[(E – 1)(E – 2)(E + 2)]¬'(1+k) = -[A(1 – A)(3 +A)]-'(1+k) %3D | A-1 (-3) (1 - A)(1+ 1½A) (1 + k) = (-/½)A-'(1+A+…)(1– /3A + )(1+k) (4.238) = (-1/½)A-'(1+2/3A)(1+ k) = (-1/½)A-'(s/3 + k) = (-1/3)[F/3k + 1/½k(k – 1)] %3D ... %3D = (-"/½)k – (/6)k². %3D ||

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determine red

Example E
The third-order equation
Yk+3 – Yk+2 – 4yk+1 +4yk = 1+k+2*
(4.235)
can be expressed in the operator form
f(E)yk = (E – 1)(E – 2)(E + 2)yk =1+ k + 2*.
(4.236)
Consequently, the homogeneous solution is
Yk(P) = c1 + c22k + c3(-2)*,
(4.237)
LINEAR DIFFERENCE EQUATIONS
145
where c1, c2, and c3 are constants. The particular solution will be determined
by separating the right-hand side of equation (4.236) into two terms, 1+ k
and 24, and letting f(E)-1 act on each. We have
[(E – 1)(E – 2)(E + 2)]='(1+k)
=-LA(1 – A)(3+ A))-"(1+ k)
A-1
(1+k)
(1 – A)(1+ /3A)
= (-1/½)A-'(1+ A+ …)(1 – 1/3A + · )(1+ k) (4.238)
= (-1/½)A-'(1+ 2/3A)(1+k)
= (-1/3)A-'(5/3+ k)
= (-1/3)[F/3k + 1/½k(k – 1)]
= (-7/½)k – (16)k².
Lines two and three on the right-hand side of equation (4.238) follow from
equations (4.209) and (4.212).
Also, from Theorem 4.9', we have
k2k-1
k2k
[(E – 2)(E – 1)(E + 2)]¯12*
(4.239a)
(1 · 4)(1!)
8
Therefore, the particular solution is
k2
:-7/gk
6
k2*
(P)
(4.239b)
8
consequently, the general solution of equation (4.235) is given by the expres-
sion
7
k2
k2k
Yk = C1 + c22* + c3(-2)*
(4.240)
8.
12
Transcribed Image Text:Example E The third-order equation Yk+3 – Yk+2 – 4yk+1 +4yk = 1+k+2* (4.235) can be expressed in the operator form f(E)yk = (E – 1)(E – 2)(E + 2)yk =1+ k + 2*. (4.236) Consequently, the homogeneous solution is Yk(P) = c1 + c22k + c3(-2)*, (4.237) LINEAR DIFFERENCE EQUATIONS 145 where c1, c2, and c3 are constants. The particular solution will be determined by separating the right-hand side of equation (4.236) into two terms, 1+ k and 24, and letting f(E)-1 act on each. We have [(E – 1)(E – 2)(E + 2)]='(1+k) =-LA(1 – A)(3+ A))-"(1+ k) A-1 (1+k) (1 – A)(1+ /3A) = (-1/½)A-'(1+ A+ …)(1 – 1/3A + · )(1+ k) (4.238) = (-1/½)A-'(1+ 2/3A)(1+k) = (-1/3)A-'(5/3+ k) = (-1/3)[F/3k + 1/½k(k – 1)] = (-7/½)k – (16)k². Lines two and three on the right-hand side of equation (4.238) follow from equations (4.209) and (4.212). Also, from Theorem 4.9', we have k2k-1 k2k [(E – 2)(E – 1)(E + 2)]¯12* (4.239a) (1 · 4)(1!) 8 Therefore, the particular solution is k2 :-7/gk 6 k2* (P) (4.239b) 8 consequently, the general solution of equation (4.235) is given by the expres- sion 7 k2 k2k Yk = C1 + c22* + c3(-2)* (4.240) 8. 12
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