Solving $x_{n} - 3x_{n-1} = -8$ with $n\geq 1$ and $x_0 = 2$ I tried two methods which gave different answers:
Method 1:  
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}{3-1}\right)\\ = 2\times 3^n - 4(3^n - 1) \\ = -2\times 3^n +4.$$  
Method 2:
Solving the homogenous equation using $x_n = r^n$,
$$r^n - 3r^{n-1} = 0 \\ \implies r = 3$$
So the homogenous solution is $h_n = a\times 3^n$ for some $a\in\mathbb{R}$.
With the initial condition, $h_n = 2\times 3^n$.  
Also, by guessing the particular solution with $x_n = C$,
$$C - 3C = -8 \implies C = 4$$.  
So, the final solution is
$$x_n = 2\times 3^n + 4.$$  
I'm unsure why the two methods differ. It looks obvious that the first one gives the correct result.

Safety way 
Set
$v_n =x_n-a$ such that $v_n$ satisfies $$ v_{n+1} =3v_n\Longleftrightarrow  x_{n+1}-a =3(x_n-a)$\Longleftrightarrow  3x_{n}-8-a =3(x_n-a)\Longleftrightarrow  a=4$$
Then 
$$v_n =3^nv_0\Longleftrightarrow  x_n-4 = 3^n(x_0-4)\Longleftrightarrow  \color{red}{x_n = -2\cdot3^n +4}$$
