(a) yk+1 + Yk = 2+ k,

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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2.1. Solve the following difference equations:
Yk+1 + Yk
2+ k,
a
(b) Yk+1 – 2yk = k°,
(с) Ук+1
- 3k yk = 0,
Yk = 1/k(k +1),
(e) Yk+1+ Yk = 1/k(k + 1),
(f) (k +2)yk+1 – (k +1)yk = 5 + 2k – k²,
(g) Yk+1 + Yk = k + 2 · 3k,
ke*,
(d) Yk+1
(h) Yk+1 – Yk
(1) Yk+1 – aa2kyk
(G) Yk+1
Bak
ayk
cos(bk),
Transcribed Image Text:2.1. Solve the following difference equations: Yk+1 + Yk 2+ k, a (b) Yk+1 – 2yk = k°, (с) Ук+1 - 3k yk = 0, Yk = 1/k(k +1), (e) Yk+1+ Yk = 1/k(k + 1), (f) (k +2)yk+1 – (k +1)yk = 5 + 2k – k², (g) Yk+1 + Yk = k + 2 · 3k, ke*, (d) Yk+1 (h) Yk+1 – Yk (1) Yk+1 – aa2kyk (G) Yk+1 Bak ayk cos(bk),
2.2.7 Example G
The inhomogeneous equation
Yk+1 – Yk =
= ek
(2.43)
has
1,
Ik = ek.
(2.44)
Pk =
Therefore,
k-1
Il Pi
(2.45)
= 1
i=1
and
k-1
i
k-1
ek
- e
qi
(2.46)
е — 1
i=1
r=1
Thus, the general solution of equation (2.43) is
ek
A +
е — I
Yk =
(2.47)
where A is an arbitrary constant.
Transcribed Image Text:2.2.7 Example G The inhomogeneous equation Yk+1 – Yk = = ek (2.43) has 1, Ik = ek. (2.44) Pk = Therefore, k-1 Il Pi (2.45) = 1 i=1 and k-1 i k-1 ek - e qi (2.46) е — 1 i=1 r=1 Thus, the general solution of equation (2.43) is ek A + е — I Yk = (2.47) where A is an arbitrary constant.
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