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)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F284e70ee-eed6-4dfe-aa18-32502476453c%2Fff23cb8e-f826-40d8-ae1a-eddd794eb3cd%2F2u7j2u_processed.jpeg&w=3840&q=75)
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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