Explanation Given: The given differential equation is, y ′ = 1 − y y ( 0 ) = 0 The actual solution is, y = 1 − e − x Formula used: The recursive formulas for Taylor method of order second are, x n + 1 = x n + h y n + 1 = y n + h f ( x n , y n ) + h 2 2 f 2 ( x n , y n ) The recursive formulas for Taylor method of order fourth are, x n + 1 = x n + h y n + 1 = y n + h f ( x n , y n ) + h 2 2 f 2 ( x n , y n ) + h 3 3 ! f 3 ( x n , y n ) + h 4 4 ! f 4 ( x n , y n ) Calculation: Consider the differential equation. y ′ = 1 − y …… ( 1 ) The differential equation is the function of x and y so, y ′ = f ( x , y ) , y ( x 0 ) = y 0 …… ( 2 ) Compare equation ( 1 ) and equation ( 2 ) . f ( x , y ) = 1 − y x 0 = 0 y o = 0 Differentiate the function f ( x , y ) = 1 − y with respect to x , f x ( x , y ) = 0 Differentiate the function f ( x , y ) = 1 − y with respect to x , f y ( x , y ) = − 1 To calculate, f 2 ( x , y ) = f x + f y ( f ( x , y ) ) …… ( 3 ) Substitute 0 for f x , − 1 for f y and x − y for f ( x , y ) in equation ( 3 ) . f 2 ( x , y ) = y − 1 Consider the recursive formulas for Taylor method of order two. y n + 1 = y n + h f ( x n , y n ) + h 2 2 f 2 ( x n , y n ) …… ( 4 ) Substitute 1 − y n for f ( x n , y n ) , y n − 1 for f 2 ( x n , y n ) in equation ( 4 ) . y n + 1 = y n + h ( 1 − y n ) + h 2 2 ( y n − 1 ) …… ( 5 ) Consider the recursive formulas for Taylor method of order four. y n + 1 = y n + h f ( x n , y n ) + h 2 2 f 2 ( x n , y n ) + h 3 3 ! f 3 ( x n , y n ) + h 4 4 ! f 4 ( x n , y n ) …… ( 6 ) f 3 ( x , y ) = y ‴ = d f 2 d x …… ( 7 ) f 4 ( x , y ) = d f 3 d x …… ( 8 ) Differentiate equation ( 7 ) with respect to x . f 3 ( x , y ) = d d x ( − y + 1 ) = − y + 1 Differentiate equation ( 8 ) with respect to x . f 4 ( x , y ) = d d x ( 1 − y ) = y − 1 Substitute 1 − y n for f ( x n , y n ) , y n − 1 for f 2 ( x n , y n ) , 1 − y n for f 3 ( x n , y n ) and y n − 1 for f 4 ( x n , y n ) in equation ( 6 )

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 3.7, Problem 6E

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Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

Ch. 3.2 - A brine solution of salt flows at a constant rate...Ch. 3.2 - Prob. 2E Ch. 3.2 - Prob. 3E Ch. 3.2 - A brine solution of salt flows at a constant rate...Ch. 3.2 - A swimming pool whose volume is 10,000gal contains...Ch. 3.2 - The air in a small room 12ft by 8ft by 8ft is 3...Ch. 3.2 - Beginning at time t=0, fresh water is pumped at...Ch. 3.2 - A tank initially contains S0lb of salt dissolved...Ch. 3.2 - In 1990 the Department of Natural Resources...Ch. 3.2 - Prob. 10E

Ch. 3.2 - Prob. 11E Ch. 3.2 - For the logistic curve15, assume pa:=p(ta) and...Ch. 3.2 - In Problem 9, suppose we have the additional...Ch. 3.2 - Prob. 14E Ch. 3.2 - Prob. 15E Ch. 3.2 - 16 Show that for a differentiable function p(t),...Ch. 3.2 - Prob. 18E Ch. 3.2 - Prob. 19E Ch. 3.2 - Prob. 20E Ch. 3.2 - A snowball melts in such a way that the rate of...Ch. 3.2 - Prob. 22E Ch. 3.2 - Prob. 23E Ch. 3.2 - Prob. 24E Ch. 3.2 - Prob. 25E Ch. 3.2 - Prob. 26E Ch. 3.2 - Prob. 27E Ch. 3.3 - Prob. 1E Ch. 3.3 - Prob. 2E Ch. 3.3 - Prob. 3E Ch. 3.3 - Prob. 4E Ch. 3.3 - Prob. 5E Ch. 3.3 - Prob. 6E Ch. 3.3 - Prob. 7E Ch. 3.3 - Prob. 8E Ch. 3.3 - Prob. 9E Ch. 3.3 - Early Monday morning, the temperature in the...Ch. 3.3 - During the summer the temperature inside a van...Ch. 3.3 - Prob. 12E Ch. 3.3 - Prob. 13E Ch. 3.3 - Prob. 14E Ch. 3.3 - Prob. 15E Ch. 3.3 - Prob. 16E Ch. 3.4 - Prob. 1E Ch. 3.4 - Prob. 2E Ch. 3.4 - Prob. 3E Ch. 3.4 - Prob. 4E Ch. 3.4 - Unless otherwise stated, in the following problems...Ch. 3.4 - Unless otherwise stated, in the following problems...Ch. 3.4 - Prob. 7E Ch. 3.4 - Unless otherwise stated, in the following problems...Ch. 3.4 - Prob. 9E Ch. 3.4 - Unless otherwise stated, in the following problems...Ch. 3.4 - Prob. 11E Ch. 3.4 - Prob. 12E Ch. 3.4 - Prob. 13E Ch. 3.4 - Prob. 14E Ch. 3.4 - Prob. 15E Ch. 3.4 - Prob. 16E Ch. 3.4 - In Problem 16, let I=50 kg-m2 and the retarding...Ch. 3.4 - Prob. 18E Ch. 3.4 - Prob. 19E Ch. 3.4 - Prob. 20E Ch. 3.4 - Prob. 21E Ch. 3.4 - Prob. 22E Ch. 3.4 - Prob. 23E Ch. 3.4 - Rocket Flight. A model rocket having initial mass...Ch. 3.4 - Escape Velocity. According to Newtons law of...Ch. 3.5 - An RL circuit with a 5- resistor and a 0.05-H...Ch. 3.5 - Prob. 2E Ch. 3.5 - The pathway for a binary electrical signal between...Ch. 3.5 - If the resistance in the RL circuit of Figure...Ch. 3.5 - Prob. 5E Ch. 3.5 - 6. Derive a power balance equation for the RL and...Ch. 3.5 - 7. An industrial electromagnet can be modeled as...Ch. 3.5 - 8. A 108F capacitor 10 nanofarads is charged to 50...Ch. 3.6 - Prob. 1E Ch. 3.6 - Prob. 2E Ch. 3.6 - Prob. 3E Ch. 3.6 - In Example 1, page 126, the improved Eulers method...Ch. 3.6 - Prob. 5E Ch. 3.6 - Prob. 6E Ch. 3.6 - Prob. 7E Ch. 3.6 - Use the improved Eulers method subroutine with...Ch. 3.6 - Prob. 9E Ch. 3.6 - Prob. 10E Ch. 3.6 - Use the improved Eulers method with tolerance to...Ch. 3.6 - Prob. 12E Ch. 3.6 - Prob. 13E Ch. 3.6 - Prob. 14E Ch. 3.6 - The solution to the initial value problem...Ch. 3.6 - Prob. 16E Ch. 3.6 - Prob. 17E Ch. 3.6 - Prob. 18E Ch. 3.6 - Prob. 20E Ch. 3.7 - Determine the recursive formulas for the Taylor...Ch. 3.7 - Determine the recursive formulas for the Taylor...Ch. 3.7 - Prob. 3E Ch. 3.7 - Prob. 4E Ch. 3.7 - Prob. 5E Ch. 3.7 - Prob. 6E Ch. 3.7 - Prob. 7E Ch. 3.7 - Prob. 8E Ch. 3.7 - Prob. 9E Ch. 3.7 - Prob. 10E Ch. 3.7 - Prob. 11E Ch. 3.7 - Prob. 12E Ch. 3.7 - Prob. 13E Ch. 3.7 - Prob. 14E Ch. 3.7 - Prob. 15E Ch. 3.7 - Prob. 16E Ch. 3.7 - The Taylor method of order 2 can be used to...Ch. 3.7 - Prob. 18E Ch. 3.7 - Prob. 19E Ch. 3.7 - Prob. 20E Ch. 3.7 - Prob. 21E

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The solution for the Taylor method of order 2 and order 4 .

Solution Summary: The author explains that the solution for Taylor method of order 2 and order 4 is y(1)=0.62747.

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The solution for the Taylor method of order 2 and order 4 .

Solution Summary: The author explains that the solution for Taylor method of order 2 and order 4 is y(1)=0.62747.

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To find: The solution for the Taylor method of order 2 and order 4.

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The solution for the Taylor method of order 2 and order 4.