General solution of recurrence relation if two equal roots Consider the recurrence relation
$$
ax_{n+1}+bx_n+cx_{n-1}=0
$$
If the characteristic equation
$$
a\lambda^2+b\lambda+c=0
$$
has two equal roots, then the general solution is given by
$$
x_n=(A+nB)\alpha^n.
$$
Could you please explain that to me, I do not see that!

First divide through by $a$ to get the form $x_{n+1}+px_n+qx_{n-1}=0$
$\lambda$ will still be a root of $\lambda^2+p\lambda+q=0$, and it will be a double root if $p=-2\lambda, q=\lambda^2$
Now let $y_n=x_n-\lambda x_{n-1}$
We have $0=x_{n+1}-2\lambda x_n + x_{n-1}=(x_{n+1}-\lambda x_n)-\lambda (x_n-\lambda x_{n-1})=y_{n+1}-\lambda y_n$
And this means $y_{n+1}=\lambda y_n =\dots = \lambda^{n+1}y_0 = B\lambda^{n+1}$
Now consider $$y_n+\lambda y_{n-1}+\lambda ^2 y_{n-2}+\dots \lambda ^{n-1}y_1=x_n-\lambda x_{n-1}+\lambda x_{n-1}-\lambda^2 x_{n-2} + \dots-\lambda^nx_0$$
The left-hand side with $n$ equal terms is just $nB\lambda^n$. The right-hand side cancels except for the first and last term to give $x_n-\lambda^nx_0=x_n-A\lambda^n$. Putting this together and rearranging gives $$x_n=(A+nB)\lambda^n$$
