<s>[INST] Asymptotic solution of the equation $\gamma_{i+2} + 4\gamma_{i+1} + \gamma_{i} = \frac{Kh^2}{12}$ I'm struggling with the following equation, I'm interested in an asymptotic solution:
$$\gamma_{i+2} + 4\gamma_{i+1} + \gamma_{i} = \frac{Kh^2}{12}$$
Where $K$ is known constant, when $h \rightarrow 0$ I guess I can assume the equation is something like:
$$\gamma_{i+2} + 4\gamma_{i+1} + \gamma_{i} = K'$$
where $K' << 1$, much smaller than $1$.
Is there a way to solve such equation, I want to find the forced response, I've tried to use the z transform but somehow I end up with divergent solutions, because of the eigenvalues calculation.
I'm quite sure there's an easier way to study the equation.
If it is of any interest I could report the derivation of such equation.
Update:
Here is my new attempt
$$
\left\{
\begin{array}{l}
\gamma_{i+2} + 4\gamma{i+1} + \gamma_{i} = K' \\
\gamma_{i+3} + 4\gamma{i+2} + \gamma_{i+1} = K'
\end{array}
\right. \Rightarrow \Delta \gamma_{i+2} + 4 \Delta \gamma_{i+1} + \Delta \gamma_{i} = 0
$$
Using the characteristic equation I find:
$$
\begin{multline}
\Delta \gamma_{k} = c_0 \left(2 - \sqrt{3} \right)^k + c_1 \left(2 + \sqrt{3} \right)^k \Rightarrow \\ 
\sum_{j=0}^{k-1} \Delta \gamma_j = c_0 \sum_{j=0}^{k-1} \left(2 - \sqrt{3} \right)^j + c_1 \sum_{j=0}^{k-1} \left(2 + \sqrt{3} \right)^j \Rightarrow \\
\gamma_k - \gamma_0 = c_0 \left[ \frac{\left(2 - \sqrt{3} \right)^k - 1}{1- \sqrt{3}} \right] + c_1 \left[ \frac{\left(2 + \sqrt{3} \right)^k - 1}{1+ \sqrt{3}} \right] \Rightarrow \\
\gamma_k = c_0 \left[ \frac{\left(2 - \sqrt{3} \right)^k - 1}{1- \sqrt{3}} \right] + c_1 \left[ \frac{\left(2 + \sqrt{3} \right)^k - 1}{1+ \sqrt{3}} \right] + \gamma_0
\end{multline}
$$
Since $\gamma_k = \gamma_k(h)$ we have
$$
\gamma_k(h) = c_0(h) \left[ \frac{\left(2 - \sqrt{3} \right)^k - 1}{1- \sqrt{3}} \right] + c_1(h) \left[ \frac{\left(2 + \sqrt{3} \right)^k - 1}{1+ \sqrt{3}} \right] + \gamma_0(h)
$$
Is it correct so far?
 [/INST] Assuming $h \neq h(i)$, i.e. $h$ is not a function of $i$, you can start by solving a difference equation: write out an expression for $ \gamma_{i+1}$ and subtract it from the one you have. Thus you will get rid of the constant term and work with $b_{i+2} = \gamma_{i+2} - \gamma_{i+1}$. From there you can solve it using GFs or characteristic equation. 
EDIT: you need two boundary values: $a_0$ and $a_1$.
$$
a_{n+2} + 4 a_{n+1} + a_{n} = C \\
a_{n+1} + 4 a_n + a_{n-1} = C\\
b_{n+2} + 4 b_{n+1} + b_n = 0
$$
Use generating function to solve for $b_n$: 
$$
\frac{G(z) - b_1 z -b_2 z^2}{z^2} + \frac{G(z) - b_1 z }{z} + G(z) = 0
$$
Solve this for $G(z)$ and extract expression for $b_n$. From that, since $b_n = a_n - a_{n-1}$, use a telescoping sum to get on LHS ana expression for $a_n$, and on RHS some sum that you will need to solve for.  
</s>