Explanation Given: The values h = 0.2 and h = 0.1 along with the differential d y d x = y . The initial condition is ( 0 , 3 ) with exact solution y = 3 e x . Formula Used: Euler’s method, x n = x n − 1 + h and y n = y n − 1 + h F ( x n − 1 , y n − 1 ) . Calculation: Exact solution of differential equation, y = 3 e x At x = 0 , it is found that: y = 3 e x = 3 e 0 = 3 At x = 0.2 , it is found that y = 3 e x = 3 e 0.2 = 3.6642 Continuing the calculation in for different value, then get the table: x 0 0.2 0.4 0.6 0.8 1 y ( x ) ( exact ) 3 3.6642 4.4755 5.4663 6.6766 8.1548 The differential equation, d y d x = y By the use of h = 0.2 , x 0 = 0 , y 0 = 3 , and F ( x , y ) = y , we have x 0 = 0 , x 1 = 0.2 , x 2 = 0.4 , x 3 = 0.6 , x 4 = 0.8 and x 5 = 1.0 . The first three approximations are First approximation: y 1 = y 0 + h F ( x 0 , y 0 ) = 3 + 0.2 F ( 0 , 3 ) = 3 + 0.2 { 3 } = 3.6 Second approximation: y 2 = y 1 + h F ( x 1 , y 1 ) = 3.6 + 0.2 F ( 0.2 , 3.6 ) = 3.6 + 0.2 { 3.6 } = 4.32 Third approximation: y 3 = y 2 + h F ( x 2 , y 2 ) = 4

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ISBN: 9780131569898

Author: C. Henry Edwards, David E. Penney

Publisher: PEARSON

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Chapter 6.1, Problem 79E

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To Calculate: Complete the table by using the exact solution of the differential equation, dydx=y and the two approximations obtained using Euler’s method.

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Chapter 6.1, Problem 78E

Chapter 6.1, Problem 80E

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Complete the table by using the exact solution of the differential equation, d y d x = y and the two approximations obtained using Euler’s method.

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To Calculate: Complete the table by using the exact solution of the differential equation, dydx=y and the two approximations obtained using Euler’s method.

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Chapter 6 Solutions