Explanation The given differential equation is y ″ − y ′ − 2 y = 2 e − t . “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 − r − 6 = 0 . r 2 − r − 6 = 0 r 2 + r − 2 r − 2 = 0 r ( r + 1 ) − 2 ( r + 1 ) = 0 ( r + 1 ) ( r − 2 ) = 0 On further simplification, r − 2 = 0 or r + 1 = 0 r = 2 or r = − 1 The characteristic roots are r 1 = − 1 or r 2 = 2 . Thus, the complementary solution is y c ( t ) = c 1 e − t + c 2 e 2 t . Note that, the functions y 1 = e 2 t and y 2 = e − t are the fundamental set of solutions of the homogeneous equation. Find the wronskian of the functions y 1 = e 2 t and y 2 = e − t . W [ y 1 , y 2 ] = | y 1 y 2 y ′ 1 y ′ 2 | = | e 2 t e − t 2 e 2 t − e − t | = e 2 t ⋅ − e − t − 2 e 2 t ⋅ e − t = − e t − 2 e t = − 3 e t Obtain the value of u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t . u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t = − ∫ e − t ( 2 e − t ) − 3 e t d t = − ∫ 2 e − 2 t − 3 e t d t = 2 3 ∫ e − 3 t d t On further simplification, u 1 ( t ) = 2 3 ∫ e − 3 t − 3 d t = − 2 9 e − 3 t Obtain the value of u 2 ( t ) = − ∫ y 1 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3.6, Problem 2P

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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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