Consider the three term boundary value problem defined by the recurrence $$ ay_{n+1}+by_n+cy_{n-1} = d_n \qquad (n = 1,2,\dots,N-1;\ y_0 = y_N = 0)\tag{$\ast$} $$ where the positive integer $N$, the constants $a,b,c$ and the sequence $\{d_n\}_{n=1}^{N-1}$ are given in advance. Suppose further that the polynomial $a + bx + cx^2$ has a repeated root $r$, so that $b^2 = 4ac$. Assume that $a,c\ne 0$, so that $b \ne 0$. I have solved this in the particular case $r = 0$, so assume further that $r \ne 0$. The main difficulty is apparently determining $y_1$ and $y_{N-1}$.
It's plain from the analysis that Wilf makes that we can solve for $y_1$ in terms of $y_{N-1}$: $$ y_1 = -a^{-1}\{D(r) + cr^Ny_{N-1}\}, \qquad (D(r) = \sum_{j=1}^{N-1}d_jr^{j}) $$ and I've tried to solve for $y_{N-1}$ using $(\ast)$, but didn't make much progress. If we know $y_1$ or $y_{N-1}$, then all the other $y_j$'s fall out of the recurrence relation.
So how do we solve for either $y_1$ or $y_{N-1}$?
For context, this is coming straight from section 1.4 of Wilf's generatingfunctionology (freely available here), where he treats separately the case in which $a + bx + cx^2$ has two distinct roots $r_{\pm}$. I won't accept any answer that uses material treated after this section, since I would like to see how this can be solved with only the ideas used up to that point.
Following the presentation in generatingfunctionology, define $$Y(x) = \sum_{j=0}^N y_j x^j$$ Then the recurrence implies $$\{a + bx + c x^2\} Y(x) = x \{D(x) + a y_1 + c y_{N-1} x^N\}$$ If $a + bx + c x^2=0$ has a repeated root $r$, then $a + bx + c x^2=c (x-r)^2$, so $$c (x-r)^2\; Y(x) = x \{D(x) + a y_1 + c y_{N-1} x^N\} \tag{*}$$ If we let $x=r$ then we have one equation involving our unknowns $y_1$ and $y_{N-1}$: $$0 = D(r) + a y_1 + c y_{N-1} r^N \tag{**}$$ To get an additional equation, differentiate (*) with respect to $x$ $$c(x-r)^2 \; Y'(x) + 2c(x-r) \; Y(x) = x \{D'(x) +cNy_{N-1} x^{N-1}\} + D(x) + a y_1 + c y_{N-1} x^N$$ and again set $x=r$: $$0 = r \{D'(r) +cNy_{N-1} r^{N-1}\} + D(r) + a y_1 + c y_{N-1} r^N$$ which in combination with (**) implies $$0 = D'(r) +c N y_{N-1} r^{N-1}$$ Now we have two equations in two unknowns, which we can solve for $y_1$ and $y_{N-1}$.