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

The general solution to the differential equation y ″ + 9 y = sec 2 ( 3 t ) . The general solution to the differential equation y ″ + 9 y = sec 2 ( 3 t ) is y ( t ) = c 1 cos 3 t + c 2 sin 3 t − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 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 ″ + 9 y = sec 3 ( 3 t ) . Compare y ″ + 9 y = sec 2 ( 3 t ) with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 9 and f = sec 2 ( 3 t ) . The corresponding auxiliary equation is r 2 + 9 = 0 . Simplify r 2 + 9 = 0 as, r 2 = − 9 r = ± − 9 r = ± 3 i ( ∵ i = − 1 ) That is, the roots are r 1 = 3 i and r 2 = − 3 i . Here, α = 0 and β = 3 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos 3 t + c 2 sin 3 t . That is, the two independent solutions to the homogenous equation y ″ + 4 y = 0 are cos 3 t and sin 3 t . Let y 1 ( t ) = cos 3 t and y 2 ( t ) = sin 3 t . Then, y ′ 1 ( t ) = − 3 sin 3 t and y ′ 2 ( t ) = 3 cos 3 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 3 t + v 2 ( t ) sin 3 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 3 t , y 2 ( t ) = sin 3 t , y ′ 1 ( t ) = − 3 sin 3 t , y ′ 2 ( t ) = 3 cos 3 t and f = sec 2 ( 3 t ) becomes, cos 3 t ( v ′ 1 ) + sin 3 t ( v ′ 2 ) = 0 … ( 1 ) − 3 sin 3 t ( v ′ 1 ) + 3 cos 3 t ( v ′ 2 ) = sec 2 ( 3 t ) … ( 2 ) Multiply equation (1) by 3 sin 3 t and equation (2) by cos 3 t ; and then add as follows: 3 sin 3 t cos 3 t ( v ′ 1 ) + 3 ( sin 3 t ) 2 ( v ′ 2 ) = 0 + − 3 sin 3 t cos 3 t ( v ′ 1 ) + 3 ( cos 3 t ) 2 ( v ′ 2 ) = sec 3 t _ 0 + 3 ( sin 3 t ) 2 ( v ′ 2 ) + 3 ( cos 3 t ) 2 ( v ′ 2 ) = sec 3 t Simplify 3 ( sin 3 t ) 2 ( v ′ 2 ) + 3 ( cos 3 t ) 2 ( v ′ 2 ) = sec 3 t as, 3 ( v ′ 2 ) [ ( sin 3 t ) 2 + ( cos 3 t ) 2 ] = sec 3 t 3 ( v ′ 2 ) [ 1 ] = sec 3 t ( ∵ ( sin 3 t ) 2 + ( cos 3 t ) 2 = 1 ) v ′ 2 = sec 3 t 3 Integrating v ′ 2 ( t ) = sec 3 t 3 gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ sec 3 t 3 d t = 1 9 ln | sec 3 t + tan 3 t | + C 1 Substitute v ′ 2 ( t ) = sec 3 t 3 in equation (1), cos 3 t ( v ′ 1 ) + sin 3 t ( v ′ 2 ) = 0 cos 3 t ( v ′ 1 ) + sin 3 t ( sec 3 t 3 ) = 0 cos 3 t ( v ′ 1 ) + 1 3 sin 3 t sec 3 t = 0 v ′ 1 = − 1 3 sin 3 t sec 3 t cos 3 t v ′ 1 = − 1 3 tan 3 t sec 3 t Integrating v ′ 1 ( t ) = − 1 3 tan 3 t sec 3 t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ − 1 3 tan 3 t sec 3 t d t = − 1 9 sec 3 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 9 sec 3 t and v 2 ( t ) = 1 9 ln | sec 3 t + tan 3 t | . Thus, the particular solution y p ( t ) = v 1 ( t ) cos 3 t + v 2 ( t ) sin 3 t becomes, y p ( t ) = − 1 9 sec 3 t cos 3 t + 1 9 ln | sec 3 t + tan 3 t | sin 3 t = − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | Combine the general solution y ( t ) = c 1 cos 3 t + c 2 sin 3 t of the corresponding homogenous equation and the particular solution y p ( t ) = − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | to obtain y ( t ) = c 1 cos 3 t + c 2 sin 3 t − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | . Therefore, the general solution to the differential equation y ″ + 9 y = sec 2 ( 3 t ) is y ( t ) = c 1 cos 3 t + c 2 sin 3 t − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | _ .

The general solution to the differential equation y ″ + 9 y = sec 2 ( 3 t ) . The general solution to the differential equation y ″ + 9 y = sec 2 ( 3 t ) is y ( t ) = c 1 cos 3 t + c 2 sin 3 t − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 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 ″ + 9 y = sec 3 ( 3 t ) . Compare y ″ + 9 y = sec 2 ( 3 t ) with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 9 and f = sec 2 ( 3 t ) . The corresponding auxiliary equation is r 2 + 9 = 0 . Simplify r 2 + 9 = 0 as, r 2 = − 9 r = ± − 9 r = ± 3 i ( ∵ i = − 1 ) That is, the roots are r 1 = 3 i and r 2 = − 3 i . Here, α = 0 and β = 3 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos 3 t + c 2 sin 3 t . That is, the two independent solutions to the homogenous equation y ″ + 4 y = 0 are cos 3 t and sin 3 t . Let y 1 ( t ) = cos 3 t and y 2 ( t ) = sin 3 t . Then, y ′ 1 ( t ) = − 3 sin 3 t and y ′ 2 ( t ) = 3 cos 3 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 3 t + v 2 ( t ) sin 3 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 3 t , y 2 ( t ) = sin 3 t , y ′ 1 ( t ) = − 3 sin 3 t , y ′ 2 ( t ) = 3 cos 3 t and f = sec 2 ( 3 t ) becomes, cos 3 t ( v ′ 1 ) + sin 3 t ( v ′ 2 ) = 0 … ( 1 ) − 3 sin 3 t ( v ′ 1 ) + 3 cos 3 t ( v ′ 2 ) = sec 2 ( 3 t ) … ( 2 ) Multiply equation (1) by 3 sin 3 t and equation (2) by cos 3 t ; and then add as follows: 3 sin 3 t cos 3 t ( v ′ 1 ) + 3 ( sin 3 t ) 2 ( v ′ 2 ) = 0 + − 3 sin 3 t cos 3 t ( v ′ 1 ) + 3 ( cos 3 t ) 2 ( v ′ 2 ) = sec 3 t _ 0 + 3 ( sin 3 t ) 2 ( v ′ 2 ) + 3 ( cos 3 t ) 2 ( v ′ 2 ) = sec 3 t Simplify 3 ( sin 3 t ) 2 ( v ′ 2 ) + 3 ( cos 3 t ) 2 ( v ′ 2 ) = sec 3 t as, 3 ( v ′ 2 ) [ ( sin 3 t ) 2 + ( cos 3 t ) 2 ] = sec 3 t 3 ( v ′ 2 ) [ 1 ] = sec 3 t ( ∵ ( sin 3 t ) 2 + ( cos 3 t ) 2 = 1 ) v ′ 2 = sec 3 t 3 Integrating v ′ 2 ( t ) = sec 3 t 3 gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ sec 3 t 3 d t = 1 9 ln | sec 3 t + tan 3 t | + C 1 Substitute v ′ 2 ( t ) = sec 3 t 3 in equation (1), cos 3 t ( v ′ 1 ) + sin 3 t ( v ′ 2 ) = 0 cos 3 t ( v ′ 1 ) + sin 3 t ( sec 3 t 3 ) = 0 cos 3 t ( v ′ 1 ) + 1 3 sin 3 t sec 3 t = 0 v ′ 1 = − 1 3 sin 3 t sec 3 t cos 3 t v ′ 1 = − 1 3 tan 3 t sec 3 t Integrating v ′ 1 ( t ) = − 1 3 tan 3 t sec 3 t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ − 1 3 tan 3 t sec 3 t d t = − 1 9 sec 3 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 9 sec 3 t and v 2 ( t ) = 1 9 ln | sec 3 t + tan 3 t | . Thus, the particular solution y p ( t ) = v 1 ( t ) cos 3 t + v 2 ( t ) sin 3 t becomes, y p ( t ) = − 1 9 sec 3 t cos 3 t + 1 9 ln | sec 3 t + tan 3 t | sin 3 t = − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | Combine the general solution y ( t ) = c 1 cos 3 t + c 2 sin 3 t of the corresponding homogenous equation and the particular solution y p ( t ) = − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | to obtain y ( t ) = c 1 cos 3 t + c 2 sin 3 t − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | . Therefore, the general solution to the differential equation y ″ + 9 y = sec 2 ( 3 t ) is y ( t ) = c 1 cos 3 t + c 2 sin 3 t − 1 9 + sin 3 t 9 ln | sec 3 t + tan 3 t | _ .

Solution Summary: The author describes the steps to obtain a particular solution to the differential equation y′′+9y=mathrmsec2(3t).

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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)5. The operator T on a vector space V over Q has as its sequence of invariant factors the sequence (x-1)3(x-2)5(x-3)2, (x-1)(x-2)(x-3)2, (x-3). Find the Jordan normal form of T.

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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

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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