Explanation Formula used: (1) The next value ( x n , y n ) of the approximate solution to a differential equation of the form d y d x = f ( x , y ) , that passes through a known point ( x 0 , y 0 ) is obtained using the formulas y n + 1 = y n + f ( x n , y n ) Δ x and x n + 1 = x n + Δ x . (2) If the linear differential equation of the first order and is of the form d y + P y d x = Q d x , then the solution of differential equation is y . e ∫ P d x = ∫ Q e ∫ P d x d x + c where e ∫ P d x is called integrating factor. Calculation: The approximate solution of a differential equation passes through ( x 0 , y 0 ) = ( 0 , 1 ) , write the differential equation d y d x = x + y as d y = ( x + y ) d x and then approximate d y as y 1 − y 0 , and d x is replaced as Δ x . Therefore, write the differential equation as ( y 1 − y 0 ) = ( x + y ) Δ x . On transposing y 0 to right hand side then get ( y 1 ) = y 0 + ( x 0 + y 0 ) Δ x and x 1 = x 0 + Δ x . On putting values of x 0 , y 0 and Δ x in above two equation. y 1 = y 0 + ( x 0 + y 0 ) Δ x = 1 + ( 0 + 1 ) × 1 = 2 And same as value of x 1 is calculated as: x 1 = x 0 + Δ x = 0 + 1 = 1 Thus, the curve passes through ( x 1 , y 1 ) = ( 1 , 2 ) . Now use point ( x 1 , y 1 ) to find the next point on the curve, which is ( x 2 , y 2 ) . y 2 = y 1 + ( x 1 + y 1 ) Δ x = 2 + ( 1 + 2 ) × 1 = 5 And same as value of x 2 is calculated as: x 2 = x 1 + Δ x = 1 + 1 = 2 Thus, the curve passes through ( x 2 , y 2 ) = ( 2 , 5 )

11th Edition

ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Differential Equations

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Chapter 31.5, Problem 15E

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To compare: The values of y found using exact solution method with the approximate method.

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