Explanation Given: In the interval of [ 0 , 1 ] the function have 5 subintervals as h = 0.2 . The differential equation is d y d x = 3 + y with y ( 0 ) = 0 . 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 . 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 0.02 for h to obtain the value of x 1 as follows: x 1 = x 0 + h = 0 + 0.2 = 0.2 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 ) = 3 + y 0 Substitute 0 for y 0 in g ( x 0 , y 0 ) as follows: g ( x 0 , y 0 ) = 3 + 0 = 3 Now substitute 0 for y 0 , 3 for g ( x 0 , y 0 ) and 0.2 for h to obtain the value of y 1 as follows: y 1 = y 0 + g ( x 0 , y 0 ) ⋅ h = 0 + 3 × 0

Calculus With Applications, Books a la Carte Plus MyLab Math Package (11th Edition)

11th Edition

ISBN: 9780133886849

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Calculus With Applications, Books a la Carte Plus MyLab Math Package (11th Edition)

Ch. 10.1 - Find all solutions of the differential equation . Ch. 10.1 - Prob. 2YT Ch. 10.1 - Prob. 3YT Ch. 10.1 - Prob. 4YT Ch. 10.1 - Prob. 1WE Ch. 10.1 - Prob. 2WE Ch. 10.1 - Prob. 3WE Ch. 10.1 - Prob. 4WE Ch. 10.1 - Prob. 5WE Ch. 10.1 - Find the general solution for each differential...

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