(k + 4)yk+2 + Yk+1 – (k + 1)yk = 0. %3D | These functions have the Casoratian (-1)*+1 (k + 2)(k + 3)(k + 4)" C(k + 1) =
(k + 4)yk+2 + Yk+1 – (k + 1)yk = 0. %3D | These functions have the Casoratian (-1)*+1 (k + 2)(k + 3)(k + 4)" C(k + 1) =
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.6: Variation
Problem 8E
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Show me the step of determine please
![Example C
The second-order, inhomogeneous equation
(k + 4)yk+2 + Yk+1 – (k + 1)yk
1
(3.97)
has the following two solutions:
.(1)
1
(3.98)
(k +1)(k +2)'
(-1)k+1(2k + 3)
4(k + 1)(k + 2)
(2)
(3.99)
to its associated homogeneous equation
(k + 4)yk+2 + Yk+1
(k + 1)yk = 0.
(3.100)
These functions have the Casoratian
(-1)k+1
(k +2)(k + 3)(k + 4)'
C(k + 1) =
(3.101)
The particular solution takes the form
Yk = c1 (k)y + c2(k)y.
(1)
(3.102)
Direct calculation shows that c1(k) and c2(k) satisfy the equations
Acı (k) = 1/4(2k + 5),
Ac2(k) = (-1)*+1.
(3.103)
Summing these expressions gives
k
c1 (k) = 14 (2i + 5) + c1 = /¼(k +1)² + c1
(3.104)
i=0
and
k
c2 (k) = -(-1)* = -\½[1+ (-1)*)+ c2,
(3.105)
i=0
Ст and
(3.104), and (3.105) into equation (3.102) and dropping the terms that contain
the arbitrary constants gives
where
c2 are arbitrary constants. Substituting equations (3.98), (3.99),
k +1
Yk
4(k + 2)
(2k + 3)[1+ (-1)*]
8(k + 1)(k + 2)
(3.106)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9396dbf6-1a2a-41f6-9ea8-33e71c92a8b7%2F3cc9f48c-195d-4e93-81c0-95388df4ddd8%2F0ajobvo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example C
The second-order, inhomogeneous equation
(k + 4)yk+2 + Yk+1 – (k + 1)yk
1
(3.97)
has the following two solutions:
.(1)
1
(3.98)
(k +1)(k +2)'
(-1)k+1(2k + 3)
4(k + 1)(k + 2)
(2)
(3.99)
to its associated homogeneous equation
(k + 4)yk+2 + Yk+1
(k + 1)yk = 0.
(3.100)
These functions have the Casoratian
(-1)k+1
(k +2)(k + 3)(k + 4)'
C(k + 1) =
(3.101)
The particular solution takes the form
Yk = c1 (k)y + c2(k)y.
(1)
(3.102)
Direct calculation shows that c1(k) and c2(k) satisfy the equations
Acı (k) = 1/4(2k + 5),
Ac2(k) = (-1)*+1.
(3.103)
Summing these expressions gives
k
c1 (k) = 14 (2i + 5) + c1 = /¼(k +1)² + c1
(3.104)
i=0
and
k
c2 (k) = -(-1)* = -\½[1+ (-1)*)+ c2,
(3.105)
i=0
Ст and
(3.104), and (3.105) into equation (3.102) and dropping the terms that contain
the arbitrary constants gives
where
c2 are arbitrary constants. Substituting equations (3.98), (3.99),
k +1
Yk
4(k + 2)
(2k + 3)[1+ (-1)*]
8(k + 1)(k + 2)
(3.106)
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