Difference Equations Example D The system of equations (E – 1)uk + 2Evk 0, (4.349) -2uk + (E – 1)vk = a", %3D when solved for vk, gives the equation (E +1)²vk = (a – 1)a*, (4.350) the solution of which is (cı + c2k)(-1)* + (4.351) Uk = (a + 1)2" Now, from the second of equations (4.349), we have Ux = 1½(E – 1)vk – 1/2a*. (4.352) - The substitution of vk from equation (4.351) into the right-hand side of tion (4.351) gives equa- 2ak+1 Uk = (-1/½)[(2c1 + c2) + 2c2k](-1)* – (4.353) (a + 1)2 °

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Show me the steps of determine red

(a +1)2ak.
Difference Equations
Example D
The system of equations
(E – 1)uk + 2EVK = 0,
-2uk + (E – 1)Uk = a",
(4.349)
when solved for vk, gives the equation
(E +1)²vr = (a – 1)a*,
(4.350)
the solution of which is
а —
1
(c1 + c2k)(-1)* +
(4.351)
Uk
Now, from the second of equations (4.349), we have
Ux =1/½(E – 1)vk – 1/2a*.
(4.352)
The substitution of vk from equation (4.351) into the right-hand side of
tion (4.351) gives
equa-
2ak+1
Uk = (-/½)[(2c1 + c2) + 2c2k](-1)k
(4.353)
(a + 1)2
Transcribed Image Text:(a +1)2ak. Difference Equations Example D The system of equations (E – 1)uk + 2EVK = 0, -2uk + (E – 1)Uk = a", (4.349) when solved for vk, gives the equation (E +1)²vr = (a – 1)a*, (4.350) the solution of which is а — 1 (c1 + c2k)(-1)* + (4.351) Uk Now, from the second of equations (4.349), we have Ux =1/½(E – 1)vk – 1/2a*. (4.352) The substitution of vk from equation (4.351) into the right-hand side of tion (4.351) gives equa- 2ak+1 Uk = (-/½)[(2c1 + c2) + 2c2k](-1)k (4.353) (a + 1)2
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