(c) 2yk+1 5yk+1Yk + 2y = 2, |

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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2.21. Plot the following curves, locate any fixed points, and obtain asymp-
totic expansions to their solution as k
Vk + a²,
Yk (y%+3)
3y+1
(c) 2y+1 – 5yk+1Yk + 2y% = 2,
(a) Yk+1 =
real number,
a =
(b) Ук+1
(d) Yk+1 =
ayk+1
Yk+a ?
5y-6yk+2
6y -8yk+3'
(e) Yk+1 =
(f) 5yk+1
5y +6yk+19
y+5
(g) yk+1
Yk + 6.
Transcribed Image Text:2.21. Plot the following curves, locate any fixed points, and obtain asymp- totic expansions to their solution as k Vk + a², Yk (y%+3) 3y+1 (c) 2y+1 – 5yk+1Yk + 2y% = 2, (a) Yk+1 = real number, a = (b) Ук+1 (d) Yk+1 = ayk+1 Yk+a ? 5y-6yk+2 6y -8yk+3' (e) Yk+1 = (f) 5yk+1 5y +6yk+19 y+5 (g) yk+1 Yk + 6.
2.2.3
Example C
The inhomogeneous equation
Yk+1 – kyj = 1
(2.21)
has Pk
k and qk
1. Since
%3D
k
IIi = k!,
(2.22)
i=1
then
k-1
k-1
1
qi
Pr
(2.23)
i=1
r=1
and equation (2.21) has the solution
k-1
1
A(k – 1)! + (k – 1)!
(2.24)
Yk =
i=1
where A is an arbitrary constant.
IWI
Transcribed Image Text:2.2.3 Example C The inhomogeneous equation Yk+1 – kyj = 1 (2.21) has Pk k and qk 1. Since %3D k IIi = k!, (2.22) i=1 then k-1 k-1 1 qi Pr (2.23) i=1 r=1 and equation (2.21) has the solution k-1 1 A(k – 1)! + (k – 1)! (2.24) Yk = i=1 where A is an arbitrary constant. IWI
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