Difference Equations Example E Consider the third-order difference equation Yk+3 – 7yk+2 + 16yk+1 – 12yk = k2*. (4.174) Its characteristic equation is p3 – 7r2 + 16r – 12 = (r – 2) (r – 3) = 0, (4.175) and the corresponding homogeneous solution is (H) (cı + c2k)2* + c33k, (4.176) || k2k is where c1, c2, and c3 are arbitrary constants. The family of Rk [2*, k2*]. Note that both members occur in the homogeneous solution; there- fore, we must multiply the family by k2 to obtain a new family that does not contain any function that appears in the homogeneous solution. The new family is [k2*, k³2*]. Thus, the particular solution is || (P) = (Ak² + Bk³)2*, (4.177) where A and B are to be determined. The substitution of equation (4.177) into equation (4.174) gives 2*(-8A+24B) + k2*(-24B) = k2*, (4.178) or 8A – 24B = 0, -24B = 1, (4.179) and A = -1/8, B = -1/24. (4.180) Therefore, the particular solution is (P) Yk -1/24(3 + k)k²2k, (4.181) and the general solution is Yk = (c1 + c2k)2* + c33k – 1/24(3 + k)k²2*. (4.182)

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Chapter2: Second-order Linear Odes
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Explain the determine

Difference Equations
Example E
Consider the third-order difference equation
7yk+2 + 16yk+1 – 12yk = k2*.
(4.174)
Yk+3
Its characteristic equation is
pi3 – 7n2 + 16r – 12 = (r – 2)2 (r – 3) = 0,
(4.175)
%3D
and the corresponding homogeneous solution is
(Н)
= (c1 + c2k)2* + c33*,
(4.176)
k2k is
where c1, c2, and c3 are arbitrary constants. The family of Rk
[2k, k2k]. Note that both members occur in the homogeneous solution; there-
fore, we must multiply the family by k? to obtain a new family that does
not contain any function that appears in the homogeneous solution. The new
family is [k?2*, k³2*]. Thus, the particular solution is
P) = (Ak² + Bk³)2*,
(4.177)
where A and B are to be determined. The substitution of equation (4.177)
into equation (4.174) gives
2* (-8A+24B) + k2* (-24B) = k2*,
(4.178)
or
8A – 24B = 0,
-24B = 1,
(4.179)
and
А
-1/s,
B = -/24.
(4.180)
Therefore, the particular solution is
(P)
:-/24(3 + k)k²2*,
(4.181)
and the general solution is
Yk = (c1 + c2k)2k + c33k – 1/24(3+ k)k²2*.
(4.182)
Transcribed Image Text:Difference Equations Example E Consider the third-order difference equation 7yk+2 + 16yk+1 – 12yk = k2*. (4.174) Yk+3 Its characteristic equation is pi3 – 7n2 + 16r – 12 = (r – 2)2 (r – 3) = 0, (4.175) %3D and the corresponding homogeneous solution is (Н) = (c1 + c2k)2* + c33*, (4.176) k2k is where c1, c2, and c3 are arbitrary constants. The family of Rk [2k, k2k]. Note that both members occur in the homogeneous solution; there- fore, we must multiply the family by k? to obtain a new family that does not contain any function that appears in the homogeneous solution. The new family is [k?2*, k³2*]. Thus, the particular solution is P) = (Ak² + Bk³)2*, (4.177) where A and B are to be determined. The substitution of equation (4.177) into equation (4.174) gives 2* (-8A+24B) + k2* (-24B) = k2*, (4.178) or 8A – 24B = 0, -24B = 1, (4.179) and А -1/s, B = -/24. (4.180) Therefore, the particular solution is (P) :-/24(3 + k)k²2*, (4.181) and the general solution is Yk = (c1 + c2k)2k + c33k – 1/24(3+ k)k²2*. (4.182)
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