Difference Equation Numerical Example

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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.

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