Solve finite differences linear equation Given
$$(\textrm{Pe}-1)u_{i+1} + 2u_i - (\textrm{Pe}-1)u_{i-1} = \frac{h^2}{\mu}$$
where $\textrm{Pe}$ stands for Péclet number, $h$ is the step size of the mesh and $\mu$ is the diffusion (constant)
How to solve such equation?
I'm trying to solve it to understand why there are no oscillations in a finite element method approximation with the upwind schema.
UPDATE 1
The roots I got are:
$r_1 = 1$ and $r_2 = \frac{1+\textrm{Pe}}{1-\textrm{Pe}}$
So I have the following guess for the general solution
$$u_i = C_1r_1^i + C_2r_2^i$$
To find the constants $C_1$ and $C_2$ I need to initial conditions. What are those? Are they related to the boundary conditions? I have $u(0) = 1$ and $u'(1) = 1$. How to proceed?
UPDATE 2
According to this text, I replace the roots in my guess for the solution to the homogeneous part.
$$u_i = C_1(1)^i + C_2\Big(\frac{1+\textrm{Pe}}{1-\textrm{Pe}}\Big)^i$$
Applying the initial conditions I have
$$u_0 = 1 = C_1(1)^0 + C_2\Big(\frac{1+\textrm{Pe}}{1-\textrm{Pe}}\Big)^0$$
$$1-C_1 = C2$$
for the other initial condition, what should I do, I mean I don't know how to proceed with that Neumann condition there.
UPDATE 3
Before applying the initial conditions to find the values of $C_1$ and $C_2$ I tried to solve for the particular case following the example on the information of this site
So I came up with using a guess like $f_i = ki$. Replacing on the original equation I have
$$(\textrm{Pe}-1)k(i+1) + 2ki - (\textrm{Pe}-1)k(i-1) = \frac{h^2}{\mu}$$
which after rearranging gives me
$$\big[(\textrm{Pe}-1)+2-(\textrm{Pe}-1)\big]ki + \big[(\textrm{Pe}-1)+(\textrm{Pe}+1)\big]k = \frac{h^2}{\mu}$$
finally $$k = \frac{h^2}{\mu \big[(\textrm{Pe}-1)+(\textrm{Pe}+1)\big]}$$
Hence, the complete (I think) equation is:
$$u_i = C_1(1)^i + C_2\Big(\frac{1+\textrm{Pe}}{1-\textrm{Pe}}\Big)^i+\frac{h^2 i}{\mu \big[(\textrm{Pe}-1)+(\textrm{Pe}+1)\big]}$$
Am I right? I just need to apply initial conditions. The $u(0) = 1$ is easy but the other one $u'(1) = 1$ how to apply that?
 A: It depends.  Do A and B depend on the index $i$?  If A and B are constant, then you can split the solution up into a homogeneous solution $u_{i}^{(H)}$ and an inhomogeoneous solution $u_{i}^{(I)}$.  The homogeneous solution satisfies
$$A u_{i+1}^{(H)} + 2 u_{i}^{(H)} + B u_{i-1}^{(H)} = 0 $$
with initial conditions such as $u_{0}^{(H)} = u_0$ and $u_{1}^{(H)} = u_1$.  In this case, with constant coefficients, you can assume that $u_{i}^{(H)}$ takes the form $\exp{(i r)}$ and find $r$ from the following equation:
$$A r^2 + 2 r + B = 0$$
which I leave to you; the solutions are then $r_1$ and $r_2$, and the general solution to the homogeneous equation is $u_{i}^{(H)} = C_1 \exp{(i r_1)} + C_2 \exp{(i r_2})$.  $C_1$ and $C_2$ are found by using the initial conditions.
The inhomogeneous solution satisfies 
$$A u_{i+1}^{(I)} + 2 u_{i}^{(I)} + B u_{i-1}^{(I)} = C $$
with initial conditions such as $u_{0}^{(I)} = 0$ and $u_{1}^{(I)} = 0$.  For a constant $C$, $u_{i}^{(I)}$ can be a constant, with such constant being offset in the homogeneous initial conditions.
For the case where $A$ and/or $B$ is not constant, then methods specific to the forms of their variation should be employed.
