## Numerical Example of Affine Difference Equation

We have difference equation y(k+1) = 2 y(k) 5, what is the solution for initial value y(0) = 1 and what is the equilibrium value? How is the stability of this equilibrium value?

### Solution

k = 0: y(0) = 1 initial value

k = 1: y(1) = 2 y(0) 5 = -3

k = 2: y(2) = 2 y(1) 5 = -11

= 2 ( 2 y(0)) 5) 5 = 4 y(0) -15

k = 3: y(3) = 2 y(2) 5 = -27

= 2 (4 y(0) -15) 5 = 8 y(0) -35

and so on. For big k value, it is easier to use the

### Closed-form Solution

Rather than recursive calculation:

a = 2, b = - 5, k = 10 then

### Equilibrium Value

or

If = 5 then y(k) = 2*5 - 5 = 5 forever. It remain constant at 5 regardless whatever the value of k.

Difference equation y(k+1) = 2 y(k) 5 has solution of and equilibrium value of .

### Stability:

Since a = 2 > 1, the equilibrium is unstable. If the initial value y(0) is smaller than the equilibrium value, the graph of y(k) versus k is exponentially decreasing without bound. If the initial value of y(0) is bigger than the equilibrium value, the graph is exponentially increasing without bound.

Preferable reference for this tutorial is

Teknomo, Kardi (2015) Difference Equation Tutorial. https:\\people.revoledu.com\kardi\ tutorial\DifferenceEquation\