Example D The second-order difference equation whose fundamental system of solutions is (1) = k, -hk = e (4.102) %3D is Yk k -hk e-h(k+1) k +1 k + 2 e-h(k+2) Yk+1 = 0. (4.103) Yk+2 Expanding the determinant and simplifying the resulting expression gives [k(1 – e-h) + 1]yk+2 – [k(1 – e-2h) + 2]yk+1 +e- [k(1 – e-h) +2 – e-"]yk = 0. (4.104)

College Algebra
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ISBN:9781337282291
Author:Ron Larson
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Chapter7: Matrices And Determinants
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Explain the determine

LINEAR DIFFERENCE EQUATIONS
129
Example D
The second-order difference equation whose fundamental system of solutions
is
(1)
= k,
(2)
-hk
(4.102)
Yk
= e
is
k
-hk
e
Yk
Yk+1
k + 1
e-h(k+1)
= 0.
(4.103)
Yk+2 k +2 e-h(k+2)
Expanding the determinant and simplifying the resulting expression gives
[k(1 – e-h) + 1]yk+2 – [k(1 – e-2h)+ 2]yk+1
+e-[k(1 – e-h) + 2 – e-]yk = 0. (4.104)
Transcribed Image Text:LINEAR DIFFERENCE EQUATIONS 129 Example D The second-order difference equation whose fundamental system of solutions is (1) = k, (2) -hk (4.102) Yk = e is k -hk e Yk Yk+1 k + 1 e-h(k+1) = 0. (4.103) Yk+2 k +2 e-h(k+2) Expanding the determinant and simplifying the resulting expression gives [k(1 – e-h) + 1]yk+2 – [k(1 – e-2h)+ 2]yk+1 +e-[k(1 – e-h) + 2 – e-]yk = 0. (4.104)
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