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 ′ = 1 2 − t + 2 y , y ( 0 ) = 1 . Rewrite the differential equation as shown below. d y d t = 1 2 − t + 2 y d y d t − 2 y = 1 2 − t Compare the above linear equation with d y d t + P ( t ) y = G ( t ) and obtain P ( t ) = − 2 , G ( t ) = 1 2 − t . The general solution of d y d t + P ( t ) y = G ( t ) is e ∫ P ( t ) d t y = ∫ G ( t ) e ∫ P ( t ) d t d t + C where e ∫ P ( t ) d y is the integrating factor. Compute the integrating factor for d y d t − 2 y = 1 2 − t as follows. e ∫ P ( t ) d t = e ∫ ( − 2 ) d t = e − 2 ∫ d t = e − 2 t Now the solution of y ′ = 1 2 − t + 2 y is obtained as shown below. e − 2 t y = ∫ ( 1 2 − t ) e − 2 t d t + C = 1 2 ∫ e − 2 t d t − ∫ t e − 2 t d t + C = 1 2 ( e − 2 t − 2 ) − [ t ( e − 2 t − 2 ) + 1 2 ∫ e − 2 t d t ] + C ( Since ∫ u d v = u v − ∫ v d u ) = − 1 4 e − 2 t − ( − 1 2 t e − 2 t − 1 4 e − 2 t + C ) On further simplification, e − 2 t y = − 1 4 e − 2 t − ( − 1 2 t e − 2 t − 1 4 e − 2 t + C ) = − 1 4 e − 2 t + 1 2 t e − 2 t + 1 4 e − 2 t + C y = 1 2 t + C e 2 t Now evaluate the constant C by using the initial condition y ( 0 ) = 1 as follows. y = 1 2 t + C e 2 t 1 = 1 2 ( 0 ) + C e 0 1 = 0 + C C = 1 Therefore the solution of y ′ = 1 2 − t + 2 y becomes, y = 1 2 t + C e 2 t = 1 2 t + e 2 t Thus, the exact solution of y ′ = 2 y − 1 , y ( 0 ) = 1 is y = ϕ ( t ) = 1 2 t + e 2 t . By Euler’s formula, obtain the n t h approximate value of the solution as follows. y n + 1 = y n + f n h = y n + ( 1 2 − t n + 2 y n ) h , n = 0 , 1 , 2 , … Substitute n = 0 , in y n + 1 = y n + ( 1 2 − t n + 2 y n ) h and obtain y 1 as shown below. y 1 = y 0 + ( 1 2 − t 0 + 2 y 0 ) h = 1 + ( 1 2 − 0 + 2 ) h = 1 + 2 h + 1 2 h = ( 1 + 2 h ) + 1 2 ( t 1 − t 0 ) = ( 1 + 2 h ) + t 1 2 Substitute n = 1 , in y n + 1 = y n + ( 1 2 − t n + 2 y n ) h and obtain y 2 as shown below

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.7, Problem 23P

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To show: The approximation obtained by Euler method converges to exact solution as h→0.

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Chapter 2.7, Problem 22P

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Chapter 2 Solutions

Elementary Differential Equations

Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - In each of Problems 1 through 12: Draw a direction...Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

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The approximation obtained by Euler method converges to exact solution as h → 0 .

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To show: The approximation obtained by Euler method converges to exact solution as h→0.

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