Explanation Given: The given differential equation is, y ′ = x + 3 cos ( x y ) f ( x , y ) = x + 3 cos ( x y ) ... ( 1 ) The step size is, h = 0.2 Initial condition is, x 0 = 0 y 0 = 0 Domain of the function is [ 0 , 2 ] . Formula: The improved Euler method is given by, y n + 1 = y n + h 2 [ f ( x n , y n ) + f ( x n + h , y n + h f ( x n , y n ) ) ] ... ( 2 ) Here, n = 0 , 1 , 2 , ... Calculation: Substitute x n + 3 cos ( x n y n ) for f in equation ( 2 ) y n + 1 = y n + h 2 [ ( x n + 3 cos ( x n y n ) ) + ( x n + h + 3 cos ( ( x n + h ) ( y n + h x n + 3 h cos ( x n y n ) ) ) ) ] = y n + h 2 [ 2 x n + h + 3 cos ( x n y n ) + 3 cos ( ( x n + h ) ( y n + h x n + 3 h cos ( x n y n ) ) ) ] ... ( 3 ) Substitute 0 for n in equation ( 3 ) . y 1 = y 0 + h 2 [ 2 x 0 + h + 3 cos ( x 0 y 0 ) + 3 cos ( ( x 0 + h ) ( y 0 + h x 0 + 3 h cos ( x 0 y 0 ) ) ) ] … ( 4 ) Substitute 0.2 for h , 0 for x 0 and 0 for y 0 in equation ( 4 ) . y 1 = 0 + 0.2 2 [ 2 ( 0 ) + 0.2 + 3 cos ( 0 ⋅ 0 ) + 3 cos ( ( 0 + 0.2 ) ( 0 + 0.2 ⋅ 0 + 3 ( 0.2 ) cos ( 0 ⋅ 0 ) ) ) ] = 0.1 [ 0.2 + 3 ( 1 ) + 3 cos ( 0.2 ( 0.6 ⋅ 1 ) ) ] = 0.1 [ 3.2 + 3 cos ( 0.12 ) ] = 0.1 [ 3.2 + 3 ( 0.99280 ) ] And, y 1 = 0.1 [ 3.2 + 2.9784 ] = 0.1 [ 6.1784 ] = 0.61784 Substitute 1 for n in equation ( 3 ) . y 2 = y 1 + h 2 [ 2 x 1 + h + 3 cos ( x 1 y 1 ) + 3 cos ( ( x 1 + h ) ( y 1 + h x 1 + 3 h cos ( x 1 y 1 ) ) ) ] ... ( 5 ) Substitute 0.2 for h , 0.2 for x 1 and 0.61784 for y 1 in equation ( 5 ) . y 2 = 0.61784 + 0.2 2 [ 2 ( 0.2 ) + 0.2 + 3 cos ( 0.2 ( 0.61784 ) ) + 3 cos ( ( 0.2 + 0.2 ) ( 0.61784 + ( 0.2 ) ( 0.2 ) + 3 ( 0.2 ) cos ( ( 0.2 ) 0.61784 ) ) ) ] = 0.61784 + 0.1 [ 0.6 + 3 cos ( 0.123568 ) + 3 cos ( 0.4 ( 0.61784 + 0.04 + 0.6 cos ( 0.123568 ) ) ) ] = 0.61784 + 0.1 [ 0.6 + 3 ( 0.99237 ) + 3 cos ( 0.4 ( 0.65784 + 0.6 ( 0.99237 ) ) ) ] = 0.61784 + 0.1 [ 0.6 + 2.97711 + 3 cos ( 0.501304 ) ] And, y 2 = 0.61784 + 0.1 [ 0.6 + 2.97711 + 2.63086 ] = 0.61784 + 0.1 ( 6.20797 ) = 0.61784 + 0.620797 = 1.23864 Substitute 2 for n in equation ( 3 ) . y 3 = y 2 + h 2 [ 2 x 2 + h + 3 cos ( x 2 y 2 ) + 3 cos ( ( x 2 + h ) ( y 2 + h x 2 + 3 h cos ( x 2 y 2 ) ) ) ] ... ( 6 ) Substitute 0.2 for h , 0.4 for x 2 and 1.23864 for y 2 in equation ( 6 ) . y 3 = 1.23864 + 0.2 2 [ 2 ( 0.4 ) + 0.2 + 3 cos ( 0.4 ( 1.23864 ) ) + 3 cos ( ( 0.4 + 0.2 ) ( 1.23864 + 0.2 ( 0.4 ) + 3 ( 0.2 ) cos ( 0

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 3.6, Problem 9E

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The approximate solution to y′=x+3cos(xy) using improved Euler’s method and a rough diagram of the solution on [0,2].

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Chapter 3.6, Problem 8E

Chapter 3.6, Problem 10E

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Fundamentals of Differential Equations and Boundary Value Problems

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The approximate solution to y ′ = x + 3 cos ( x y ) using improved Euler’s method and a rough diagram of the solution on [ 0 , 2 ] .

Solution Summary: The author explains the approximate solution to yprime using improved Euler's method and a rough diagram of the solution.

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To find:

The approximate solution to y′=x+3cos(xy) using improved Euler’s method and a rough diagram of the solution on [0,2].

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The approximate solution to y ′ = x + 3 cos ( x y ) using improved Euler’s method and a rough diagram of the solution on [ 0 , 2 ] .

Solution Summary: The author explains the approximate solution to yprime using improved Euler's method and a rough diagram of the solution.

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To find: The approximate solution to y′=x+3cos(xy) using improved Euler’s method and a rough diagram of the solution on [0,2].

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Chapter 3 Solutions

To find:

The approximate solution to y′=x+3cos(xy) using improved Euler’s method and a rough diagram of the solution on [0,2].