FIRST-ORDER DIFFERENCE EQUATIONS 49 2.2.5 Example E The equation (k + 1)yk+1 – kyk = 0 (2.29) can be put in simpler form by the transformation Zk = kyk, (2.30) which gives Zk+1 - Zk = 0. (2.31) The...

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Extracted text: FIRST-ORDER DIFFERENCE EQUATIONS 49 2.2.5 Example E The equation (k + 1)yk+1 – kyk = 0 (2.29) can be put in simpler form by the transformation Zk = kyk, (2.30) which gives Zk+1 - Zk = 0. (2.31) The solution to this latter equation is Zk = A = arbitrary constant. (2.32) Therefore, Yk = = A/k. (2.33) Note that equation (2.29) corresponds to having pk = k/(k+1). Therefore, k-1 k- (k – 1)! 1 Il Pi (2.34) +1 k! k i=1 i=1 Putting this result into equation (2.4) again gives the solution of equation (2.33).