Let w2n-1 = Un, W2n-2 = Vn, 22n-1 = Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = µ, (11) Un+1 – kį Vn+1 – kị Un = µ, and Yn+1 – Vo Xn – v Yn = €, (12) Xn+1 – Vị Yn+1 - v¼ Xn = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko Un (13) (E – k1) Un – ki E V, = µ, (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – kı) and ko, respectively, (E – ki) (E – ko) Vn – (E – k1) ko Un (E – ki) µ, (15) ko (E – k1) Un – ko ki E Vn ko H. (16) %3D Thus, the combination of (15) and (16) is (E² – (ko + ki + ko ki) E + ko ki) Vn = (1– kı + ko) µ, (E² – O E + ko k1) Vn = (1 – k1 + ko) H, (17) where o = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, $ + V62 – 4 ko ki $ - Vo2 – 4 ko ki 1- k1 + ko + 1- ko - k1 Vn = A + B (18) 2 since A and B are constants. By Substituting (18) in (13), we give V 6² – 4 ko ki 62 – 4 ko ki Un = A ko - 1 0 - Vø2 – 4 ko k1 ko 1 4 ko ki (19) + B - 1 • (G-) (는)-1)~ 1 - ki + ko -- ko – k , H. ko