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Equating the coefficients of corresponding powers of t gives 2/3(13A2 + 2) = 10A2 + 1, (2.200) 2/9(35A3 + 10A2) = 10A3 + 2A2, and A2 = 1/4, A3 = 1/40. (2.201) Therefore, z(t) = 3+t+ 1/at² + 1/4ot3 + ..,

Equating the coefficients of corresponding powers of t gives 2/3(13A2 + 2) = 10A2 + 1, (2.200) 2/9(35A3 + 10A2) = 10A3 + 2A2, and A2 = 1/4, A3 = 1/40. (2.201) Therefore, z(t) = 3+t+ 1/at² + 1/4ot3 + ..,

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 22RE
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2.8.4 Example D The equation Yk+1 = 1/29k + 2 – 3/2yk (2.193) 76 Difference Equations 100 corresponds to a hyperbola in the (yk, Yk+1) plane having asymptotes yk = 0 and yk+1 = slopes at (1, 1) and (3, 3) are, respectively, 2 and 2/3. Therefore, yk = 1 is an unstable fixed point, while y = 3 is a stable fixed point. A geometric analysis indicates that arbitrary starting points lead finally to the stable fixed point. See Figure 2.10. The first approximation to the solution of equation (2.193), near yk = 3, is given by the expression 1/2Yk + 2. There are two fixed points, Yk = 1 and yk = 3. The Yk = 3+ A(?/3)*, (2.194) where A is an arbitrary constant. Therefore, if we let t = A(2/3)*, z(t) = Yk, (2.195) t%3D then equation (2.193) becomes 2t 2:(0): ) 2(t)² + 4z(t) – (2.196) - 3. 3 We assume a solution of the form 2(t) = 3+t+ A2t² + A3t³ + • . .. (2.197) Substitution of this expression into equation (2.197) gives, respectively, for the left- and right-hand sides 2t 2z(t)z = 18 + 10t + 2/3(13A2 + 2)t² 3 + 2/9(35A3 + 10A2)t² + · · · , (2.198) and z(t)? + 4z(t) – 3 = 18+ 10t + (10A2 +1)t² + (10A3 + 2A2)t³ + · · · . (2.199) Equating the coefficients of corresponding powers of t gives 2/3(13A2 + 2) = 10A2 + 1, (2.200) 2/0(35A3 + 10A2) = 10A3 + 2A2, and A2 = 1/4, A3 = '/40. (2.201) Therefore, z(t) = 3+t+ 1/4t2 + 1/40t + ·.., and Yk = 3+ A(2/3)* + 1¼A² (?/3)²k + 1/40A³ (?/3)³k + • . . . (2.202)
Transcribed Image Text:2.8.4 Example D The equation Yk+1 = 1/29k + 2 – 3/2yk (2.193) 76 Difference Equations 100 corresponds to a hyperbola in the (yk, Yk+1) plane having asymptotes yk = 0 and yk+1 = slopes at (1, 1) and (3, 3) are, respectively, 2 and 2/3. Therefore, yk = 1 is an unstable fixed point, while y = 3 is a stable fixed point. A geometric analysis indicates that arbitrary starting points lead finally to the stable fixed point. See Figure 2.10. The first approximation to the solution of equation (2.193), near yk = 3, is given by the expression 1/2Yk + 2. There are two fixed points, Yk = 1 and yk = 3. The Yk = 3+ A(?/3)*, (2.194) where A is an arbitrary constant. Therefore, if we let t = A(2/3)*, z(t) = Yk, (2.195) t%3D then equation (2.193) becomes 2t 2:(0): ) 2(t)² + 4z(t) – (2.196) - 3. 3 We assume a solution of the form 2(t) = 3+t+ A2t² + A3t³ + • . .. (2.197) Substitution of this expression into equation (2.197) gives, respectively, for the left- and right-hand sides 2t 2z(t)z = 18 + 10t + 2/3(13A2 + 2)t² 3 + 2/9(35A3 + 10A2)t² + · · · , (2.198) and z(t)? + 4z(t) – 3 = 18+ 10t + (10A2 +1)t² + (10A3 + 2A2)t³ + · · · . (2.199) Equating the coefficients of corresponding powers of t gives 2/3(13A2 + 2) = 10A2 + 1, (2.200) 2/0(35A3 + 10A2) = 10A3 + 2A2, and A2 = 1/4, A3 = '/40. (2.201) Therefore, z(t) = 3+t+ 1/4t2 + 1/40t + ·.., and Yk = 3+ A(2/3)* + 1¼A² (?/3)²k + 1/40A³ (?/3)³k + • . . . (2.202)
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