Example B Consider the inhomogeneous equation Yk+1 – 2yk coS Ø + Yk-1 = Rx, sin ø # 0, %3D
Example B Consider the inhomogeneous equation Yk+1 – 2yk coS Ø + Yk-1 = Rx, sin ø # 0, %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Explain the determine
![Example B
Consider the inhomogeneous equation
Yk+1 – 2yk coS Ø + Yk–1 = Rx, sinø # 0,
(3.89)
%3D
96
Difference Equations
where o is a constant and R is an arbitrary function of k. The fundamental
set of solutions to the homogeneous equation is
.(1)
%3D
= cos(kø), Yk
(2)
sin(kø).
(3.90)
The unknown functions in the particular solution expression
Yk = C1(k) cos(kø) + C2(k) sin(kø)
(3.91)
satisfy the following equations:
[cos(k+1)ø]AC1(k) + [sin(k + 1)ø]AC2(k) = 0,
[cos(k+ 2)¢]AC1(k) + [sin(k + 2)ø]AC2(k) = Rx+1-
(3.92)
%3D
Solving for AC1(k) and AC2(k) gives
AC: (k) :
Rk+1 sin(k +1)ø
sin o
(3.93)
Rk+1 Cos(k + 1)¢
sin ø
AC2(k) =
Therefore,
k
R; sin(iø)
C1(k) = C1 – E
sin ø
i=1
(3.94)
k
C2(k) = C2 +
R; cos(iø)
sin ø
i=1
where C1 and C2 are arbitrary constants. Substitution of equations (3.94)
into equation (3.91) and retaining the arbitrary constants gives the general
solution to equation (3.89):
k
cos(kø)
sin ø
R; sin(iø)
Yk =
C1 cos(kø) + C2 cos(kø) -
i=1
k
sin(kø)
sin ø
R; cos(iø). (3.95)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0083eca2-b9bf-4164-b4a6-ff40b8bc4111%2Fd621f0f6-c7c0-4d8f-858f-9b19f230168b%2Frhy0my8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example B
Consider the inhomogeneous equation
Yk+1 – 2yk coS Ø + Yk–1 = Rx, sinø # 0,
(3.89)
%3D
96
Difference Equations
where o is a constant and R is an arbitrary function of k. The fundamental
set of solutions to the homogeneous equation is
.(1)
%3D
= cos(kø), Yk
(2)
sin(kø).
(3.90)
The unknown functions in the particular solution expression
Yk = C1(k) cos(kø) + C2(k) sin(kø)
(3.91)
satisfy the following equations:
[cos(k+1)ø]AC1(k) + [sin(k + 1)ø]AC2(k) = 0,
[cos(k+ 2)¢]AC1(k) + [sin(k + 2)ø]AC2(k) = Rx+1-
(3.92)
%3D
Solving for AC1(k) and AC2(k) gives
AC: (k) :
Rk+1 sin(k +1)ø
sin o
(3.93)
Rk+1 Cos(k + 1)¢
sin ø
AC2(k) =
Therefore,
k
R; sin(iø)
C1(k) = C1 – E
sin ø
i=1
(3.94)
k
C2(k) = C2 +
R; cos(iø)
sin ø
i=1
where C1 and C2 are arbitrary constants. Substitution of equations (3.94)
into equation (3.91) and retaining the arbitrary constants gives the general
solution to equation (3.89):
k
cos(kø)
sin ø
R; sin(iø)
Yk =
C1 cos(kø) + C2 cos(kø) -
i=1
k
sin(kø)
sin ø
R; cos(iø). (3.95)
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