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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.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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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 Yk (k – 1)!. We wish to find a second linearly independent solution. From equation (3.146), we see that Ik = -k(k + 1). (3.147) Substitution of this into equations (3.111) and (3.113) gives = (-1)* (k – 1)!. (2) (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) thus showing that Yk * and y are linearly independent.
Transcribed Image Text: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 Yk (k – 1)!. We wish to find a second linearly independent solution. From equation (3.146), we see that Ik = -k(k + 1). (3.147) Substitution of this into equations (3.111) and (3.113) gives = (-1)* (k – 1)!. (2) (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) thus showing that Yk * and y are linearly independent.
Therefore, k-1 C(k) = AQk = A 1| 4i, (3.111) i=1 where A is an arbitrary, nonzero constant. Now (1)(2) (2),(1) (2) C(k) .(1),,(1) Yk Yk+1 Yk Yk+1 Yk Yk+1 (3.112) (1)„(1) .(1) Applying A-1 to both sides gives C(k) (1),(1) Yk Yk+1 (1)A- Aye (2) .(1) -1 Yk = Yk A (3.113) (1),,(1) Yk Yk+1
Transcribed Image Text:Therefore, k-1 C(k) = AQk = A 1| 4i, (3.111) i=1 where A is an arbitrary, nonzero constant. Now (1)(2) (2),(1) (2) C(k) .(1),,(1) Yk Yk+1 Yk Yk+1 Yk Yk+1 (3.112) (1)„(1) .(1) Applying A-1 to both sides gives C(k) (1),(1) Yk Yk+1 (1)A- Aye (2) .(1) -1 Yk = Yk A (3.113) (1),,(1) Yk Yk+1
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