Sorry for this dumb question, but I need help finding an error in my proof.
Assume you have a difference equation $$u_{n+1} + au_n + bu_{n-1} = 0.$$ We assume the solution is $$A \lambda^{n+1} + aA\lambda^n + bA\lambda^{n-1}.$$ Then the auxiliary equation is $$\lambda^2 + a\lambda + b = 0.$$ Assume this has repeated roots $$\lambda = \lambda_1 = \lambda_2.$$ Then the textbook tells me that the solution is of the form $$u_n = (A+Bn)\lambda^n.$$ I tried proving this but I got stuck:
\begin{align} (A + B(n+1))\lambda^{n+1} + a(A+Bn)\lambda^n + b(A+B(n-1))\lambda^{n-1} &= 0 \\ (A + Bn + B)\lambda^{n+1} + a(A + Bn)\lambda^n + b(A + Bn - B)\lambda^{n-1} &= 0 \\ (A + Bn)\lambda^{n+1} + B\lambda^{n+1} + a(A + Bn)\lambda^n + b(A + Bn)\lambda^{n-1} - bB\lambda^{n-1}&= 0 \\ \lambda^{n-1}(A+Bn)(\lambda^2 + a\lambda + b) + B\lambda^{n+1} -bB\lambda^{n-1} &= 0 \\ B\lambda^{n+1} -bB\lambda^{n-1} &= 0 \\ \lambda^2 &= b \\ \end{align}
This is clearly not correct. I'm feeling really dumb but I'm just stuck and I don't see what I did wrong up there ...
You haven’t done anything wrong. You simply haven’t applied the fact that the auxiliary equation has a repeated root. It is therefore a perfect square, so factors as $(\lambda\pm\sqrt b)^2$, from which $\lambda^2=b$. (Note that this also means that $b\ge0$ for the equation to have a repeated root.)