Explain the determine LINEAR DIFFERENCE EQUATIONS137Example DThe second-order difference equationYk+2 – 3yk+1 + 2yk = 2 sin(3k)(4.164)has the homogeneous solutionyk(H)= c1 + c22*,(4.165)2 sin(3k) iswhere 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(Р)A sin(3k) + B cos(3k).(4.166)Substitution of this last equation into equation (4.164) and using the trigono-metric relationssin(01 + 02)sin 01 cos 02+ cos 01 sin 02,(4.167)cos(01 + 02) = cos 01 cos 02 - sin 01 sin 02gives{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 equationsA[cos(6) – 3 cos(3) + 2] + B[– sin(6) + 3 sin(3)] = 2,(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 argumentsare in radians) givescos(3) = -0.991,sin(3) = 0.139,(4.170)cos(6) = 0.964, sin(6) = –0.276.Thus, the solution to equation (4.169) for A and B isA = 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) isYk = C1 + c22* + 0.337 sin(3k) + 0.039 cos(3k).(4.173)

Name: Explain the determine LINEAR DIFFERENCE EQUATIONS137Example DThe secon
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Description: Explain the determine LINEAR DIFFERENCE EQUATIONS137Example DThe second-order difference equationYk+2 – 3yk+1 + 2yk = 2 sin(3k)(4.164)has the homogeneous solutionyk(H)= c1 + c22*,(4.165)2 sin(3k) iswhere c1 and c2 are arbitrary constants. The family