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 ′ = x sin y , y ( 0 ) = 1 and d x = 1 3 . Since y ( 0 ) = 1 , x 0 = 0 and y 0 = 1 . Evaluate the value of y ( 0 ) or first approximation as follows. y 1 = y 0 + f ( x 0 , y 0 ) d x = y 0 + x 0 sin y 0 d x = 1 + ( 0 ) ( sin 1 ) ( 1 3 ) ( ∵ x 0 = 0 , y 0 = 1 and d x = 1 3 ) = 1 Thus, the first approximation is 1. Evaluate the value of y ( 1 3 ) or second approximation as follows. y 2 = y 1 + f ( x 1 , y 1 ) d x = y 1 + x 1 sin y 1 d x = 1 + ( 1 3 ⋅ sin 1 ) ( 1 3 ) ( ∵ x 1 = x 0 + d x , y 1 = 1 and d x = 1 3 ) = 1 + 0.9350 = 1.9350 Thus, the second approximation is 1.9350. Evaluate the value of y ( 2 3 ) or third approximation as follows. y 3 = y 2 + f ( x 2 , y 2 ) d x = y 2 + x 2 sin y 2 d x = 1.09350 + ( 2 3 ⋅ sin 1.09350 ) ( 1 3 ) = 1.09350 + 0.19739 = 1.29089 Thus, the third approximation is 1.29089. Evaluate the value of y ( 3 3 ) or fourth approximation as follows. y 4 = y 3 + f ( x 3 , y 3 ) d x = y 3 + x 3 sin y 3 d x = 1.29089 + ( 3 3 ⋅ sin 1.29089 ) ( 1 3 ) = 1.29089 + 0.32036 = 1.61125 Thus, the fourth approximation is 1.61125. Evaluate the value of y ( 4 3 ) or fifth approximation as follows. y 5 = y 4 + f ( x 4 , y 4 ) d x = y 4 + x 4 sin y 4 d x = 1.61125 + ( 4 3 ⋅ sin 1.61125 ) ( 1 3 ) = 1.61125 + 0.44408 = 2.05533 Thus, the fifth approximation is 2.05533. Evaluate the value of y ( 5 3 ) or sixth approximation as follows. y 6 = y 5 + f ( x 5 , y 5 ) d x = y 5 + x 5 sin y 5 d x = 2

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Chapter 9.1, Problem 24E

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Chapter 9.1, Problem 23E

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The estimated value of y ( 2 ) by using Euler method and also to find the exact value of y ( 2 ) .

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The estimated value of y(2) by using Euler method and also to find the exact value of y(2).

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