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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

The general solution to the differential equation y ″ + 4 y ′ + 4 y = e − 2 t ln t . The general solution to the differential equation y ″ + 4 y ′ + 4 y = e − 2 t ln t is y ( t ) = c 1 e − 2 t + c 2 t e − 2 t + ( 2 ln t − 3 4 ) t 2 e − 2 t _ . 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, y ″ + 4 y ′ + 4 y = e − 2 t ln t . Compare y ″ + 4 y ′ + 4 y = e − 2 t ln t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 4 , c = 4 and f = e − 2 t ln t . The corresponding auxiliary equation is r 2 + 4 r + 4 = 0 . Simplify r 2 + 4 r + 4 = 0 as, r 2 + 2 r + 2 r + 4 = 0 ( r + 2 ) ( r + 2 ) = 0 r = − 2 , − 2 That is, the repeated roots are r 1 = − 2 and r 2 = − 2 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 e − 2 t + c 2 t e − 2 t . That is, the two independent solutions to the homogenous equation y ″ + 4 y ′ + 4 y = 0 are e − 2 t and t e − 2 t . Let y 1 ( t ) = e − 2 t and y 2 ( t ) = t e − 2 t . Then, y ′ 1 ( t ) = − 2 e − t and y ′ 2 ( t ) = − 2 t e − 2 t + e − 2 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 − 2 t + v 2 ( t ) t e − 2 t . Thus, the system of equations y 1 v ′ 1 + y 2 v ′ 2 = 0 and y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a for a = 1 , y 1 ( t ) = e − 2 t , y 2 ( t ) = t e − 2 t , y ′ 1 ( t ) = − 2 e − 2 t , y ′ 2 ( t ) = − 2 t e − 2 t + e − 2 t and f = e − 2 t ln t becomes, e − 2 t ( v ′ 1 ) + t e − 2 t ( v ′ 2 ) = 0 … ( 1 ) − 2 e − 2 t ( v ′ 1 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t … ( 2 ) Multiply equation (1) by 2 and then add the resultant equations: 2 e − 2 t ( v ′ 1 ) + 2 t e − 2 t ( v ′ 2 ) = 0 + − 2 e − 2 t ( v ′ 1 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t _ 2 t e − 2 t ( v ′ 2 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t Simplify 2 t e − 2 t ( v ′ 2 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t as, 2 t e − 2 t ( v ′ 2 ) − 2 t e − t ( v ′ 2 ) + e − 2 t ( v ′ 2 ) = e − 2 t ln t e − 2 t ( v ′ 2 ) = e − 2 t ln t v ′ 2 = ln t Integrating v ′ 2 ( t ) = ln t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ ln t d t = t ln t − t + C 1 Substitute v ′ 2 ( t ) = ln t in equation (1), 2 e − 2 t ( v ′ 1 ) + 2 t e − 2 t ( v ′ 2 ) = 0 2 e − 2 t ( v ′ 1 ) + 2 t e − 2 t ln t = 0 2 e − 2 t ( v ′ 1 ) = − 2 t e − 2 t ln t v ′ 1 = − t ln t Integrating v ′ 1 ( t ) = − t ln t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ − t ln t d t = − [ ∫ t ln t d t ] = − [ ln t ∫ t d t − ∫ t 2 2 d t t ] ( using integration by parts ) Simplify further as, v 1 ( t ) = ∫ t 2 d t − ln t ∫ t d t = t 2 4 − t 2 2 ln t + C 2 Since particular solution is to be obtained, take both C 1 and C 2 zero for simplicity. That is, v 1 ( t ) = t 2 4 − t 2 2 ln t and v 2 ( t ) = t ln t − t . Thus, the particular solution y p ( t ) = v 1 ( t ) e − 2 t + v 2 ( t ) t e − 2 t becomes, y p ( t ) = v 1 ( t ) e − 2 t + v 2 ( t ) t e − 2 t = ( t 2 4 − t 2 2 ln t ) e − 2 t + ( t ln t − t ) t e − 2 t = t 2 4 e − 2 t − t 2 2 e − 2 t ln t + t 2 e − 2 t ln t − t 2 e − 2 t = t 2 e − 2 t ln t − t 2 2 e − 2 t ln t + t 2 4 e − 2 t − t 2 e − 2 t Simplify further as, y p ( t ) = t 2 2 e − 2 t ln t − 3 4 t 2 e − 2 t = ( ln t 2 − 3 4 ) t 2 e − 2 t = ( 2 ln t − 3 4 ) t 2 e − 2 t Combine the general solution y ( t ) = c 1 e − 2 t + c 2 t e − 2 t of the corresponding homogenous equation and the particular solution y p ( t ) = ( 2 ln t − 3 4 ) t 2 e − 2 t to obtain y ( t ) = c 1 e − 2 t + c 2 t e − 2 t + ( 2 ln t − 3 4 ) t 2 e − 2 t . Therefore, the general solution to the differential equation y ″ + 4 y ′ + 4 y = e − 2 t ln t is y ( t ) = c 1 e − 2 t + c 2 t e − 2 t + ( 2 ln t − 3 4 ) t 2 e − 2 t _ .

The general solution to the differential equation y ″ + 4 y ′ + 4 y = e − 2 t ln t . The general solution to the differential equation y ″ + 4 y ′ + 4 y = e − 2 t ln t is y ( t ) = c 1 e − 2 t + c 2 t e − 2 t + ( 2 ln t − 3 4 ) t 2 e − 2 t _ . 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, y ″ + 4 y ′ + 4 y = e − 2 t ln t . Compare y ″ + 4 y ′ + 4 y = e − 2 t ln t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 4 , c = 4 and f = e − 2 t ln t . The corresponding auxiliary equation is r 2 + 4 r + 4 = 0 . Simplify r 2 + 4 r + 4 = 0 as, r 2 + 2 r + 2 r + 4 = 0 ( r + 2 ) ( r + 2 ) = 0 r = − 2 , − 2 That is, the repeated roots are r 1 = − 2 and r 2 = − 2 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 e − 2 t + c 2 t e − 2 t . That is, the two independent solutions to the homogenous equation y ″ + 4 y ′ + 4 y = 0 are e − 2 t and t e − 2 t . Let y 1 ( t ) = e − 2 t and y 2 ( t ) = t e − 2 t . Then, y ′ 1 ( t ) = − 2 e − t and y ′ 2 ( t ) = − 2 t e − 2 t + e − 2 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 − 2 t + v 2 ( t ) t e − 2 t . Thus, the system of equations y 1 v ′ 1 + y 2 v ′ 2 = 0 and y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a for a = 1 , y 1 ( t ) = e − 2 t , y 2 ( t ) = t e − 2 t , y ′ 1 ( t ) = − 2 e − 2 t , y ′ 2 ( t ) = − 2 t e − 2 t + e − 2 t and f = e − 2 t ln t becomes, e − 2 t ( v ′ 1 ) + t e − 2 t ( v ′ 2 ) = 0 … ( 1 ) − 2 e − 2 t ( v ′ 1 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t … ( 2 ) Multiply equation (1) by 2 and then add the resultant equations: 2 e − 2 t ( v ′ 1 ) + 2 t e − 2 t ( v ′ 2 ) = 0 + − 2 e − 2 t ( v ′ 1 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t _ 2 t e − 2 t ( v ′ 2 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t Simplify 2 t e − 2 t ( v ′ 2 ) + ( − 2 t e − t + e − 2 t ) ( v ′ 2 ) = e − 2 t ln t as, 2 t e − 2 t ( v ′ 2 ) − 2 t e − t ( v ′ 2 ) + e − 2 t ( v ′ 2 ) = e − 2 t ln t e − 2 t ( v ′ 2 ) = e − 2 t ln t v ′ 2 = ln t Integrating v ′ 2 ( t ) = ln t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ ln t d t = t ln t − t + C 1 Substitute v ′ 2 ( t ) = ln t in equation (1), 2 e − 2 t ( v ′ 1 ) + 2 t e − 2 t ( v ′ 2 ) = 0 2 e − 2 t ( v ′ 1 ) + 2 t e − 2 t ln t = 0 2 e − 2 t ( v ′ 1 ) = − 2 t e − 2 t ln t v ′ 1 = − t ln t Integrating v ′ 1 ( t ) = − t ln t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ − t ln t d t = − [ ∫ t ln t d t ] = − [ ln t ∫ t d t − ∫ t 2 2 d t t ] ( using integration by parts ) Simplify further as, v 1 ( t ) = ∫ t 2 d t − ln t ∫ t d t = t 2 4 − t 2 2 ln t + C 2 Since particular solution is to be obtained, take both C 1 and C 2 zero for simplicity. That is, v 1 ( t ) = t 2 4 − t 2 2 ln t and v 2 ( t ) = t ln t − t . Thus, the particular solution y p ( t ) = v 1 ( t ) e − 2 t + v 2 ( t ) t e − 2 t becomes, y p ( t ) = v 1 ( t ) e − 2 t + v 2 ( t ) t e − 2 t = ( t 2 4 − t 2 2 ln t ) e − 2 t + ( t ln t − t ) t e − 2 t = t 2 4 e − 2 t − t 2 2 e − 2 t ln t + t 2 e − 2 t ln t − t 2 e − 2 t = t 2 e − 2 t ln t − t 2 2 e − 2 t ln t + t 2 4 e − 2 t − t 2 e − 2 t Simplify further as, y p ( t ) = t 2 2 e − 2 t ln t − 3 4 t 2 e − 2 t = ( ln t 2 − 3 4 ) t 2 e − 2 t = ( 2 ln t − 3 4 ) t 2 e − 2 t Combine the general solution y ( t ) = c 1 e − 2 t + c 2 t e − 2 t of the corresponding homogenous equation and the particular solution y p ( t ) = ( 2 ln t − 3 4 ) t 2 e − 2 t to obtain y ( t ) = c 1 e − 2 t + c 2 t e − 2 t + ( 2 ln t − 3 4 ) t 2 e − 2 t . Therefore, the general solution to the differential equation y ″ + 4 y ′ + 4 y = e − 2 t ln t is y ( t ) = c 1 e − 2 t + c 2 t e − 2 t + ( 2 ln t − 3 4 ) t 2 e − 2 t _ .

Solution Summary: The author explains the steps to obtain a particular solution to the differential equation y′′+4

Fundamentals of Differential Equations (9th Edition)

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 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

Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 3E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 5E Ch. 4.2 - Prob. 6E Ch. 4.2 - Prob. 7E Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Prob. 10E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 15E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - Prob. 19E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Prob. 24E Ch. 4.2 - Prob. 25E Ch. 4.2 - Prob. 26E Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - Prob. 31E Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Using the definition in Problem 35, prove that if...Ch. 4.2 - Prob. 37E Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.2 - In Problems 3741, find three linearly independent...Ch. 4.2 - Prob. 42E Ch. 4.2 - Prob. 43E Ch. 4.2 - Solve the initial value problem: y 2y y + 2y =...Ch. 4.2 - Prob. 45E Ch. 4.2 - Prob. 46E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - Prob. 5E Ch. 4.3 - Prob. 6E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 920, find a general solution. 9. y 8y...Ch. 4.3 - In Problems 920, find a general solution. 10. y +...Ch. 4.3 - Prob. 11E Ch. 4.3 - Prob. 12E Ch. 4.3 - In Problems 920, find a general solution. 13. y ...Ch. 4.3 - Prob. 14E Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - In Problems 920, find a general solution. 17. y y...Ch. 4.3 - In Problems 920, find a general solution. 18. 2y +...Ch. 4.3 - Prob. 19E Ch. 4.3 - In Problems 920, find a general solution. 20. y y...Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 25E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 27E Ch. 4.3 - Prob. 28E Ch. 4.3 - Prob. 29E Ch. 4.3 - Prob. 30E Ch. 4.3 - Prob. 31E Ch. 4.3 - Prob. 32E Ch. 4.3 - Prob. 33E Ch. 4.3 - Prob. 34E Ch. 4.3 - Swinging Door. 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Derivatives of Trigonometric Functions - Product Rule Quotient & Chain Rule - Calculus Tutorial; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=_niP0JaOgHY;License: Standard YouTube License, CC-BY