Example A Assume that we know that one solution of the equation Yk+2 – k(k + 1)yk = 0 (3.146) LINEAR DIFFERENCE EQUATIONS 103 (1) is y = (k – 1)!. We wish to find a second linearly independent solution. From equation (3.146), we see that qk = -k(k + 1). (3.147) Substitution of this into equations (3.111) and (3.113) gives Yk = (-1)*(k – 1)!. (3.148) Therefore, the general solution to equation (3.146) is Yk = [c1 +c2(-1)*](k – 1)!, (3.149) where C1 and C2 are arbitrary constants. Note that the Casoratian is C(k) = (-1)*+1(k -– 1)!]²(2k), (3.150) (2) (1) and Yk thus showing that Yk are linearly independent.

Example A Assume that we know that one solution of the equation Yk+2 – k(k + 1)yk = 0 (3.146) LINEAR DIFFERENCE EQUATIONS 103 (1) is y = (k – 1)!. We wish to find a second linearly independent solution. From equation (3.146), we see that qk = -k(k + 1). (3.147) Substitution of this into equations (3.111) and (3.113) gives Yk = (-1)(k – 1)!. (3.148) Therefore, the general solution to equation (3.146) is Yk = [c1 +c2(-1)](k – 1)!, (3.149) where C1 and C2 are arbitrary constants. Note that the Casoratian is C(k) = (-1)*+1(k -– 1)!]²(2k), (3.150) (2) (1) and Yk thus showing that Yk are linearly independent.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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