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 ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t . The two independent solutions to the homogenous equation t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t are y 1 = 5 t − 1 and y 2 = e − 5 t . The general solution to the homogenous equation will be, y ( t ) = c 1 ( 5 t − 1 ) + c 2 ( e − 5 t ) . The derivatives is, y ′ 1 ( t ) = 5 and y ′ 2 ( t ) = − 5 e − 5 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 ) ( 5 t − 1 ) + v 2 ( t ) ( e − 5 t ) . Obtain the Wronskian y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) as, W = | 5 t − 1 e − 5 t 5 − 5 e − 5 t | = ( 5 t − 1 ) ( − 5 e − 5 t ) − 5 e − 5 t = − 25 t e − 5 t + 5 e − 5 t − 5 e − 5 t = − 25 t e − 5 t Comparing t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t with the general form a y ″ + b y ′ + c y = f implies a = t , b = 5 t − 1 , c = 5 and f = t 2 e − 5 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 25E

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The general solution to the differential equation ty″+(5t−1)y′−5y=t2e−5t, where y1=5t−1 and y2=e−5t are linearly independent solutions to the corresponding homogenous equation for t>0.

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Chapter 4.6, Problem 24E

Chapter 4.7, Problem 1E

Chapter 4 Solutions

Fundamentals of Differential Equations (9th Edition)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

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The general solution to the differential equation t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t , where y 1 = 5 t − 1 and y 2 = e − 5 t are linearly independent solutions to the corresponding homogenous equation for t &gt; 0 .

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The general solution to the differential equation ty″+(5t−1)y′−5y=t2e−5t, where y1=5t−1 and y2=e−5t are linearly independent solutions to the corresponding homogenous equation for t>0.

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The general solution to the differential equation t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t , where y 1 = 5 t − 1 and y 2 = e − 5 t are linearly independent solutions to the corresponding homogenous equation for t > 0 .