ssume that we know that one solution of the equation Yk+2 – k(k + 1)yk %3D | LINEAR DIFFERENCE EQUATIONS „(1) = (k – 1)!. We wish to find a second linearly independen rom equation (3.146), we see that Ik -k(k + 1). ubstitution of this into equations (3.111) and (3.113) gives (2) y = (-1)* (k – 1)!. 'herefore, the general solution to equation (3.146) is Yk = [c1 + c2(-1)*](k – 1)!, here C1 and C2 are arbitrary constants. Note that the Casoratian C(k) = (-1)*+'[(k – 1)!]²(2k), ,(1) (2) hus showing that and are linearly independent.
ssume that we know that one solution of the equation Yk+2 – k(k + 1)yk %3D | LINEAR DIFFERENCE EQUATIONS „(1) = (k – 1)!. We wish to find a second linearly independen rom equation (3.146), we see that Ik -k(k + 1). ubstitution of this into equations (3.111) and (3.113) gives (2) y = (-1)* (k – 1)!. 'herefore, the general solution to equation (3.146) is Yk = [c1 + c2(-1)*](k – 1)!, here C1 and C2 are arbitrary constants. Note that the Casoratian C(k) = (-1)*+'[(k – 1)!]²(2k), ,(1) (2) hus showing that 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
Problem 1RQ
Related questions
Question
Explain the determaine please and use equations 3.111 and 3.113
!,
(3.149)
where c1 and c2 are arbitrary constants. Note that the Casoratian is
C(k) = (-1)*+(k – 1)!1°(2k),
k+1
(3.150)
thus showing that y and Y
.(1)
(2)
are linearly independent.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0083eca2-b9bf-4164-b4a6-ff40b8bc4111%2Feb1ee4c6-33cd-40a2-91bf-970f6a255a8a%2Fs6gwqi_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 Yk
= (k – 1)!. We wish to find a second linearly independent solution.
From equatiom (3.146), we see that
= -k(k + 1).
(3.147)
Substitution of this into equations (3.111) and (3.113) gives
(2)
Y = (-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)!1°(2k),
k+1
(3.150)
thus showing that y and Y
.(1)
(2)
are linearly independent.
Transcribed Image Text:Therefore,
k-1
C(k) = AQk = A II 4i,
(3.111)
i=1
where A is an arbitrary, nonzero constant. Now
(1)(2)
(2),(1)
(2)
Yk
.(1)
C(k)
(3.112)
(1)(1)
Yk Yk+1
(1) „(1)
Yk Yk+1
Applying A-1 to both sides gives
C(k)
(1),(1)
Yk Yk+1
(2)
(1)
Yk
-1
Ay
(3.113)
(1),,(1)
Yk Yk+1
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