, i.e., $f(x_0) = \int f(x)\delta(x-x_0)dx$. So, what is the reasoning behind this solution, how does one find this solution and how to check if it's a solution of the equation, since I don't know how to take the derivative of a delta.
For $c>0$, they say that the solution is $y_c(x) = a_+\delta(x-1) + a_-\delta(x+1) + \phi_c(x)\theta(x+1)\theta(x-1)$, being $a_+$ and $a_-$ constants and $\theta(x)$ the Heavisde function. They plug this ansatz in the original equation and find an equal one but for the function $\phi_c$. What happened with the $\delta$s? Besides, they claim that there are now two extra conditions $\lim_{ x\rightarrow \pm 1} \phi_c = - \frac{c}{2} a_{\pm}$. My question here is how to plug all these non-derivable functions in the original ODE and why this extra condition is necessary. Thanks!
All this treatment with delta functions is surprising to me and before entering to solve explicitly for $\phi_c(x)$ I wanted a bit of comprehension.