Explanation The given differential equation is y ″ − 2 y ′ + y = e t ( 1 + t 2 ) . “Note that, the general solution of the differential equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) is in the form of Y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) , where y 1 ( t ) and y 2 ( t ) form a fundamental set of solution of the homogenous differential equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = 0 , here u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t and u 2 ( t ) = ∫ y 1 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t .” Obtain the complementary solution of the given differential equation. The characteristic equation is r 2 − 2 r + 1 = 0 . r 2 − 2 r + 1 = 0 r 2 − r − r + 1 = 0 r ( r − 1 ) − 1 ( r − 1 ) = 0 ( r − 1 ) ( r − 1 ) = 0 On further simplification, r − 1 = 0 or r − 1 = 0 r = 1 or r = 1 The characteristic roots are r 1 = 1 and r 2 = 1 . Since the roots are repeated, the complementary solution is y c ( t ) = c 1 e t + c 2 t e t . Note that, the functions y 1 = e t and y 2 = t e t are the fundamental set of solutions of the homogeneous equation. Find the wronskian of the functions y 1 = e t and y 2 = t e t . W [ y 1 , y 2 ] = | y 1 y 2 y ′ 1 y ′ 2 | = | e t t e t e t e t + t e t | = e t ( e t + t e t ) − ( e t ) ( t e t ) = e 2 t + t e 2 t − t e 2 t = e 2 t Obtain the value of u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t

Elementary Differential Equations And Boundary Value Problems

11th Edition

ISBN: 9781119256007

Author: Boyce, William E.

Publisher: John Wiley & Sons, Incorporated

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Question

Chapter 3.6, Problem 8P

To determine

The particular solution of the given differential equation by using the method of variation of parameters and check the result by using the method of undetermined coefficient.

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Chapter 3.6, Problem 7P

Chapter 3.6, Problem 9P

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Chapter 3 Solutions

Elementary Differential Equations And Boundary Value Problems

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