Explanation Procedure used: Method of variation of parameters: The following are the steps to obtain a particular solution to a y ″ + b y ′ + c y = f . (i) Find two linearly independent solutions { y 1 ( t ) , y 2 ( t ) } to the corresponding homogenous equation and take y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . (ii) Determine v 1 ( t ) and v 2 ( t ) by solving the below system for v ′ 1 ( t ) and v ′ 2 ( t ) and integrating. y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a This can be simplified using Cramer’s rule as, v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] and v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] (iii) Substitute v 1 ( t ) and v 2 ( t ) into the expression for y p ( t ) to obtain a particular solution. Calculation: The given differential equation is, t y ″ + ( 1 − 2 t ) y ′ + ( t − 1 ) y = t e t . The two independent solutions to the homogenous equation t y ″ + ( 1 − 2 t ) y ′ + ( t − 1 ) y = t e t are y 1 = e t and y 2 = e t ln t . The general solution to the homogenous equation will be, y ( t ) = c 1 ( e t ) + c 2 ( e t ln t ) . The derivatives is, y ′ 1 ( t ) = e t and y ′ 2 ( t ) = e t t + e t ln t . Set the particular solution y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) as y p ( t ) = v 1 ( t ) e t + v 2 ( t ) ( e t ln t ) . Obtain the Wronskian y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) as, W = | e t e t ln t e t e t t + e t ln t | = ( e t t + e t ln t ) ( e t ) − ( e t ) ( e t ln t ) = e 2 t t + e 2 t ln t − e 2 t ln t = e 2 t t Comparing t y ″ + ( 1 − 2 t ) y ′ + ( t − 1 ) y = t e t with the general form a y ″ + b y ′ + c y = f implies a = t , b = 1 − 2 t , c = t − 1 and f = t e t

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4.6, Problem 24E

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The general solution to the differential equation ty″+(1−2t)y′+(t−1)y=tet, where y1=et and y2=etlnt are linearly independent solutions to the corresponding homogenous equation for t>0.

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Chapter 4.6, Problem 23E

Chapter 4.6, Problem 25E

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Fundamentals of Differential Equations (9th Edition)

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The general solution to the differential equation t y ″ + ( 1 − 2 t ) y ′ + ( t − 1 ) y = t e t , where y 1 = e t and y 2 = e t ln t are linearly independent solutions to the corresponding homogenous equation for t > 0 .

Solution Summary: The author explains the steps to obtain a particular solution to the differential equation ty′′+(1-2t

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The general solution to the differential equation ty″+(1−2t)y′+(t−1)y=tet, where y1=et and y2=etlnt are linearly independent solutions to the corresponding homogenous equation for t>0.

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Chapter 4 Solutions

The general solution to the differential equation t y ″ + ( 1 − 2 t ) y ′ + ( t − 1 ) y = t e t , where y 1 = e t and y 2 = e t ln t are linearly independent solutions to the corresponding homogenous equation for t &gt; 0 .

Solution Summary: The author explains the steps to obtain a particular solution to the differential equation ty′′+(1-2t

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The general solution to the differential equation ty″+(1−2t)y′+(t−1)y=tet, where y1=et and y2=etlnt are linearly independent solutions to the corresponding homogenous equation for t>0.

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Chapter 4 Solutions

The general solution to the differential equation t y ″ + ( 1 − 2 t ) y ′ + ( t − 1 ) y = t e t , where y 1 = e t and y 2 = e t ln t are linearly independent solutions to the corresponding homogenous equation for t > 0 .