Explanation Given: The initial value problem y ′′ − y = 1 t y ( 1 ) = 0 y ′ ( 1 ) = − 2 Approach: The differential equation y ′′ − y = 1 t (1) is non-homogeneous. The associated homogeneous differential equation to (1) is y ′′ − 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 1 t y 2 ( t ) = e r 2 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 , 0 for b and − 1 for c in the characteristic equation, a r 2 + b r + c = 0 r 2 − 1 = 0 r 1 = 1 r 2 = − 1 ( 6 ) From (6) substitute 1 for r 1 and − 1 for r 2 in (3) y 1 ( t ) = e t y 2 ( t ) = e − t Substitute y 1 ( t ) and y 2 ( t ) in Y p ( t ) in (4) Y p ( t ) = v 1 ( t ) e t + v 2 ( t ) e − 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 t for y 1 , e − t for y 2 , e t for y ′ 1 , − e − t for y ′ 2 , 1 t for f ( t ) and 1 for a in (5) e t v ′ 1 + e − t v ′ 2 = 0 e t v ′ 1 − e − t v ′ 2 = 1 t Solve for v ′ 1 and v ′ 2 , v ′ 1 ( t ) = e − t 2 t v ′ 2 ( t ) = − e t 2 t Integrate, v 1 ( t ) = ∫ e − t 2 t d t v 2 ( t ) = − ∫ e t 2 t d t Substitute v 1 and v 2 in (7) and simplify, Y p ( t ) = v 1 ( t ) e t + v 2 ( t ) e − t = ( ∫ e − t 2 t d t ) e t + ( − ∫ e t 2 t d t ) e − t = ( ∫ e − t 2 t d t ) e t − ( ∫ e t 2 t d t ) e − 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 19E

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The solution of the given initial value problem using definite integrals and an approximation for y(2) to 2 decimal places using numerical integration.

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The solution of the given initial value problem using definite integrals and an approximation for y ( 2 ) to 2 decimal places using numerical integration .

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The solution of the given initial value problem using definite integrals and an approximation for y(2) to 2 decimal places using numerical integration.

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The solution of the given initial value problem using definite integrals and an approximation for y ( 2 ) to 2 decimal places using numerical integration .

Solution Summary: The author explains the solution of the initial value problem using definite integrals and an approximation for y(2) to 2 decimal places using numerical integration

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To find: The solution of the given initial value problem using definite integrals and an approximation for y(2) to 2 decimal places using numerical integration.

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Chapter 4 Solutions

To find:

The solution of the given initial value problem using definite integrals and an approximation for y(2) to 2 decimal places using numerical integration.