Explanation Given: The differential equation is y ' = ( x − y ) 2 . Concept used: Runga-Kutta method and Adams-Bashforth-Moulton method: The value of y 1 , y 2 and y 3 is calculated by the Runga-Kutta 4 t h order method and with the initial values x 0 and y 0 . The general formulas used in the Runga-Kutta 4 t h order method are as follows: k 1 = h F ( x n , y n ) k 2 = h F ( x n + h 2 , y n + k 1 2 ) k 3 = h F ( x n + h 2 , y n + k 2 2 ) k 4 = h F ( x n + h , y n + k 3 ) And y n + 1 = y n + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) . The value of y 4 is calculated using the Adams-Bashforth-Moulton method using the following formulas as follows: y ′ n = f ( x n , y n ) y ′ n − 1 = f ( x n − 1 , y n − 1 ) y ′ n − 2 = f ( x n − 2 , y n − 2 ) y ′ n − 3 = f ( x n − 3 , y n − 3 ) Then, y 4 ∗ = y 3 + 0.2 24 ( 55 y ′ 3 − 59 y ′ 2 + 37 y ′ 1 − 9 y ′ 0 ) , y ′ 4 = f ( x 4 , y 4 ∗ ) , y 4 = y 3 + 0.2 24 ( 9 y ′ 4 + 19 y ′ 3 − 5 y ′ 2 + y ′ 1 ) Calculation: The given differential equation is written as follows: y ' = F ( x , y ) F ( x , y ) = ( x − y ) 2 The boundary conditions is given as follows: y 0 = f ( x 0 ) x 0 = 0 y 0 = 0 First obtain the approximation with the step size h = 0.2 . For n = 0 , substitute the value of x 0 , y 0 and h in the equation k 1 = h F ( x 0 , y 0 ) . The value of k 1 is calculated as follows: k 1 = h F ( x 0 , y 0 ) = 0.2 × ( 0 − 0 ) 2 = 0.2 × ( 0 ) = 0 For n = 0 , substitute the value of x 0 , y 0 and h in the equation k 2 = h F ( x 0 + h 2 , y 0 + k 1 2 ) . The value of k 2 is calculated as follows: k 2 = h F ( x 0 + h 2 , y 0 + k 1 2 ) = 0.2 × F ( 0 + 0.2 2 , 0 + 0 2 ) = 0.2 × F ( 0.1 , 0 ) = 0.002 For n = 0 , substitute the value of x 0 , y 0 and h in the equation k 3 = h F ( x 0 + h 2 , y 0 + k 2 2 ) . The value of k 3 is calculated as follows: k 3 = h F ( x 0 + h 2 , y 0 + k 2 2 ) = 0.2 × F ( 0 + 0.2 2 , 0 + 0.002 2 ) = 0.2 × F ( 0.1 , 0.001 ) = 0.00196 For n = 0 , substitute the value of x 0 , y 0 and h in the equation k 4 = h F ( x 0 + h , y 0 + k 3 ) . The value of k 4 is calculated as follows: k 4 = h F ( x 0 + h , y 0 + k 3 ) = 0.2 × F ( 0 + 0.2 , 0 + 0.00196 ) = 0.2 × F ( 0.2 , 0.00196 ) = 0.0078 Now substitute the value of k 1 , k 2 , k 3 and k 4 in y 1 = y 0 + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) . y 1 = y 0 + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) = 0 + 1 6 ( 0 + ( 2 × 0.002 ) + ( 2 × 0.00196 ) + 0.0078 ) = 0 + 1 6 × ( 0 + 0.004 + 0.00392 + 0.0078 ) = 0.00262 For n = 1 , the value of x 1 is calculated as follows: x 1 = x 0 + h = 0 + 0.2 = 0.2 Similarly, the value of y 2 and y 3 is calculated to be 0.0200 and 0.06296 respectively. The value of x 2 and x 3 is calculated as follows: x 2 = x 1 + h = 0.4 x 3 = x 2 + h = 0.6 Now calculate the value of y 4 using the Adams-Bashforth-Moulton method. Substitute the values of x 0 and y 0 in y ′ 0 = f ( x 0 , y 0 ) . y ′ 0 = f ( x 0 , y 0 ) = f ( 0 , 1 ) = 1 Substitute the values of x 1 and y 1 in y ′ 1 = f ( x 1 , y 1 ) . y ′ 1 = f ( x 1 , y 1 ) = f ( 0.2 , 0.00262 ) = 0.03896 Substitute the values of x 2 and y 2 in y ′ 2 = f ( x 2 , y 2 ) . y ′ 2 = f ( x 2 , y 2 ) = f ( 0.4 , 0.0200 ) = 0.1444 Substitute the values of x 3 and y 3 in y ′ 3 = f ( x 3 , y 3 ) . y ′ 3 = f ( x 3 , y 3 ) = f ( 0.6 , 0.06296 ) = 0.2884 Substitute the values y ′ 0 , y ′ 1 , y ′ 2 , y ′ 3 and y 3 in y 4 ∗ = y 3 + h 24 ( 55 y ′ 3 − 59 y ′ 2 + 37 y ′ 1 − 9 y ′ 0 ) . y 4 ∗ = y 3 + 0.2 24 ( 55 y ′ 3 − 59 y ′ 2 + 37 y ′ 1 − 9 y ′ 0 ) = 0.06296 + 0.2 24 ( 55 × ( 0.2884 ) − 59 × ( 0.1444 ) + 37 × ( 0.03896 ) − 9 × ( 1 ) ) = 0.06296 + 0.2 24 ( − 0.2161 ) = 0.061159 Substitute the value of x 4 and y 4 * in y ′ 4 = f ( x 4 , y 4 ∗ ) . y ′ 4 = f ( x 4 , y 4 ∗ ) = f ( 0.8 , 0.061159 ) = 0.5459 The value of y 4 is calculated as follows: y 4 = y 3 + 0.2 24 ( 9 y ′ 4 + 19 y ′ 3 − 5 y ′ 2 + y ′ 1 ) = 0

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 9.3, Problem 7E

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The approximated value of y(1) by the Adams-Bashforth-Moulton method with the first step size h=0.2 and then h=0.1 and use Runga-Kutta method of 4th order to compute y1,y2 and y3.

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Chapter 9.3, Problem 6E

Chapter 9.3, Problem 8E

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A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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The approximated value of y ( 1 ) by the Adams-Bashforth-Moulton method with the first step size h = 0.2 and then h = 0.1 and use Runga-Kutta method of 4 t h order to compute y 1 , y 2 and y 3 .

Solution Summary: The author describes the Runga-Kutta method and the Adams-Bashforth-Moulton method.

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The approximated value of y(1) by the Adams-Bashforth-Moulton method with the first step size h=0.2 and then h=0.1 and use Runga-Kutta method of 4th order to compute y1,y2 and y3.

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