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!

If equation has two equal roots then we can write $D=0$ or $b^2=4ac$ or $c=b^2/4a$ .Then we write $ax_{n+1}+bx_n+b^2/4a*x_{n−1}=0$ or $ax_{n+1}=-bx_n-b^2/4a*x_{n−1}$ or $x_{n+1}=-b/a(x_n+(b/4a)*x_{n−1})$. Let $a_1=u$ and $a_2=w$ .Let $-a/b=t$ Then we get $a_3=t(x_2-1/4tx_1)=t(w-1/4*ut)$. Then we get $a_4=t(x_3-1/4*tx_2)=t^2(w-1/4*ut)-1/4*w)$ . We can show by induction that all next numbers will be shown using $t,u$ and $w$ and then we get answer
