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)
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)
Chapter7: Matrices And Determinants
Section: Chapter Questions
Problem 13PS
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