tions, and we conclude that V1(r:) $1(r:) (4.336) Consequently, the solution to the system given by equation (4.329) is vk and Uk as expressed in equations (4.333) and (4.334), with the conditions between

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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functions, and we conclude that
/1(ri)
-Ci.
(4.336)
01(ri)
Consequently, the solution to the system given by equation (4.329) is vk and
Uk as expressed in equations (4.333) and (4.334), with the conditions between
the coefficients given by equation (4.336). Thus, the solution contains only n
arbitrary constants.
4.8.1
Example A
Consider the equations
(4E – 17)ur + (E – 4)vk = 0,
(4.337)
(2E – 1)uk + (E – 2)vk
0.
Solving for v gives
(E² – 8E + 15)vk = (E – 3)(E – 5)vk = 0,
(4.338)
LINEAR DIFFERENCE EQUATIONS
159
which has the solution
Vk =
c13k + c25k.
(4.339)
Using equation (4.336), we obtain
3 – 4
C1
- 17
C1
С1,
12
(4.340a)
1
-C2 = –
17
1
C2
20
3
Therefore, the solution for Uk
is
Uk
713k + ē25k = -1/sc,3* – 1/3c25*.
(4.340b)
Transcribed Image Text:functions, and we conclude that /1(ri) -Ci. (4.336) 01(ri) Consequently, the solution to the system given by equation (4.329) is vk and Uk as expressed in equations (4.333) and (4.334), with the conditions between the coefficients given by equation (4.336). Thus, the solution contains only n arbitrary constants. 4.8.1 Example A Consider the equations (4E – 17)ur + (E – 4)vk = 0, (4.337) (2E – 1)uk + (E – 2)vk 0. Solving for v gives (E² – 8E + 15)vk = (E – 3)(E – 5)vk = 0, (4.338) LINEAR DIFFERENCE EQUATIONS 159 which has the solution Vk = c13k + c25k. (4.339) Using equation (4.336), we obtain 3 – 4 C1 - 17 C1 С1, 12 (4.340a) 1 -C2 = – 17 1 C2 20 3 Therefore, the solution for Uk is Uk 713k + ē25k = -1/sc,3* – 1/3c25*. (4.340b)
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