Explanation Given: y satisfies the equation y ′′ + 10 y ′ + 25 y = e t 3 subject to y ( 0 ) = 1 y ′ ( 0 ) = − 5 Approach: The differential equation y ′′ + 10 y ′ + 25 y = e t 3 ... ( 1 ) is non-homogeneous. The associated homogeneous differential equation to (1) is y ′′ + 10 y ′ + 25 y = 0 ... ( 2 ) The characteristic equation to (2) is given by a r 2 + b r + c = 0 a is a coefficient of y ′′ , b is a coefficient of y ′ and, c is a coefficient of y . Linearly independent solutions of (2) are y 1 ( t ) = e r t y 2 ( t ) = t e r t ... ( 3 ) The general solution to (1) is given by y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + Y p ( t ) Y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) ... ( 4 ) Such that v 1 and v 2 satisfies the following equations y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f ( t ) a ... ( 5 ) Y p ( t ) is a particular solution to (1) solved by the method of variation of parameters, v 1 ( t ) and v 2 ( t ) are unknown functions of t , f ( t ) is a function of t in (1), a is a coefficient of y ′′ in (1) and c 1 and c 2 are constants. Calculation: From (2) substitute 1 for a , 10 for b and 25 for c in the characteristic equation, a r 2 + b r + c = 0 r 2 + 10 r + 25 = 0 r 1 = r 2 = − 5 ... ( 6 ) From (6) substitute − 5 for r 1 and r 2 in (3) y 1 ( t ) = e − 5 t y 2 ( t ) = t e − 5 t Substitute y 1 ( t ) and y 2 ( t ) in Y p ( t ) in (4) Y p ( t ) = v 1 ( t ) e − 5 t + v 2 ( t ) t e − 5 t ... ( 7 ) From (5) v 1 and v 2 should satisfy the following equations y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f ( t ) a Substitute e − 5 t for y 1 ; t e − 5 t y 2 ; − 5 e − 5 t for y ′ 1 ; e − 5 t − 5 t e − 5 t y ′ 2 ; e t 3 for f ( t ) and 1 for a in (5) e − 5 t v ′ 1 + t e − 5 t v ′ 2 = 0 − 5 e − 5 t v ′ 1 + ( e − 5 t − 5 t e − 5 t ) v ′ 2 = e t 3 Solve for v ′ 1 and v ′ 2 , v ′ 1 ( t ) = − t e ( t 3 + 5 t ) v ′ 2 ( t ) = e ( t 3 + 5 t ) Integrate, v 1 ( t ) = − ∫ t e ( t 3 + 5 t ) d t v 2 ( t ) = ∫ e ( t 3 + 5 t ) d t Substitute v 1 and v 2 in (7) and simplify, Y p ( t ) = v 1 ( t ) e − 5 t + v 2 ( t ) t e − 5 t = − e − 5 t ∫ t e ( t 3 + 5 t ) d t + t e − 5 t ∫ e ( t 3 + 5 t ) d t Substitute values of y 1 ( t ) , y 2 ( t ) , and Y p ( t ) in (4)

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 4.6, Problem 21E

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The estimation of y(0.2) to within ±0.0001 by numerically approximating the integrals in the variation of parameters formula.

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Chapter 4.6, Problem 20E

Chapter 4.6, Problem 22E

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Fundamentals of Differential Equations and Boundary Value Problems

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The estimation of y ( 0.2 ) to within ± 0.0001 by numerically approximating the integrals in the variation of parameters formula.

Solution Summary: The author evaluates the estimation of y(0.2) to within pm 0.0001 by numerically approximating the integrals in the variation of parameters formula.

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To find:

The estimation of y(0.2) to within ±0.0001 by numerically approximating the integrals in the variation of parameters formula.

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Chapter 4 Solutions

The estimation of y ( 0.2 ) to within ± 0.0001 by numerically approximating the integrals in the variation of parameters formula.

Solution Summary: The author evaluates the estimation of y(0.2) to within pm 0.0001 by numerically approximating the integrals in the variation of parameters formula.

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To find: The estimation of y(0.2) to within ±0.0001 by numerically approximating the integrals in the variation of parameters formula.

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Chapter 4 Solutions

To find:

The estimation of y(0.2) to within ±0.0001 by numerically approximating the integrals in the variation of parameters formula.