Using trigonometric substitution, this latter equation can be written 1 Yk = C1 cos(kø) + C2 sin(kø) + ERi sin(k – i)ø. (3.96) sin o i=1

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Example B
Consider the inhomogeneous equation
Yk+1 –
2yk coS O + Yk–1= Rµ, sin ø + 0,
(3.89)
96
Difference Equations
where o is a constant and Rk is an arbitrary function of k. The fundamental
set of solutions to the homogeneous equation is
.(1)
(2)
cos(kø), y = sin(kø).
(3.90)
%3D
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)ø]AC1(k) + [sin(k + 2)ø]AC2(k) = Rx+1-
(3.92)
Solving for AC1(k) and AC2(k) gives
Rk+1 sin(k +1)O
AC: (k) =
sin ø
(3.93)
Rk+1 COs(k + 1)¢
sin ø
AC2(k) =
Therefore,
k
C1 (k) = C1 – E
R; sin(iø)
sin ø
i=1
(3.94)
k
R; cos(iø)
C2(k) = C2 +
sin ø
i=1
where C1
into equation (3.91) and retaining the arbitrary constants gives the general
solution to equation (3.89):
C2 are arbitrary constants. Substitution of equations (3.94)
Yk = C1 cos(kø) + C2 cos(kø) –
cos(kø)
sin ø
ERi sin(iø)
i=1
k
sin(kø)
Rị cos(io). (3.95)
sin ø
i=1
Using trigonometric substitution, this latter equation can be written
1
Yk = C1 cos(kø)+C2 sin(kø) +
R; sin(k – i)ø.
(3.96)
sin o
i=1
Transcribed Image Text:Example B Consider the inhomogeneous equation Yk+1 – 2yk coS O + Yk–1= Rµ, sin ø + 0, (3.89) 96 Difference Equations where o is a constant and Rk is an arbitrary function of k. The fundamental set of solutions to the homogeneous equation is .(1) (2) cos(kø), y = sin(kø). (3.90) %3D 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)ø]AC1(k) + [sin(k + 2)ø]AC2(k) = Rx+1- (3.92) Solving for AC1(k) and AC2(k) gives Rk+1 sin(k +1)O AC: (k) = sin ø (3.93) Rk+1 COs(k + 1)¢ sin ø AC2(k) = Therefore, k C1 (k) = C1 – E R; sin(iø) sin ø i=1 (3.94) k R; cos(iø) C2(k) = C2 + sin ø i=1 where C1 into equation (3.91) and retaining the arbitrary constants gives the general solution to equation (3.89): C2 are arbitrary constants. Substitution of equations (3.94) Yk = C1 cos(kø) + C2 cos(kø) – cos(kø) sin ø ERi sin(iø) i=1 k sin(kø) Rị cos(io). (3.95) sin ø i=1 Using trigonometric substitution, this latter equation can be written 1 Yk = C1 cos(kø)+C2 sin(kø) + R; sin(k – i)ø. (3.96) sin o i=1
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