(a) Explanation Calculation: The given initial value problem is: y ' = − sin y + 1 , y ( 0 ) = 0 Each iteration in the method of successive approximation is calculated by ϕ n + 1 ( t ) = ϕ 0 + ∫ t 0 t f ( x , ϕ n ) d x where y ' = f ( t , y ) . To match the initial condition, let ϕ 0 ( t ) = 0 . Write f in terms of its Taylor expansion: f = 1 − sin y = 1 − ∑ n = 0 ∞ ( − 1 ) n y 2 n + 1 ( 2 n + 1 ) ! = 1 − y + y 3 3 ! − y 5 5 ! + O ( y 7 ) Take t 0 = 0 for convenience. Calculate the first value of the approximation ϕ 1 is as follows: ϕ 1 = ϕ 0 + ∫ t 0 t f ( x , ϕ 0 ) d x = 0 + ∫ 0 t ( − sin 0 + 1 ) d x = [ x ] 0 t = t Similarly calculate the second value of the approximation ϕ 2 is as follows: ϕ 2 = ϕ 1 + ∫ t 0 t f ( x , ϕ 1 ) d x = 0 + ∫ 0 t ( − sin x + 1 ) d x = [ cos x + x ] 0 t = cos t + t − 1 Write ϕ 2 in terms of its Taylor expansion: ϕ 2 = − 1 + t + ∑ n = 0 ∞ ( − 1 ) n t 2 n ( 2 n ) ! = − 1 + t + 1 − t 2 2 ! + t 4 4 ! − t 6 6 ! + O ( t 8 ) = t − t 2 2 ! + t 4 4 ! − t 6 6 ! + O ( t 8 ) Similarly calculate the third value of the approximation ϕ 3 is as follows: ϕ 3 = ∫ t 0 t [ 1 − ϕ 2 + 1 3 ! ϕ 2 3 − 1 5 ! ϕ 2 5 + O ( ϕ 2 7 ) ] d x = ∫ 0 t [ 1 − x + 1 2 x 2 − 1 4 ! x 4 + 1 3 ! ( x 3 − 3 2 x 4 + 3 4 (b) To determine To plot: The function ϕ 1 ( t ) , ... , ϕ 4 ( t ) and observe whether the iterates appear to be converging.

Elementary Differential Equations and Boundary Value Problems, 11e WileyPLUS Registration Card + Loose-leaf Print Companion

11th Edition

ISBN: 9781119336617

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley (WileyPLUS Products)

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Question

Chapter 2.8, Problem 9P

(a)

To determine

The value of the ϕ1(t),...,ϕ4(t) for Taylor approximation and keep the terms up to six order.

(b)

To determine

To plot: The function ϕ1(t),...,ϕ4(t) and observe whether the iterates appear to be converging.

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Chapter 2.8, Problem 8P

Chapter 2.8, Problem 10P

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Chapter 2 Solutions

Elementary Differential Equations and Boundary Value Problems, 11e WileyPLUS Registration Card + Loose-leaf Print Companion

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

Ch. 2.1 - Prob. 11P Ch. 2.1 - Prob. 12P Ch. 2.1 - Prob. 13P Ch. 2.1 - Prob. 14P Ch. 2.1 - Prob. 15P Ch. 2.1 - Prob. 16P Ch. 2.1 - Prob. 17P Ch. 2.1 - Prob. 18P Ch. 2.1 - Prob. 19P Ch. 2.1 - Prob. 20P Ch. 2.1 - Prob. 21P Ch. 2.1 - Prob. 22P Ch. 2.1 - Prob. 23P Ch. 2.1 - Prob. 24P Ch. 2.1 - Prob. 25P Ch. 2.1 - Prob. 26P Ch. 2.1 - Prob. 27P Ch. 2.1 - Variation of Parameters. Consider the following...Ch. 2.1 - Prob. 29P Ch. 2.1 - In each of Problems 29 and 30, use the method of...Ch. 2.2 - Prob. 1P Ch. 2.2 - Prob. 2P Ch. 2.2 - Prob. 3P Ch. 2.2 - Prob. 4P Ch. 2.2 - Prob. 5P Ch. 2.2 - Prob. 6P Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - Prob. 9P Ch. 2.2 - Prob. 10P Ch. 2.2 - Prob. 11P Ch. 2.2 - Prob. 12P Ch. 2.2 - Prob. 13P Ch. 2.2 - Prob. 14P Ch. 2.2 - Prob. 15P Ch. 2.2 - Prob. 16P Ch. 2.2 - Prob. 17P Ch. 2.2 - Prob. 18P Ch. 2.2 - Prob. 19P Ch. 2.2 - Prob. 20P Ch. 2.2 - Prob. 21P Ch. 2.2 - Prob. 22P Ch. 2.2 - Prob. 23P Ch. 2.2 - Prob. 24P Ch. 2.2 - Prob. 25P Ch. 2.2 - Prob. 26P Ch. 2.2 - Prob. 27P Ch. 2.2 - Prob. 28P Ch. 2.2 - Prob. 29P Ch. 2.2 - Prob. 30P Ch. 2.2 - Prob. 31P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P Ch. 2.3 - Prob. 3P Ch. 2.3 - Prob. 4P Ch. 2.3 - Prob. 5P Ch. 2.3 - Prob. 6P Ch. 2.3 - Prob. 7P Ch. 2.3 - Prob. 8P Ch. 2.3 - Prob. 9P Ch. 2.3 - Prob. 10P Ch. 2.3 - Prob. 11P Ch. 2.3 - Prob. 12P Ch. 2.3 - Prob. 13P Ch. 2.3 - Prob. 14P Ch. 2.3 - Prob. 15P Ch. 2.3 - Prob. 16P Ch. 2.3 - Assume that the conditions are as in Problem 16...Ch. 2.3 - Prob. 18P Ch. 2.3 - Prob. 19P Ch. 2.3 - Prob. 20P Ch. 2.3 - Prob. 21P Ch. 2.3 - Prob. 22P Ch. 2.3 - Prob. 23P Ch. 2.3 - Prob. 24P Ch. 2.4 - In each of Problems 1 through 6, determine...Ch. 2.4 - Prob. 2P Ch. 2.4 - Prob. 3P Ch. 2.4 - Prob. 4P Ch. 2.4 - Prob. 5P Ch. 2.4 - Prob. 6P Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - Prob. 9P Ch. 2.4 - Prob. 10P Ch. 2.4 - Prob. 11P Ch. 2.4 - Prob. 12P Ch. 2.4 - Prob. 13P Ch. 2.4 - Prob. 14P Ch. 2.4 - Prob. 15P Ch. 2.4 - Prob. 16P Ch. 2.4 - Prob. 17P Ch. 2.4 - Prob. 18P Ch. 2.4 - Prob. 19P Ch. 2.4 - Prob. 20P Ch. 2.4 - Prob. 21P Ch. 2.4 - Prob. 22P Ch. 2.4 - Prob. 23P Ch. 2.4 - Prob. 24P Ch. 2.4 - Prob. 25P Ch. 2.4 - Prob. 26P Ch. 2.4 - Prob. 27P Ch. 2.5 - Prob. 1P Ch. 2.5 - Prob. 2P Ch. 2.5 - Prob. 3P Ch. 2.5 - Prob. 4P Ch. 2.5 - Prob. 5P Ch. 2.5 - Prob. 6P Ch. 2.5 - Prob. 7P Ch. 2.5 - Prob. 8P Ch. 2.5 - Prob. 9P Ch. 2.5 - Prob. 10P Ch. 2.5 - Prob. 11P Ch. 2.5 - Prob. 12P Ch. 2.5 - Prob. 13P Ch. 2.5 - Prob. 14P Ch. 2.5 - Prob. 15P Ch. 2.5 - Prob. 16P Ch. 2.5 - Prob. 17P Ch. 2.5 - Prob. 18P Ch. 2.5 - Prob. 19P Ch. 2.5 - Prob. 20P Ch. 2.5 - Prob. 21P Ch. 2.5 - Prob. 22P Ch. 2.5 - Prob. 23P Ch. 2.5 - Prob. 24P Ch. 2.5 - Prob. 25P Ch. 2.5 - Prob. 26P Ch. 2.5 - Prob. 27P Ch. 2.6 - Prob. 1P Ch. 2.6 - Prob. 2P Ch. 2.6 - Prob. 3P Ch. 2.6 - Prob. 4P Ch. 2.6 - Prob. 5P Ch. 2.6 - Prob. 6P Ch. 2.6 - Prob. 7P Ch. 2.6 - Prob. 8P Ch. 2.6 - Prob. 9P Ch. 2.6 - Prob. 10P Ch. 2.6 - Prob. 11P Ch. 2.6 - Prob. 12P Ch. 2.6 - Prob. 13P Ch. 2.6 - Prob. 14P Ch. 2.6 - Prob. 15P Ch. 2.6 - Prob. 16P Ch. 2.6 - Prob. 17P Ch. 2.6 - Prob. 18P Ch. 2.6 - Prob. 19P Ch. 2.6 - Prob. 20P Ch. 2.6 - Prob. 21P Ch. 2.6 - Prob. 22P Ch. 2.7 - Prob. 1P Ch. 2.7 - Prob. 2P Ch. 2.7 - Prob. 3P Ch. 2.7 - Prob. 4P Ch. 2.7 - Prob. 5P Ch. 2.7 - Prob. 6P Ch. 2.7 - Prob. 7P Ch. 2.7 - Prob. 8P Ch. 2.7 - Prob. 9P Ch. 2.7 - Prob. 10P Ch. 2.7 - Prob. 11P Ch. 2.7 - Prob. 12P Ch. 2.7 - Prob. 14P Ch. 2.7 - Prob. 15P Ch. 2.7 - Prob. 16P Ch. 2.7 - Prob. 17P Ch. 2.8 - Prob. 1P Ch. 2.8 - Prob. 2P Ch. 2.8 - Prob. 3P Ch. 2.8 - Prob. 4P Ch. 2.8 - Prob. 5P Ch. 2.8 - Prob. 6P Ch. 2.8 - Prob. 7P Ch. 2.8 - Prob. 8P Ch. 2.8 - Prob. 9P Ch. 2.8 - Prob. 10P Ch. 2.8 - Prob. 11P Ch. 2.8 - Prob. 12P Ch. 2.8 - Prob. 13P Ch. 2.8 - Prob. 14P Ch. 2.8 - Prob. 15P Ch. 2.8 - Prob. 16P Ch. 2.8 - Prob. 17P Ch. 2.8 - Prob. 18P Ch. 2.9 - Prob. 1P Ch. 2.9 - Prob. 2P Ch. 2.9 - Prob. 3P Ch. 2.9 - Prob. 4P Ch. 2.9 - Prob. 5P Ch. 2.9 - Prob. 6P Ch. 2.9 - Prob. 7P Ch. 2.9 - Prob. 8P Ch. 2.9 - Prob. 9P Ch. 2.9 - Prob. 10P Ch. 2 - Prob. 1MP Ch. 2 - Prob. 2MP Ch. 2 - Prob. 3MP Ch. 2 - Prob. 4MP Ch. 2 - Prob. 5MP Ch. 2 - Prob. 6MP Ch. 2 - Prob. 7MP Ch. 2 - Prob. 8MP Ch. 2 - Prob. 9MP Ch. 2 - Prob. 10MP Ch. 2 - Prob. 11MP Ch. 2 - Prob. 12MP Ch. 2 - Prob. 13MP Ch. 2 - Prob. 14MP Ch. 2 - Prob. 15MP Ch. 2 - Prob. 16MP Ch. 2 - Prob. 17MP Ch. 2 - Prob. 18MP Ch. 2 - Prob. 19MP Ch. 2 - Prob. 20MP Ch. 2 - Prob. 21MP Ch. 2 - Prob. 22MP Ch. 2 - Prob. 23MP Ch. 2 - Prob. 24MP Ch. 2 - Prob. 25MP Ch. 2 - Prob. 28MP Ch. 2 - Prob. 29MP Ch. 2 - Prob. 31MP Ch. 2 - Prob. 32MP Ch. 2 - Prob. 33MP Ch. 2 - Prob. 34MP Ch. 2 - Prob. 35MP Ch. 2 - Prob. 36MP Ch. 2 - Prob. 37MP

Knowledge Booster

The value of the ϕ 1 ( t ) , ... , ϕ 4 ( t ) for Taylor approximation and keep the terms up to six order.

The value of the ϕ1(t),...,ϕ4(t) for Taylor approximation and keep the terms up to six order.

To plot: The function ϕ1(t),...,ϕ4(t) and observe whether the iterates appear to be converging.

Want to see the full answer?

Chapter 2 Solutions