Explanation Given: The linear differential equation is y ′ = 4 x − 2 y and y ( 0 ) = 2 . Concept used: Runga-Kutta method: The solution of a linear differential equation of the form y ′ = f ( x , y ) is given as follows: y n + 1 = y n + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) , here k 1 = f ( x n , y n ) k 2 = f ( ( x n + h 2 ) , ( y n + h k 1 2 ) ) k 3 = f ( ( x n + h 2 ) , ( y n + h k 2 2 ) ) k 4 = f ( ( x n + h ) , ( y n + h k 3 ) ) Adams-Bashforth-Moulton method: Adams-Bashforth-Moulton method is a predictor-corrector method. It is a multistep method. The predictor formula is as follows: y n + 1 * = y n + h 24 ( 55 y ′ n − 59 y ′ n − 1 + 37 y ′ n − 2 − 9 y ′ n − 3 ) , here 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 ) The Moulton corrector formula is as follows: y n + 1 = y n + h 24 ( 9 y ′ n + 1 + 19 y ′ n − 5 y ′ n − 1 + y ′ n − 2 ) y ′ n + 1 = f ( x n + 1 , y n + 1 ) Calculation: The linear differential equation is given as follows: y ′ = 4 x − 2 y It is given that for Runga-Kutta method of fourth order the value of h is 0.1 . As per the Runga-Kutta method of fourth order the value of y n + 1 is given as follows: y n + 1 = y n + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) … … ( 1 ) Here, the value of k 1 , k 2 , k 3 and k 4 is given as follows: k 1 = f ( x n , y n ) k 2 = f ( ( x n + h 2 ) , ( y n + h k 1 2 ) ) k 3 = f ( ( x n + h 2 ) , ( y n + h k 2 2 ) ) k 4 = f ( ( x n + h ) , ( y n + h k 3 ) ) The differential equation y ′ = 4 x − 2 y is of the form y ′ = f ( x , y ) . This implies that f ( x , y ) = 4 x − 2 y . It is given that the initial value is y ( 0 ) = 2 . So, the value of x 0 is 0 and the value of y 0 is 2 . To calculate the value of y 1 substitute n = 0 in equation ( 1 ) . y 1 = y 0 + h 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) … … ( 2 ) Calculate the value of k 1 , k 2 , k 3 and k 4 as follows: k 1 = f ( x 0 , y 0 ) = ( ( 4 ⋅ 0 ) − ( 2 ⋅ 2 ) ) = ( 0 − 4 ) = − 4 Therefore, the value of k 1 is − 4 . k 2 = f ( ( x 0 + 0.1 2 ) , ( y 0 + ( 0.1 ) ⋅ − 4 2 ) ) = f ( ( 0.05 ) , ( 1.8 ) ) = ( 0.2 − 3.6 ) = − 3.4 Therefore, the value of k 2 is − 3.4 . k 3 = f ( ( x 0 + 0.1 2 ) , ( y 0 + ( − 3.4 ) ⋅ ( 0.1 ) 2 ) ) = f ( ( 0.05 ) , ( 2 − 0.17 ) ) = f ( 0.05 , 1.83 ) = − 3.46 Therefore, the value of k 3 is − 3.46 . k 4 = f ( ( x 0 + 0.1 ) , ( y 0 + ( 0.1 ) ( − 3.46 ) ) ) = f ( ( 0.1 ) , ( 1.654 ) ) = ( 0.4 − 3.308 ) = − 2.908 Substitute the value k 1 , k 2 , k 3 and k 4 in equation ( 2 ) . y 1 = 2 + 0.1 6 ( − 4 + 2 ⋅ ( − 3.4 ) + 2 ⋅ ( − 3.46 ) − 2.908 ) = 2 + 0.1 6 ( − 4 − 6.8 − 6.92 − 2.908 ) = 1.6562 Therefore, the value of y 1 is 1.6562 . The value of x 1 is calculated as follows: x 1 = x 0 + h = 0 + 0.1 = 0.1 To calculate the value of y 2 substitute n = 1 in equation ( 1 ) . y 2 = y 1 + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) … … ( 3 ) Calculate the value of k 1 , k 2 , k 3 and k 4 as follows: k 1 = f ( x 1 , y 1 ) = f ( 0.1 , 1.6562 ) = ( ( 4 ⋅ 0.1 ) − ( 2 ⋅ 1.6562 ) ) = − 2.9124 Therefore, the value of k 1 is − 2.9124 . k 2 = f ( ( x 1 + 0.1 2 ) , ( y 1 + ( 0.1 ) ⋅ ( − 2.9124 ) 2 ) ) = f ( ( 0.15 ) , ( 1.51058 ) ) = − 2.42116 Therefore, the value of k 2 is − 2.42116 . k 3 = f ( ( x 1 + 0.1 2 ) , ( y 1 + ( − 2.42116 ) ⋅ ( 0.1 ) 2 ) ) = f ( ( 0.15 ) , ( 1.534942 ) ) = − 2.469884 Therefore, the value of k 3 is − 2.469884 . k 4 = f ( ( x 1 + 0.1 ) , ( y 1 + ( 0.1 ) ( − 2.469884 ) ) ) = f ( ( 0.2 ) , ( 1.4092116 ) ) = − 2.0184232 Substitute the value k 1 , k 2 , k 3 and k 4 in equation ( 3 ) . y 2 = 1.6562 + 0.1 6 ( − 2.9124 + 2 ⋅ ( − 2.42116 ) + 2 ⋅ ( − 2.469884 ) − 2.0184232 ) = 1.6562 − 0.245215 = 1.41098 Therefore, the value of y 2 is 1.41098 . The value of x 2 is calculated as follows: x 2 = x 1 + h = 0.1 + 0.1 = 0.2 To calculate the value of y 3 substitute n = 2 in equation ( 1 ) . y 3 = y 2 + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) … … ( 4 ) Calculate the value of k 1 , k 2 , k 3 and k 4 as follows: k 1 = f ( x 2 , y 2 ) = f ( 0.2 , 1.41098 ) = ( ( 4 ⋅ 0.2 ) − ( 2 ⋅ 1.41098 ) ) = − 2.02196 Therefore, the value of k 1 is − 2.02196 . k 2 = f ( ( x 2 + 0.1 2 ) , ( y 2 + ( 0.1 ) ⋅ ( − 2

9th Edition

ISBN: 9781337810906

Author: ZILL

Publisher: CENGAGE L

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Chapter 9, Problem 6RE

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The approximated value of y(0.4) for the differential equation y′=4x−2y such that y(0)=2 by Adams-Bashforth-Moulton method and the value of y1,y2 and y3 by Runga-Kutta method of fourth order with h=0.1.

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Chapter 9, Problem 5RE

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Ch. 9.1 - In Problems 110 use the improved Eulers method to...Ch. 9.1 - In Problems 1–10 use the improved Euler’s method...Ch. 9.1 - In Problems 1–10 use the improved Euler’s method...Ch. 9.1 - In Problems 1–10 use the improved Euler’s method...Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - In Problems 1–10 use the improved Euler’s method...Ch. 9.1 - In Problems 1–10 use the improved Euler’s method...Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E

Ch. 9.1 - Prob. 11E Ch. 9.1 - Prob. 13E Ch. 9.1 - Prob. 14E Ch. 9.1 - Prob. 15E Ch. 9.1 - Prob. 16E Ch. 9.1 - Consider the initial-value problem y = 2x 3y + 1,...Ch. 9.1 - Prob. 18E Ch. 9.1 - Prob. 19E Ch. 9.1 - Repeat Problem 19 using the improved Euler’s...Ch. 9.1 - Prob. 21E Ch. 9.2 - Use the RK4 method with h = 0.1 to approximate...Ch. 9.2 - Prob. 2E Ch. 9.2 - Prob. 3E Ch. 9.2 - Prob. 4E Ch. 9.2 - Prob. 5E Ch. 9.2 - Prob. 6E Ch. 9.2 - Prob. 7E Ch. 9.2 - Prob. 8E Ch. 9.2 - Prob. 9E Ch. 9.2 - Prob. 10E Ch. 9.2 - In Problems 312 use the RK4 method with h = 0.1 to...Ch. 9.2 - Prob. 12E Ch. 9.2 - Prob. 13E Ch. 9.2 - Consider the initial-value problem y′ = 2y, y(0) =...Ch. 9.2 - Prob. 17E Ch. 9.2 - Consider the initial-value problem y′ = 2x – 3y +...Ch. 9.2 - Prob. 19E Ch. 9.2 - Prob. 20E Ch. 9.3 - Prob. 1E Ch. 9.3 - Prob. 3E Ch. 9.3 - Prob. 4E Ch. 9.3 - Prob. 5E Ch. 9.3 - Prob. 6E Ch. 9.3 - Prob. 7E Ch. 9.3 - Prob. 8E Ch. 9.4 - Use Euler’s method to approximate y(0.2), where...Ch. 9.4 - Prob. 2E Ch. 9.4 - Prob. 3E Ch. 9.4 - Prob. 4E Ch. 9.4 - Prob. 5E Ch. 9.5 - Prob. 1E Ch. 9.5 - Prob. 2E Ch. 9.5 - Prob. 3E Ch. 9.5 - Prob. 4E Ch. 9.5 - Prob. 5E Ch. 9.5 - In Problems 1-18 use Definition 7.1.1 to find ℒ{f...Ch. 9.5 - Prob. 7E Ch. 9.5 - Prob. 8E Ch. 9.5 - Prob. 9E Ch. 9.5 - Prob. 10E Ch. 9.5 - Prob. 11E Ch. 9.5 - The electrostatic potential u between two...Ch. 9.5 - Consider the boundary-value problem y″ + xy = 0,...Ch. 9 - Prob. 1RE Ch. 9 - Prob. 2RE Ch. 9 - In Problems 1–4 construct a table comparing the...Ch. 9 - In Problems 1–4 construct a table comparing the...Ch. 9 - Prob. 5RE Ch. 9 - Prob. 6RE Ch. 9 - Prob. 7RE Ch. 9 - Prob. 8RE

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The approximated value of y ( 0.4 ) for the differential equation y ′ = 4 x − 2 y such that y ( 0 ) = 2 by Adams-Bashforth-Moulton method and the value of y 1 , y 2 and y 3 by Runga-Kutta method of fourth order with h = 0.1 .

Solution Summary: The author illustrates the Runga-Kutta method's solution of a linear differential equation of the form yprime

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The approximated value of y(0.4) for the differential equation y′=4x−2y such that y(0)=2 by Adams-Bashforth-Moulton method and the value of y1,y2 and y3 by Runga-Kutta method of fourth order with h=0.1.

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Chapter 9 Solutions