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.

Alternatively, you can always double check the result with generating functions (like here and here for more examples):
$$f(t)=\sum\limits_{n=0}x_n\cdot t^{n}=2 + \sum\limits_{n=1}x_n\cdot t^{n}= 2 + \sum\limits_{n=1}(3x_{n-1}-8)\cdot t^{n}=\\
2 + 3\sum\limits_{n=1}x_{n-1}\cdot t^{n}-8\sum\limits_{n=1} t^{n}=2+3t\sum\limits_{n=1}x_{n-1}t^{n-1}-\frac{8}{1-t}+8$$
or
$$f(t)=10+3tf(t)-\frac{8}{1-t} \Rightarrow f(t)=\frac{10}{1-3t}-\frac{8}{(1-t)(1-3t)}=\\
\frac{10}{1-3t}+\frac{4}{1-t} -\frac{12}{1-3t}=\frac{4}{1-t}-\frac{2}{1-3t}$$
or
$$f(x)=4\sum\limits_{n=0}t^n-2\sum\limits_{n=0}(3t)^n=\sum\limits_{n=0}\left(\color{red}{4-2\cdot3^n}\right)t^n$$
Thus $$x_n=4-2\cdot3^n$$
Some of the shortcuts are explained here. 
