Explanation Formula used: To find an 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 ) . Then the next value of ( x n , y n ) is y n + 1 = y n + f ( x n , y n ) Δ x and x n + 1 = x n + Δ x . 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 + 1 as d y = ( x + 1 ) 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 + 1 ) Δ x . On transposing y 0 to right hand side then get ( y 1 ) = y 0 + ( x 0 + 1 ) 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 + 1 ) Δ x = 1 + ( 0 + 1 ) 0.1 = 1.1 The value of x 1 is calculated as: x 1 = x 0 + Δ x = 0 + 0.1 = 0.1 Thus, the curve passes through ( x 1 , y 1 ) = ( 0.1 , 1.1 ) . 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 + 1 ) Δ x = 1.1 + ( 0.1 + 1 ) 0.1 = 1.21 The value of x 2 is calculated as: x 2 = x 1 + Δ x = 0.1 + 0.1 = 0.2 Thus, the curve passes through ( x 2 , y 2 ) = ( 0

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 31.5, Problem 5E

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The y-values of the solution for the values of x and Δx by using of Euler’s method, and plot the graphs of the solutions.

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The y -values of the solution for the values of x and Δ x by using of Euler’s method, and plot the graphs of the solutions.

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The y-values of the solution for the values of x and Δx by using of Euler’s method, and plot the graphs of the solutions.

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