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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 + +...

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
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 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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