First order difference equation. Solve $u_{n+1}=3u_n+2$. First order difference equation. Solve $u_{n+1}=3u_n+2$ with $u_0=0$

My notes are very sparse on this topic, so I need some help solving what should be an easy question. I would really appreciate some going through this basic question thoroughly so that I can study it and answer the other question with less help.
In particular, how do you solve for the homogeneous equation. One example that I am given is that for the general form of this question, i.e. $u_{n+1}=au_n+b$, the homogeneous equation is $w_{n+1}=aw_n$. The solution is found by "putting it into itself", and we get
$w_n=aw_{n-1}=...=a^nw_0=Aa^n$
for some constant $A$.
Could someone help me out here, because none of this makes much sense to me
 A: First Step
Write the equation in standard form.
$$u_{n+1} - 3u_n = 2$$
Then write down the characteristic equation of the associated homogenous equation.
$$r - 3 = 0$$
$$\iff r = 3$$
Thus we know the general form of the associated homogeneous equation is:
$$u_n = c_13^n $$
Second Step
Write down the form of the particular equation. 
$$u_p = c_2$$
To solve for $c_2$, replace in the equation.
$$c_2 - 3c_2 = 2$$
$$-2c_2 = 2$$
$$c_2 = -1$$
Thus $u_n = c_13^n -1$
Last Step
Use the initial conditions. 
$$u_0 = 0$$
$$c_1 -1 = 0$$
$$c_1=1$$
Thus we get:
$$u_n = 3^n -1$$
A: Note that $$u_{n+1}=3u_n+2$$ implies $$u_n=3u_{n-1}+2$$
Subtract the second from the first to obtain $$u_{n+1}-u_n=3u_n-3u_{n-1}=3(u_n-u_{n-1})$$ 
This tells us that $v_n:= u_{n+1}-u_{n}=3^nv_0$ (using what you have already been given with $v_n$ and with $v_0$ being a constant depending on the initial value) we obtain $$u_{n+1}=u_n+3^nv_0$$
Now subtract this equation from the first one to obtain $$2u_n=3^nv_0-2$$
Now $u_0=0$ tells us that $v_0=2$ and $u_n=3^n-1$

This is not necessarily the recommended general method, but it shows why some of the more general methods work, and it uses the hint you have been given and not much else.
Note that $v_1=3v_0$; $v_2=3v_1=9v_0$; $v_3=3v_2=27v_0$ if you need to get a handle on what is happening with $w_n$ in your question.
A: You want the characteristic equation. For $u_{n+1}=au_n$ this gives $u_{n+1}-au_n=0$. This gives $x-a=0$ for the characteristic equation, found by replacing $u_{n+1}$ by $x$ and $u_n$ by 1, solving gives $x=a$, so the homogeneous equation gives $Aa^n$.
To find the particular solution, you want $u_n$ to have the same form as the inhomogeneous part, so in this case, a constant $C$. This gives $C=3C+2$, or $C=-1$.
Therefore the general equation has the form $A3^n-1$. We know that $u_0=0$, so $A-1=0$, and $A=1$. 
Finally the equation is $u_n=3^n-1$. 
