where z(t) has the expansion z(t) =t+ Ao + (2.170) Substitution of this result into equation (2.169) gives, respectively, for the left- and right-hand sides z(t)z(2t) = 2t² + 3Aot + (Až + /½A1) + ¾A2 + 3½A0A1 %3D (2.171) 17/s A3 + 5/¾A0A2 +½A? t2 + +...

Advanced Engineering Mathematics
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ISBN:9780470458365
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determaine and the Q is complete
2.8.1 Example A
The equation
YKYK+1 = 2y +1
(2.164)
can be written as
Yk+1 = 2yk +1/yk.
(2.165)
In this form, it is easily seen that the curve in the (yk, Yk+1) plane is a hyper-
bola with asymptotes yk = 0 and yk+1 = 2yk. There are no real fixed points
since a? = 2a² + 1. Consequently, the curve does not intersect yk+1 = Yk. Se
Figure 2.5.
The asymptotic behavior (i.e., k → 0) is determined by
Yk+1 = 2yk;
(2.166)
the solution of which is
Yk = A2k,
(2.167)
where A is an arbitrary constant. Therefore, if we let
t = A2k and z(t) = Yk;
(2.168)
equation (2.164) becomes
z(t)z(2t) = 2z(t)² + 1,
(2.169)
where 2(t) has the expansion
A1
2(t) = t + Ao +
A2
+
t2
Аз
+...
13
(2.170)
t
Substitution of this result into equation (2.169) gives, respectively, for the left-
and right-hand sides
4A2 + 3/2A0A1
z(t)z(2t) = 2t2 + 3Aot + (Až + 5/2A1) +
(2.171)
17/s A3 + 5/4A0 A2 + ½A?
+...
t2
Transcribed Image Text:2.8.1 Example A The equation YKYK+1 = 2y +1 (2.164) can be written as Yk+1 = 2yk +1/yk. (2.165) In this form, it is easily seen that the curve in the (yk, Yk+1) plane is a hyper- bola with asymptotes yk = 0 and yk+1 = 2yk. There are no real fixed points since a? = 2a² + 1. Consequently, the curve does not intersect yk+1 = Yk. Se Figure 2.5. The asymptotic behavior (i.e., k → 0) is determined by Yk+1 = 2yk; (2.166) the solution of which is Yk = A2k, (2.167) where A is an arbitrary constant. Therefore, if we let t = A2k and z(t) = Yk; (2.168) equation (2.164) becomes z(t)z(2t) = 2z(t)² + 1, (2.169) where 2(t) has the expansion A1 2(t) = t + Ao + A2 + t2 Аз +... 13 (2.170) t Substitution of this result into equation (2.169) gives, respectively, for the left- and right-hand sides 4A2 + 3/2A0A1 z(t)z(2t) = 2t2 + 3Aot + (Až + 5/2A1) + (2.171) 17/s A3 + 5/4A0 A2 + ½A? +... t2
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