2.2.2 Example B Now consider the inhomogeneous equation Yk+1 – Byk = a, (2.12) where a and B are constants. For this case, we have pk Therefore, 3 and gk = a. 71 k-1 k-1 k-1 qi Σ. (2.13) Pr i=1 r=1...

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Extracted text: 2.2.2 Example B Now consider the inhomogeneous equation Yk+1 – Byk = a, (2.12) where a and B are constants. For this case, we have pk Therefore, 3 and gk = a. 71 k-1 k-1 k-1 qi Σ. (2.13) Pr i=1 r=1 i=1 i=1 Using the fact that k-1 pk (2.14) 1- r i=1 gives k-1 1– 3-k+1 В — 1 (2.15) Therefore, for 3+ 1, the general solution is Yk = (2.16) В — 1 where C is an arbitrary constant. When B = 1, we have Yk+1 – Yk = a, (2.17) with pr = 1 and q = a. Now k-1 k-1 SII Pk = II 1 = 1, (2.18) i=1 i=1 and k-1 /k-1 k-1 > (1) = a(k – 1). (2.19) i=1 r=1 i=1 48 Difference Equations Therefore, the general solution, for 3 = 1, is Yk = A + a(k – 1) = C + ak, (2.20) where A and C = A – a are arbitrary constants. IWI