To determine The approximate value of the indicated function y ( 0.6 ) , value of f ( 0.6 ) and then compare the result. Explanation Given: In the interval of [ 0 , 1 ] the given function have 10 subintervals as h = 0.1 . The differential equation is d y d x = x 2 y ; y ( 0 ) = 1 . Definition used: If y = f ( x ) is the solution of a differential equation d y d x = g ( x , y ) with y ( x 0 ) = y 0 , where x 0 = 0 and y 0 = 4 . Assume, for an interval [ x 0 , x n ] , x i + 1 = x i + h and y i + 1 = y i + g ( x i , y i ) h , where 0 ≤ i ≤ n − 1 . Then f ( x i + 1 ) ≈ y i + 1 . Calculation: Substitute 0 for x 0 and 0.1 for h to obtain the value of x 1 as follows: To obtain the value of x 1 , x 1 = x 0 + h = 0 + 0.1 = 0.1 Substitute x 0 for x and y 0 for y in g ( x , y ) to obtain the value of g ( x 0 , y 0 ) as follows: g ( x 0 , y 0 ) = x 0 2 y 0 = 0 Now substitute 1 for y 0 , 0 for g ( x 0 , y 0 ) and 0.1 for h to obtain the value of y 1 as follows. y 1 = y 0 + g ( x 0 , y 0 ) ⋅ h = 1 + 0 × 0.1 = 1 Substitute 1 for y 1 , 0.1 for x 1 to calculate the value of g ( x 1 , y 1 ) . g ( x 1 , y 1 ) = x 1 2 y 1 = 0.1 2 × 1 = 0.01 Now substitute 1 for y 1 , 0.01 for g ( x 1 , y 1 ) and 0.1 for h to obtain the value of y 2 as follows: Now the value of y 2 can be calculated as follows: y 2 = y 1 + g ( x 1 , y 1 ) h = 1 + 0.01 × 0

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Chapter 10.3, Problem 14E

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The approximate value of the indicated function y(0.6), value of f(0.6) and then compare the result.

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Chapter 10.3, Problem 13E

Chapter 10.3, Problem 15E

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