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

The general solution to the differential equation y ″ + 4 y = tan 2 t . The general solution to the differential equation y ″ + 4 y = tan 2 t is y ( t ) = c 1 cos 2 t + c 2 sin 2 t − 1 4 ( cos 2 t ) ln | sec 2 t + tan 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 = tan 2 t . Compare y ″ + 4 y = tan 2 t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 4 and f = tan 2 t . The corresponding auxiliary equation is r 2 + 4 = 0 . Simplify r 2 + 4 = 0 as, r 2 = − 4 r = ± − 4 r = ± 2 i ( ∵ i = − 1 ) That is, the roots are r 1 = 2 i and r 2 = − 2 i . Here, α = 0 and β = 2 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos 2 t + c 2 sin 2 t . That is, the two independent solutions to the homogenous equation y ″ + 4 y = 0 are cos 2 t and sin 2 t . Let y 1 ( t ) = cos 2 t and y 2 ( t ) = sin 2 t . Then, y ′ 1 ( t ) = − 2 sin 2 t and y ′ 2 ( t ) = 2 cos 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 ) cos 2 t + v 2 ( t ) sin 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 ) = cos 2 t , y 2 ( t ) = sin 2 t , y ′ 1 ( t ) = − 2 sin 2 t , y ′ 2 ( t ) = 2 cos 2 t and f = tan 2 t becomes, cos 2 t ( v ′ 1 ) + sin 2 t ( v ′ 2 ) = 0 … ( 1 ) − 2 sin 2 t ( v ′ 1 ) + 2 cos 2 t ( v ′ 2 ) = tan 2 t … ( 2 ) Multiply equation (1) by 2 sin 2 t and equation (2) by cos 2 t ; and then add as follows: 2 sin 2 t cos 2 t ( v ′ 1 ) + 2 ( sin 2 t ) 2 ( v ′ 2 ) = 0 + − 2 sin 2 t cos 2 t ( v ′ 1 ) + 2 ( cos 2 t ) 2 ( v ′ 2 ) = sin 2 t _ 0 + 2 ( sin 2 t ) 2 ( v ′ 2 ) + 2 ( cos 2 t ) 2 ( v ′ 2 ) = sin 2 t Simplify 2 ( sin 2 t ) 2 ( v ′ 2 ) + 2 ( cos 2 t ) 2 ( v ′ 2 ) = sin 2 t as, ( v ′ 2 ) [ 2 ( sin 2 t ) 2 + 2 ( cos 2 t ) 2 ] = sin 2 t ( v ′ 2 ) [ 2 ] = sin 2 t ( ∵ ( sin 2 t ) 2 + ( cos 2 t ) 2 = 1 ) v ′ 2 = 1 2 sin 2 t Integrating v ′ 2 ( t ) = 1 2 sin 2 t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ 1 2 sin 2 t d t = 1 2 ( − cos 2 t 2 ) + C 1 = − 1 4 cos 2 t + C 1 Substitute v ′ 2 ( t ) = 1 2 sin 2 t in equation (1), cos 2 t ( v ′ 1 ) + sin 2 t ( v ′ 2 ) = 0 cos 2 t ( v ′ 1 ) + sin 2 t ( 1 2 sin 2 t ) = 0 cos 2 t ( v ′ 1 ) + 1 2 ( sin 2 t ) 2 = 0 v ′ 1 = − 1 2 ( sin 2 t ) 2 cos 2 t Simplify further as, v ′ 1 = − 1 2 ( 1 − ( cos 2 t ) 2 cos 2 t ) = 1 2 ( ( cos 2 t ) 2 − 1 cos 2 t ) = 1 2 ( cos 2 t − 1 cos 2 t ) = 1 2 ( cos 2 t − sec 2 t ) Integrating v ′ 1 ( t ) = 1 2 ( cos 2 t − sec 2 t ) gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ 1 2 ( cos 2 t − sec 2 t ) d t = 1 2 ( sin 2 t 2 − ln | sec 2 t + tan 2 t | 2 ) + C 2 = 1 4 ( sin 2 t − ln | sec 2 t + tan 2 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 ) = 1 4 ( sin 2 t − ln | sec 2 t + tan 2 t | ) and v 2 ( t ) = − 1 4 cos 2 t . Thus, the particular solution y p ( t ) = v 1 ( t ) cos 2 t + v 2 ( t ) sin 2 t becomes, y p ( t ) = 1 4 ( sin 2 t − ln | sec 2 t + tan 2 t | ) ( cos 2 t ) − 1 4 ( cos 2 t ) ( sin 2 t ) = 1 4 sin 2 t cos 2 t − 1 4 ln | sec 2 t + tan 2 t | ( cos 2 t ) − 1 4 sin 2 t cos 2 t = − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | Combine the general solution y ( t ) = c 1 cos 2 t + c 2 sin 2 t of the corresponding homogenous equation and the particular solution y p ( t ) = − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | to obtain y ( t ) = c 1 cos 2 t + c 2 sin 2 t − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | . Therefore, the general solution to the differential equation y ″ + 4 y = tan 2 t is y ( t ) = c 1 cos 2 t + c 2 sin 2 t − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | _ .

The general solution to the differential equation y ″ + 4 y = tan 2 t . The general solution to the differential equation y ″ + 4 y = tan 2 t is y ( t ) = c 1 cos 2 t + c 2 sin 2 t − 1 4 ( cos 2 t ) ln | sec 2 t + tan 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 = tan 2 t . Compare y ″ + 4 y = tan 2 t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 4 and f = tan 2 t . The corresponding auxiliary equation is r 2 + 4 = 0 . Simplify r 2 + 4 = 0 as, r 2 = − 4 r = ± − 4 r = ± 2 i ( ∵ i = − 1 ) That is, the roots are r 1 = 2 i and r 2 = − 2 i . Here, α = 0 and β = 2 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos 2 t + c 2 sin 2 t . That is, the two independent solutions to the homogenous equation y ″ + 4 y = 0 are cos 2 t and sin 2 t . Let y 1 ( t ) = cos 2 t and y 2 ( t ) = sin 2 t . Then, y ′ 1 ( t ) = − 2 sin 2 t and y ′ 2 ( t ) = 2 cos 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 ) cos 2 t + v 2 ( t ) sin 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 ) = cos 2 t , y 2 ( t ) = sin 2 t , y ′ 1 ( t ) = − 2 sin 2 t , y ′ 2 ( t ) = 2 cos 2 t and f = tan 2 t becomes, cos 2 t ( v ′ 1 ) + sin 2 t ( v ′ 2 ) = 0 … ( 1 ) − 2 sin 2 t ( v ′ 1 ) + 2 cos 2 t ( v ′ 2 ) = tan 2 t … ( 2 ) Multiply equation (1) by 2 sin 2 t and equation (2) by cos 2 t ; and then add as follows: 2 sin 2 t cos 2 t ( v ′ 1 ) + 2 ( sin 2 t ) 2 ( v ′ 2 ) = 0 + − 2 sin 2 t cos 2 t ( v ′ 1 ) + 2 ( cos 2 t ) 2 ( v ′ 2 ) = sin 2 t _ 0 + 2 ( sin 2 t ) 2 ( v ′ 2 ) + 2 ( cos 2 t ) 2 ( v ′ 2 ) = sin 2 t Simplify 2 ( sin 2 t ) 2 ( v ′ 2 ) + 2 ( cos 2 t ) 2 ( v ′ 2 ) = sin 2 t as, ( v ′ 2 ) [ 2 ( sin 2 t ) 2 + 2 ( cos 2 t ) 2 ] = sin 2 t ( v ′ 2 ) [ 2 ] = sin 2 t ( ∵ ( sin 2 t ) 2 + ( cos 2 t ) 2 = 1 ) v ′ 2 = 1 2 sin 2 t Integrating v ′ 2 ( t ) = 1 2 sin 2 t gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ 1 2 sin 2 t d t = 1 2 ( − cos 2 t 2 ) + C 1 = − 1 4 cos 2 t + C 1 Substitute v ′ 2 ( t ) = 1 2 sin 2 t in equation (1), cos 2 t ( v ′ 1 ) + sin 2 t ( v ′ 2 ) = 0 cos 2 t ( v ′ 1 ) + sin 2 t ( 1 2 sin 2 t ) = 0 cos 2 t ( v ′ 1 ) + 1 2 ( sin 2 t ) 2 = 0 v ′ 1 = − 1 2 ( sin 2 t ) 2 cos 2 t Simplify further as, v ′ 1 = − 1 2 ( 1 − ( cos 2 t ) 2 cos 2 t ) = 1 2 ( ( cos 2 t ) 2 − 1 cos 2 t ) = 1 2 ( cos 2 t − 1 cos 2 t ) = 1 2 ( cos 2 t − sec 2 t ) Integrating v ′ 1 ( t ) = 1 2 ( cos 2 t − sec 2 t ) gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ 1 2 ( cos 2 t − sec 2 t ) d t = 1 2 ( sin 2 t 2 − ln | sec 2 t + tan 2 t | 2 ) + C 2 = 1 4 ( sin 2 t − ln | sec 2 t + tan 2 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 ) = 1 4 ( sin 2 t − ln | sec 2 t + tan 2 t | ) and v 2 ( t ) = − 1 4 cos 2 t . Thus, the particular solution y p ( t ) = v 1 ( t ) cos 2 t + v 2 ( t ) sin 2 t becomes, y p ( t ) = 1 4 ( sin 2 t − ln | sec 2 t + tan 2 t | ) ( cos 2 t ) − 1 4 ( cos 2 t ) ( sin 2 t ) = 1 4 sin 2 t cos 2 t − 1 4 ln | sec 2 t + tan 2 t | ( cos 2 t ) − 1 4 sin 2 t cos 2 t = − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | Combine the general solution y ( t ) = c 1 cos 2 t + c 2 sin 2 t of the corresponding homogenous equation and the particular solution y p ( t ) = − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | to obtain y ( t ) = c 1 cos 2 t + c 2 sin 2 t − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | . Therefore, the general solution to the differential equation y ″ + 4 y = tan 2 t is y ( t ) = c 1 cos 2 t + c 2 sin 2 t − 1 4 ( cos 2 t ) ln | sec 2 t + tan 2 t | _ .

Solution Summary: The author describes the steps to obtain a particular solution to the differential equation y′′+4y=mathrmtan2t.

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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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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Definite Integral Calculus Examples, Integration - Basic Introduction, Practice Problems; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=rCWOdfQ3cwQ;License: Standard YouTube License, CC-BY