(A[cos(6) - 3 cos(3) + 2] + B[– sin(6) +3 sin(3)] = 2, A[sin(6) + 3 sin(3)] + B[cos(6) – 3 cos(3) + 2] = 0. %3D %3D

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ISBN:9780470458365
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Chapter2: Second-order Linear Odes
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LINEAR DIFFERENCE EQUATIONS
137
Example D
The second-order difference equation
Yk+2 – 3yk+1 + 2yk = 2 sin(3k)
(4.164)
has the homogeneous solution
yk(H)
= c1 + c22*,
(4.165)
2 sin(3k) is
where c1 and c2 are arbitrary constants. The family of Rk
[sin(3k), cos(3k)] and contains no term that appears in the homogeneous so-
lution. Therefore, the particular solution to equation (4.164) can be written
(P)
= A sin(3k) + B cos(3k).
(4.166)
Substitution of this last equation into equation (4.164) and using the trigono-
metric relations
sin(01 + 02)
sin 01 cos 02 + cos 01 sin O2,
(4.167)
cos(01 + 02)
cos 01 cos 02 – sin 01 sin 02
gives
{A[cos(6) – 3 cos(3) + 2] + B[– sin(6) + 3 sin(3)]} sin(3k)
+ {A[sin(6) – 3 sin(3)] + B[cos(6) – 3 cos(3) + 2]} cos(3k)
= 2 sin(3k). (4.168)
Therefore, A and B must satisfy the two linear equations
(A[cos(6) – 3 cos(3) + 2] + B[– sin(6) + 3 sin(3)] = 2,
%3|
|
(4.169)
A[sin(6) + 3 sin(3)] + B[cos(6) – 3 cos(3) + 2] = 0.
Evaluating the trigonometric functions (using the fact that their arguments
are in radians) gives
cos(3) = -0.991, sin(3) = 0.139,
(4.170)
cos(6) = 0.964, sin(6) = -0.276.
%3D
Thus, the solution to equation (4.169) for A and B is
A = 0.337,
B = 0.039,
(4.171)
and the particular solution can be written
(P)
0.337 sin(3k) + 0.039 cos(3k).
(4.172)
Finally, the general solution of equation (4.164) is
Yk = C1 + c22* +0.337 sin(3k)+0.039 cos(3k).
(4.173)
Transcribed Image Text:LINEAR DIFFERENCE EQUATIONS 137 Example D The second-order difference equation Yk+2 – 3yk+1 + 2yk = 2 sin(3k) (4.164) has the homogeneous solution yk(H) = c1 + c22*, (4.165) 2 sin(3k) is where c1 and c2 are arbitrary constants. The family of Rk [sin(3k), cos(3k)] and contains no term that appears in the homogeneous so- lution. Therefore, the particular solution to equation (4.164) can be written (P) = A sin(3k) + B cos(3k). (4.166) Substitution of this last equation into equation (4.164) and using the trigono- metric relations sin(01 + 02) sin 01 cos 02 + cos 01 sin O2, (4.167) cos(01 + 02) cos 01 cos 02 – sin 01 sin 02 gives {A[cos(6) – 3 cos(3) + 2] + B[– sin(6) + 3 sin(3)]} sin(3k) + {A[sin(6) – 3 sin(3)] + B[cos(6) – 3 cos(3) + 2]} cos(3k) = 2 sin(3k). (4.168) Therefore, A and B must satisfy the two linear equations (A[cos(6) – 3 cos(3) + 2] + B[– sin(6) + 3 sin(3)] = 2, %3| | (4.169) A[sin(6) + 3 sin(3)] + B[cos(6) – 3 cos(3) + 2] = 0. Evaluating the trigonometric functions (using the fact that their arguments are in radians) gives cos(3) = -0.991, sin(3) = 0.139, (4.170) cos(6) = 0.964, sin(6) = -0.276. %3D Thus, the solution to equation (4.169) for A and B is A = 0.337, B = 0.039, (4.171) and the particular solution can be written (P) 0.337 sin(3k) + 0.039 cos(3k). (4.172) Finally, the general solution of equation (4.164) is Yk = C1 + c22* +0.337 sin(3k)+0.039 cos(3k). (4.173)
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