Explanation Formula used: The first approximation of Euler’s method is given by y 1 = L ( x 1 ) = y 0 + f ( x 0 , y 0 ) d x . The second approximation of Euler’s method is given by y 2 = y 1 + f ( x 1 , y 1 ) d x . The third approximation of Euler’s method is given by y 3 = y 2 + f ( x 2 , y 2 ) d x . Calculation: The function is y ′ = y , y ( 0 ) = 1 and d x = 0.2 . Since y ( 0 ) = 1 , x 0 = 0 and y 0 = 1 . Evaluate the value of y ( 0.2 ) or first approximation as follows. y 1 = y 0 + f ( x 0 , y 0 ) d x = y 0 + y 0 d x = 1 + 1 ( 0.2 ) ( ∵ x 0 = 0 , y 0 = 1 and d x = 0.2 ) = 1 + 0.2 = 1.2 Thus, the first approximation is 1.2. Evaluate the value of y ( 0.4 ) or second approximation as follows. y 2 = y 1 + f ( x 1 , y 1 ) d x = y 1 + y 1 d x = 1.2 + 1.2 ( 0.2 ) ( ∵ x 1 = x 0 + d x , y 1 = 1.2 and d x = 0.2 ) = 1.2 + 0.24 = 1.44 Thus, the second approximation is 1.44. Evaluate the value of y ( 0.6 ) or third approximation as follows. y 3 = y 2 + f ( x 2 , y 2 ) d x = y 2 + y 2 d x = 1

University Calculus: Early Transcendentals, Loose-leaf Edition (4th Edition)

4th Edition

ISBN: 9780135164860

Author: Joel R. Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 16.1, Problem 21E

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The value of y(1) and also to estimate the exact value of y(1) by using Euler method.

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Chapter 16.1, Problem 20E

Chapter 16.1, Problem 22E

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Chapter 16 Solutions

University Calculus: Early Transcendentals, Loose-leaf Edition (4th Edition)

Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 5 and 6, copy the slope fields, and...Ch. 16.1 - In Exercises 5 and 6, copy the slope fields, and...Ch. 16.1 - Prob. 7E Ch. 16.1 - Prob. 8E Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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Chapter 16 Solutions