Explanation Formula used: Euler’s formula : The Euler’s formula is given by, y n + 1 = y n + f n h , n = 0 , 1 , 2 , … , where h is the uniform size between the points t 0 , t 1 , t 2 , … , such that t n + 1 = t n + h . Calculation: The given initial value problem is y ′ = t 2 + y 2 , y ( 0 ) = 1 . By Euler’s formula, obtain the n t h approximate value of the solution with h = 0.1 as follows. y n + 1 = y n + f n h = y n + ( t n 2 + y n 2 ) ( 0.1 ) Obtain the first approximation by substituting n = 0 in y n + 1 = y n + ( t n 2 + y n 2 ) ( 0.1 ) as follows. y 1 = y 0 + ( t 0 2 + y 0 2 ) ( 0.1 ) = 1 + ( 0 2 + 1 2 ) ( 0.1 ) = 1 + ( 1 ) ( 0.1 ) = 1.1 Thus, the value of y ( 0.1 ) is 1.1. Obtain the second approximation by substituting n = 1 in y n + 1 = y n + ( t n 2 + y n 2 ) ( 0.1 ) as follows. y 2 = y 1 + ( t 1 2 + y 1 2 ) ( 0.1 ) = 1.1 + ( ( 0.1 ) 2 + ( 1.1 ) 2 ) ( 0.1 ) = 1.1 + ( 1.22 ) ( 0.1 ) = 1.1 + 0.122 = 1.222 Thus, the value of y ( 0.2 ) is 1.222. Proceed in the similar way and obtain the approximate values as shown in below table. t y (Euler) 0 1 0.1 1.1 0.2 1.222 0.3 1.375328 0.4 1.573481 0.5 1.837066 0.6 2.199547 0.7 2.719347 0.8 3.507832 0.9 4.80232 1 7.189548 From the above table, it is observed that the approximate values of solution at t = 0.8 and at t = 1 with h = 0.1 are 3.507832 , and 7 .189548 _ respectively. By Euler’s formula, obtain the n t h approximate value of the solution with h = 0.05 as follows. y n + 1 = y n + f n h = y n + ( t n 2 + y n 2 ) ( 0.05 ) Obtain the first approximation by substituting n = 0 in y n + 1 = y n + ( t n 2 + y n 2 ) ( 0.05 ) as follows. y 1 = y 0 + ( t 0 2 + y 0 2 ) ( 0.05 ) = 1 + ( 0 2 + 1 2 ) ( 0.05 ) = 1 + ( 1 ) ( 0.05 ) = 1.05 Thus, the value of y ( 0.05 ) is 1.05. Obtain the second approximation by substituting n = 1 in y n + 1 = y n + ( t n 2 + y n 2 ) ( 0.05 ) as follows. y 2 = y 1 + ( t 1 2 + y 1 2 ) ( 0.05 ) = 1.05 + ( ( 0.05 ) 2 + ( 1.05 ) 2 ) ( 0.05 ) = 1.05 + ( 1.105 ) ( 0.05 ) = 1.05 + 0.05525 = 1.10525 Thus, the value of y ( 0.1 ) is 1.10525. Proceed in the similar way and obtain the approximate values as shown in below table. t y (Euler) 0 1 0.05 1.05 0.1 1.10525 0.15 1.166829 0.2 1.236028 0.25 1.314417 0.3 1.403926 0.35 1.506977 0.4 1.626651 0.45 1.76695 0.5 1.933181 0.55 2.13254 0.6 2.375052 0.65 2.675095 0.7 3.054027 0.75 3.544881 0.8 4.201315 0.85 5.115867 0.9 6.460597 0.95 8.588063 1 12.32093 From the above table, it is observed that the approximate values of solution at t = 0.8 and at t = 1 with h = 0.05 are 4.201315 , and 12 .32093 _ respectively. By Euler’s formula, obtain the n t h approximate value of the solution with h = 0.025 as follows. y n + 1 = y n + f n h = y n + ( t n 2 + y n 2 ) ( 0.025 ) Obtain the first approximation by substituting n = 0 in y n + 1 = y n + ( t n 2 + y n 2 ) ( 0.025 ) as follows. y 1 = y 0 + ( t 0 2 + y 0 2 ) ( 0

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Chapter 2.7, Problem 12P

To determine

The best estimate values of the solution of initial value problem y′=t2+y2, y(0)=1 using Euler method with h=0.1, 0.05, 0.025, and 0.01 at t=0.8 and at t=1.

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Chapter 2.7, Problem 11P

Chapter 2.7, Problem 13P

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Chapter 2 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

Ch. 2.1 - Prob. 11P Ch. 2.1 - Prob. 12P Ch. 2.1 - Prob. 13P Ch. 2.1 - Prob. 14P Ch. 2.1 - Prob. 15P Ch. 2.1 - Prob. 16P Ch. 2.1 - Prob. 17P Ch. 2.1 - Prob. 18P Ch. 2.1 - Prob. 19P Ch. 2.1 - Prob. 20P Ch. 2.1 - Prob. 21P Ch. 2.1 - Prob. 22P Ch. 2.1 - Prob. 23P Ch. 2.1 - Prob. 24P Ch. 2.1 - Prob. 25P Ch. 2.1 - Prob. 26P Ch. 2.1 - Prob. 27P Ch. 2.1 - Variation of Parameters. Consider the following...Ch. 2.1 - Prob. 29P Ch. 2.1 - In each of Problems 29 and 30, use the method of...Ch. 2.2 - Prob. 1P Ch. 2.2 - Prob. 2P Ch. 2.2 - Prob. 3P Ch. 2.2 - Prob. 4P Ch. 2.2 - Prob. 5P Ch. 2.2 - Prob. 6P Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - Prob. 9P Ch. 2.2 - Prob. 10P Ch. 2.2 - Prob. 11P Ch. 2.2 - Prob. 12P Ch. 2.2 - Prob. 13P Ch. 2.2 - Prob. 14P Ch. 2.2 - Prob. 15P Ch. 2.2 - Prob. 16P Ch. 2.2 - Prob. 17P Ch. 2.2 - Prob. 18P Ch. 2.2 - Prob. 19P Ch. 2.2 - Prob. 20P Ch. 2.2 - Prob. 21P Ch. 2.2 - Prob. 22P Ch. 2.2 - Prob. 23P Ch. 2.2 - Prob. 24P Ch. 2.2 - Prob. 25P Ch. 2.2 - Prob. 26P Ch. 2.2 - Prob. 27P Ch. 2.2 - Prob. 28P Ch. 2.2 - Prob. 29P Ch. 2.2 - Prob. 30P Ch. 2.2 - Prob. 31P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P Ch. 2.3 - Prob. 3P Ch. 2.3 - Prob. 4P Ch. 2.3 - Prob. 5P Ch. 2.3 - Prob. 6P Ch. 2.3 - Prob. 7P Ch. 2.3 - Prob. 8P Ch. 2.3 - Prob. 9P Ch. 2.3 - Prob. 10P Ch. 2.3 - Prob. 11P Ch. 2.3 - Prob. 12P Ch. 2.3 - Prob. 13P Ch. 2.3 - Prob. 14P Ch. 2.3 - Prob. 15P Ch. 2.3 - Prob. 16P Ch. 2.3 - Assume that the conditions are as in Problem 16...Ch. 2.3 - Prob. 18P Ch. 2.3 - Prob. 19P Ch. 2.3 - Prob. 20P Ch. 2.3 - Prob. 21P Ch. 2.3 - Prob. 22P Ch. 2.3 - Prob. 23P Ch. 2.3 - Prob. 24P Ch. 2.4 - In each of Problems 1 through 6, determine...Ch. 2.4 - Prob. 2P Ch. 2.4 - Prob. 3P Ch. 2.4 - Prob. 4P Ch. 2.4 - Prob. 5P Ch. 2.4 - Prob. 6P Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - Prob. 9P Ch. 2.4 - Prob. 10P Ch. 2.4 - Prob. 11P Ch. 2.4 - Prob. 12P Ch. 2.4 - Prob. 13P Ch. 2.4 - Prob. 14P Ch. 2.4 - Prob. 15P Ch. 2.4 - Prob. 16P Ch. 2.4 - Prob. 17P Ch. 2.4 - Prob. 18P Ch. 2.4 - Prob. 19P Ch. 2.4 - Prob. 20P Ch. 2.4 - Prob. 21P Ch. 2.4 - Prob. 22P Ch. 2.4 - Prob. 23P Ch. 2.4 - Prob. 24P Ch. 2.4 - Prob. 25P Ch. 2.4 - Prob. 26P Ch. 2.4 - Prob. 27P Ch. 2.5 - Prob. 1P Ch. 2.5 - Prob. 2P Ch. 2.5 - Prob. 3P Ch. 2.5 - Prob. 4P Ch. 2.5 - Prob. 5P Ch. 2.5 - Prob. 6P Ch. 2.5 - Prob. 7P Ch. 2.5 - Prob. 8P Ch. 2.5 - Prob. 9P Ch. 2.5 - Prob. 10P Ch. 2.5 - Prob. 11P Ch. 2.5 - Prob. 12P Ch. 2.5 - Prob. 13P Ch. 2.5 - Prob. 14P Ch. 2.5 - Prob. 15P Ch. 2.5 - Prob. 16P Ch. 2.5 - Prob. 17P Ch. 2.5 - Prob. 18P Ch. 2.5 - Prob. 19P Ch. 2.5 - Prob. 20P Ch. 2.5 - Prob. 21P Ch. 2.5 - Prob. 22P Ch. 2.5 - Prob. 23P Ch. 2.5 - Prob. 24P Ch. 2.5 - Prob. 25P Ch. 2.5 - Prob. 26P Ch. 2.5 - Prob. 27P Ch. 2.6 - Prob. 1P Ch. 2.6 - Prob. 2P Ch. 2.6 - Prob. 3P Ch. 2.6 - Prob. 4P Ch. 2.6 - Prob. 5P Ch. 2.6 - Prob. 6P Ch. 2.6 - Prob. 7P Ch. 2.6 - Prob. 8P Ch. 2.6 - Prob. 9P Ch. 2.6 - Prob. 10P Ch. 2.6 - Prob. 11P Ch. 2.6 - Prob. 12P Ch. 2.6 - Prob. 13P Ch. 2.6 - Prob. 14P Ch. 2.6 - Prob. 15P Ch. 2.6 - Prob. 16P Ch. 2.6 - Prob. 17P Ch. 2.6 - Prob. 18P Ch. 2.6 - Prob. 19P Ch. 2.6 - Prob. 20P Ch. 2.6 - Prob. 21P Ch. 2.6 - Prob. 22P Ch. 2.7 - Prob. 1P Ch. 2.7 - Prob. 2P Ch. 2.7 - Prob. 3P Ch. 2.7 - Prob. 4P Ch. 2.7 - Prob. 5P Ch. 2.7 - Prob. 6P Ch. 2.7 - Prob. 7P Ch. 2.7 - Prob. 8P Ch. 2.7 - Prob. 9P Ch. 2.7 - Prob. 10P Ch. 2.7 - Prob. 11P Ch. 2.7 - Prob. 12P Ch. 2.7 - Prob. 14P Ch. 2.7 - Prob. 15P Ch. 2.7 - Prob. 16P Ch. 2.7 - Prob. 17P Ch. 2.8 - Prob. 1P Ch. 2.8 - Prob. 2P Ch. 2.8 - Prob. 3P Ch. 2.8 - Prob. 4P Ch. 2.8 - Prob. 5P Ch. 2.8 - Prob. 6P Ch. 2.8 - Prob. 7P Ch. 2.8 - Prob. 8P Ch. 2.8 - Prob. 9P Ch. 2.8 - Prob. 10P Ch. 2.8 - Prob. 11P Ch. 2.8 - Prob. 12P Ch. 2.8 - Prob. 13P Ch. 2.8 - Prob. 14P Ch. 2.8 - Prob. 15P Ch. 2.8 - Prob. 16P Ch. 2.8 - Prob. 17P Ch. 2.8 - Prob. 18P Ch. 2.9 - Prob. 1P Ch. 2.9 - Prob. 2P Ch. 2.9 - Prob. 3P Ch. 2.9 - Prob. 4P Ch. 2.9 - Prob. 5P Ch. 2.9 - Prob. 6P Ch. 2.9 - Prob. 7P Ch. 2.9 - Prob. 8P Ch. 2.9 - Prob. 9P Ch. 2.9 - Prob. 10P Ch. 2 - Prob. 1MP Ch. 2 - Prob. 2MP Ch. 2 - Prob. 3MP Ch. 2 - Prob. 4MP Ch. 2 - Prob. 5MP Ch. 2 - Prob. 6MP Ch. 2 - Prob. 7MP Ch. 2 - Prob. 8MP Ch. 2 - Prob. 9MP Ch. 2 - Prob. 10MP Ch. 2 - Prob. 11MP Ch. 2 - Prob. 12MP Ch. 2 - Prob. 13MP Ch. 2 - Prob. 14MP Ch. 2 - Prob. 15MP Ch. 2 - Prob. 16MP Ch. 2 - Prob. 17MP Ch. 2 - Prob. 18MP Ch. 2 - Prob. 19MP Ch. 2 - Prob. 20MP Ch. 2 - Prob. 21MP Ch. 2 - Prob. 22MP Ch. 2 - Prob. 23MP Ch. 2 - Prob. 24MP Ch. 2 - Prob. 25MP Ch. 2 - Prob. 28MP Ch. 2 - Prob. 29MP Ch. 2 - Prob. 31MP Ch. 2 - Prob. 32MP Ch. 2 - Prob. 33MP Ch. 2 - Prob. 34MP Ch. 2 - Prob. 35MP Ch. 2 - Prob. 36MP Ch. 2 - Prob. 37MP

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The best estimate values of the solution of initial value problem y ′ = t 2 + y 2 , y ( 0 ) = 1 using Euler method with h = 0.1 , 0 .05, 0 .025, and 0 .01 at t = 0.8 and at t = 1 .

The best estimate values of the solution of initial value problem y′=t2+y2, y(0)=1 using Euler method with h=0.1, 0.05, 0.025, and 0.01 at t=0.8 and at t=1.

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Chapter 2 Solutions