Without using characteristic equation, how can I show genenal formula of this sequence? I know that, solution of the sequence $x_n$ so that 
$$x_{n+2}-2 x_{n+1}= 2  x_{n+1} - 4 x_{n}$$
is $x_n = (A+ Bn)2^n$. I want to show this without using characteristic equation. I tried
\begin{align*}
   x_{n+2}-2 x_{n+1} &= 2 x_{n+1} - 4 x_{n} \\
   &= 2(x_{n+1}-2x_{n}) \\
   &= 2^2(x_{n}-2x_{n-1}) \\
   &= \ldots \ldots \\
   &= 2^n (x_2 - 2x_1)
\end{align*}
From here, I cannot get the solution. Without using characteristic equation, how can I show general formula of this sequence?
 A: Use induction.  
We are given a sequence $x_1,x_2,\dots$ satisfying the recursion 
$$
x_{n+2}-2 x_{n+1}= 2  x_{n+1} - 4 x_{n},
$$
and we want to show that for some $A,B$ we have $x_n = (A + Bn)2^n$ for all $n$.
Select $A,B$ such that $x_n = (A + Bn)2^n$ holds for $n = 1,2$ (or note that such an $A,B$ necessarily exist). The base case of our induction (that $x_n = (A + Bn)2^n$ holds for $n = 1,2$) is now trivial.  For the inductive step: suppose that $x_n = (A + Bn)2^n$ holds for all $n \leq k$ (where $k \geq 2$).  We find that
$$
\begin{align}
x_{k+1} &= 2x_{k} + (x_{k+1} - 2x_{k}) = 
2x_{k} + (2x_{k} - 4x_{k-1}) 
\\ & = 
2\cdot (A + Bk)2^k + (2[(A + Bk)2^k] - 4[(A + B(k-1))2^{k-1}])
\\ &=
(A + Bk)2^{k+1} + (2[(A + Bk)2^k] - 4[(A + B(k-1))2^{k-1}])
\\ & = (A + Bk) 2^{k+1} + B\cdot 2^{k+1} 
\\&= (A + B(k+1))2^{k+1}.
\end{align}
$$
We have now completed the inductive proof and reached the desired conclusion.

Using what you tried: you showed that the $x_n$ satisfies the recursion
$$
x_{n+2} = 2x_{n+1} + b\cdot 2^n, \quad n \geq 0
$$
where $b = (x_2 - 2x_1)$.  By solving this first order linear recurrence (method of your choice), we find that the general solution is
$$
x_n = 2^{n-2}(bn + C)
$$
where $C$ is an arbitrary constant.  The conclusion follows.

Another approach: suppose that $x_1,x_2,\dots$ is a sequence satisfying 
$$
x_{n+2}-2 x_{n+1}= 2  x_{n+1} - 4 x_{n}.
$$
Define the sequence $y_1,y_2,\dots$ by $y_n = 2^{-n} x_n$.  It's easy to see that
$$
y_{n+2} - y_{n+1} = y_{n+1} - y_n.
$$
Solve this to find that $y_n$ has the form $y_n = A + Bn$.  The conclusion follows.
A: $\begin{align}x_{n+2}=2x_{n+1}+2^n(x_2-2x_1)&&2x_{n+1}=2[2x_n+2^{n-1}(x_2-2x_1)]=4x_{n}+2^n(x_2-2x_1)\\
x_{n+2}=4x_n+2^n(x_2-2x_1)2&&4x_n=4[2x_{n-1}+2^{n-2}(x_2-2x_1)]=8x_{n-1}+2^n(x_2-2x_1)\\
x_{n+2}=8x_{n-1}+2^n(x_2-2x_1)3&&8x_{n-1}=8[2x_{n-2}+2^{n-3}(x_2-2x_1)]=16x_{n-2}+2^n(x_2-x_1)\end{align}$
$\dots$
$x_{n+2}=2^nx_2+2^n(x_2-2x_1)n$
i.e., $x_n=2^{n-2}x_2+2^{n-2}(x_2-2x_1)(n-2)$
