Solving an inhomogeneous difference equation Consider the following difference equation:
$$-(1+a)u_{i-1} + 2 u_i - (1-a) u_{i+1} - \frac{h^2}{\epsilon} = 0, \; i = 1, ..., N, \tag{1}$$
$$u_0 = 0, \; u_N = 1.$$
Can someone explain me how to solve this equation?
I read somewhere the following method:
1) Solve the homogeneous equation;
2) Find a particular solution of inhomogeneous equation (1);
3)The general solution of (1) is the sum of the particular solution and the solution of homogeneous equation.
Is this method right? 
I can't find a particular solution of equation (1). Can someone help me?
Thank you!
 A: Note that (1) can be written as
$$(1+a)(u_i-u_{i-1})- (1-a)(u_{i+1}-u_i) = \frac{h^2}{\epsilon}. \tag{2}$$
Let $v_i=u_i-u_{i-1}$ and then (2) becomes
$$(1+a)v_i- (1-a)v_{i+1} = \frac{h^2}{\epsilon} = 0$$
or
$$(1+a)\left(v_i-\frac{h^2}{2a\epsilon}\right)=(1-a)\left(v_{i+1}-\frac{h^2}{2a\epsilon}\right)$$
or
$$v_{i+1}-\frac{h^2}{2a\epsilon}=\frac{1+a}{1-a}\left(v_i-\frac{h^2}{2a\epsilon}\right)\tag{3}$$
From (3), one obtains
$$ v_i-\frac{h^2}{2a\epsilon}=\left(v_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}=\left(u_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}.$$
Thus
\begin{eqnarray}
u_i&=&\sum_{k=1}^i(u_k-u_{k-1})+u_0\\
&=&\sum_{k=1}^iv_i\\
&=&\sum_{k=1}^i\left[\frac{h^2}{2a\epsilon}+\left(u_1-\frac{h^2}{2a\epsilon}\right)\left(\frac{1+a}{1-a}\right)^{i-1}\right]\\
&=&\frac{h^2}{2a\epsilon}i+\left(u_1-\frac{h^2}{2a\epsilon}\right)\frac{1-\left(\frac{1+a}{1-a}\right)^{i}}{1-\frac{1+a}{1-a}}\\
&=&\frac{h^2}{2a\epsilon}i-\frac{1-a}{2a}\left(u_1-\frac{h^2}{2a\epsilon}\right)\left[1-\left(\frac{1+a}{1-a}\right)^{i}\right].
\end{eqnarray}
Using $u_N=1$, one can determine $u_1$ and I omit the detail.
