3.5.1 Example A Assume that we know that one solution of the equation Yk+2 – k(k + 1)yk (3.146) LINEAR DIFFERENCE EQUATIONS 103 (1) is y From equation (3.146), we see that (k – 1)!. We wish to find a second linearly independent solution. Ik = -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)*+[(k – 1)!?(2k), k+1 (3.150) thus showing that Yk (1) (2) and are linearly independent.
3.5.1 Example A Assume that we know that one solution of the equation Yk+2 – k(k + 1)yk (3.146) LINEAR DIFFERENCE EQUATIONS 103 (1) is y From equation (3.146), we see that (k – 1)!. We wish to find a second linearly independent solution. Ik = -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)*+[(k – 1)!?(2k), k+1 (3.150) thus showing that Yk (1) (2) and are linearly independent.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Explan the determine
!,
(3.149)
where c1 and c2 are arbitrary constants. Note that the Casoratian is
C(k) = (-1)*+1[(k – 1)!]²(2k),
(3.150)
(1)
(2)
thus showing that y and y are linearly independent.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb26a9dbb-23fe-4152-bd29-b95e4a6b8149%2F7174fd87-bed9-42a8-a5d4-a0dc5cb2ddb0%2F145drqh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3.5.1
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
From equation (3.146), we see that
(k – 1)!. We wish to find a second linearly independent solution.
Ik
-k(k + 1).
(3.147)
Substitution of this into equations (3.111) and (3.113) gives
(2)
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)
(1)
(2)
thus showing that y and y are linearly independent.
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