$a_n=2a_{n-1}-a_{n-2}$
$a_{0}=a_{1}=2$
$x^{2}=2x-1$
$x^{2}-2x+1=0$
$(x-1)^2=0$
$x=1$
$a_{n}=(α+βn) 1^{n}$
$(α+β(0)) 1^{0}=2$
$α=2$
$(α+β(1)) 1^{1}=2$
$α+β=2$
$a_{n}=2$
I can't seem to find of anything I did wrong. It seems weird that this would be the answer though, so I have a feeling I did something wrong. Just looking for some confirmation on my HW.
No, everything is fine. You got that the solution is a linear function of $n$, and since the initial slope is $0$, the solution stays horizontal, i.e., constant.