<s>[INST] 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.
 [/INST] 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. 
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