The system of difference equations for uz and vk given by (2E – 3)uk + 5vk = 2, 2uk + (E – 2)vk = 7 (4. - can be solved for uk; it satisfies the second-order equation ((2E? – 7E – 4)uk = –37, (4. %3D and has the general solution - ca(-1/6)k 1 co4k 1 37/

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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4.8.3 Example C
The system of difference equations for uk and vk given by
(2E - 3)uk + 5uк — 2,
(4.345)
2uk + (E – 2)vk
= 7
can be solved for uk; it satisfies the second-order equation
((2E? – 7E – 4)Uk = -37,
(4.346)
and has the general solution
c1(-2)* + c24* + 37/9.
(4.347)
Uk =
Using the first of equations (4.345), we obtain for v the result
Vk = -1/5(2E – 3)uk + 2/5 = 11/9 + 4/5c1(-1/2)* – c24*.
(4.348)
Transcribed Image Text:4.8.3 Example C The system of difference equations for uk and vk given by (2E - 3)uk + 5uк — 2, (4.345) 2uk + (E – 2)vk = 7 can be solved for uk; it satisfies the second-order equation ((2E? – 7E – 4)Uk = -37, (4.346) and has the general solution c1(-2)* + c24* + 37/9. (4.347) Uk = Using the first of equations (4.345), we obtain for v the result Vk = -1/5(2E – 3)uk + 2/5 = 11/9 + 4/5c1(-1/2)* – c24*. (4.348)
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