Explanation The given differential equation is y ″ + 2 y ′ + y = 3 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 + 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 or 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 . u 1 ( t ) = − ∫ y 2 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t = − ∫ t e − t ( 3 e − t ) e − 2 t d t = − ∫ 3 t e − 2 t ⋅ e 2 t d t = − 3 ∫ t d t On further simplification, u 1 ( t ) = − 3 [ t 2 2 ] = − 3 2 t 2 Obtain the value of u 2 ( t ) = − ∫ y 1 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t . u 2 ( t ) = ∫ y 1 ( t ) g ( t ) W [ y 1 , y 2 ] ( t ) d t = ∫ e − t ( 3 e − t ) e − 2 t d t = ∫ 3 e − 2 t ⋅ e 2 t d t = 3 ∫ t d t On further simplification, u 2 ( t ) = 3 ( t ) = 3 t Substitute the values u 1 ( t ) = − 3 2 t 2 , u 2 ( t ) = 3 t , y 1 ( t ) = e − t and y 2 ( t ) = t e − t in Y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t )

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Differential Equations

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Chapter 3.6, Problem 3P

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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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Chapter 3.6, Problem 2P

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Elementary Differential Equations

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