PROBLEMS Section 2.2 2.1. Solve the following difference equations: (a) Yk+1+ Yk = 2+ k, (b) ук+1 — 2ук — k3, - (c) Yk+1 3kyk = 0, - (d) Yk+1 – Yk = 1/k(k+1), (e) Yk+1+ Yk = 1/k(k+1), (f) (k+2)yk+1 – (k + 1)yk = 5 + 2* – k², (g) Yk+1+Yk = k + 2 · 3k, (h) Yk+1 – Yk = kek, = Bak®, (1) Yk+1 – aa2k, (J) Yk+1 – aYk = Cos(bk), - (k) Yk+1+Yk = (-1)*, (1) Yk+1 k+1 Yk k = k. -

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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PROBLEMS
Section 2.2
2.1. Solve the following difference equations:
(a) Yk+1+ Yk = 2+ k,
(b) Yk+1 – 2yk = k³,
(c) Yk+1
-
3kyk = 0,
(d) Yk+1 – 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,
kek,
(h) Yk+1 – Yk =
Bak?
(j) Yk+1 – ayk = cos(bk),
(k) Yk+1 +Yk = (-1)k,
* = k.
(i) Yk+1
(1)
Yk+1
k+1
Transcribed Image Text:PROBLEMS Section 2.2 2.1. Solve the following difference equations: (a) Yk+1+ Yk = 2+ k, (b) Yk+1 – 2yk = k³, (c) Yk+1 - 3kyk = 0, (d) Yk+1 – 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, kek, (h) Yk+1 – Yk = Bak? (j) Yk+1 – ayk = cos(bk), (k) Yk+1 +Yk = (-1)k, * = k. (i) Yk+1 (1) Yk+1 k+1
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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