k c2(k) = -(-1)* = -1½[1+ (–1)*] + c2, i=0

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 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)*
(2)
(-1)k+1|
k+1(2k + 3)
(3.99)
4(k + 1)(k + 2)
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
(1)
Yk = c1 (k)y + c2(k)y,.
(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) +¢1 = /4(k + 1)² + c1
(3.104)
i=0
and
k
c2(k)
E(-1)' = -1/½[1 + (-1)*]+c2,
(3.105)
i=0
where ci
and
c2 are arbitrary constants. Substituting equations (3.98), (3.99),
(3.104), and (3.105) into equation (3.102) and dropping the terms that contain
the arbitrary constants gives
k +1
4(k + 2)
(2k + 3)[1+(-1)*]
8(k + 1)(k + 2)
(3.106)
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)* (2) (-1)k+1| k+1(2k + 3) (3.99) 4(k + 1)(k + 2) 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 (1) Yk = c1 (k)y + c2(k)y,. (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) +¢1 = /4(k + 1)² + c1 (3.104) i=0 and k c2(k) E(-1)' = -1/½[1 + (-1)*]+c2, (3.105) i=0 where ci and c2 are arbitrary constants. Substituting equations (3.98), (3.99), (3.104), and (3.105) into equation (3.102) and dropping the terms that contain the arbitrary constants gives k +1 4(k + 2) (2k + 3)[1+(-1)*] 8(k + 1)(k + 2) (3.106)
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