How this general term is attained from the default linear differnce equation?

Question

The sequence $\{a\}$ holds the below equation.

$$a_{i-1}+a_{i+1}=2\cosh(2\kappa)*a_{i}$$

$$\kappa~\text{ is a constant.}$$

The each element of the sequence only contains $i$ as a variable(single variable function).

We can ignore imaginary numbers for this problem.

The textbook states that assuming $a_i=\alpha^{i}$ and then $a_i$ can be represented as the below equation.

$$a_i=A*\cosh(2i\kappa)+B*\sinh(2i\kappa)$$

$$A,B~~~\text{are constants.}$$

Currently I can't get the below $3$ things .

1. derivation of $a_i=A*\cosh(2i\kappa)+B*\sinh(2i\kappa)$
2. why the textbook asserted that the ith term can be represented in the exponential form of $i$
3. whether $\alpha$ contains $i$ as a variable.

Answer 1

Hint:

From the theory of linear difference equations with constant coefficients, the solution can be expressed as a sum of exponentials.

If we assume $a_i=\alpha^i$ and plug this in the equation, we get

$$\alpha^{i+1}+\alpha^{i-1}=2\cosh(2\kappa)\alpha^i$$ or, after simplification,

$$\alpha^2-2\cosh(2\kappa)\,\alpha+1=0.$$

Now solve this quadratic equation to get the possible $\alpha$'s.