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The particular solution to the differential equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 , where f ( t ) = t is a solution to the corresponding homogeneous equation. The particular solution to the differential equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 is y p ( t ) = t 2 + t + 1 _ . Theorem used: 1. Reduction of order: If y 1 ( t ) is a solution, not identically zero, to the homogenous differential equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = 0 in an interval I , then y 2 ( t ) = y 1 ( t ) ∫ e − ∫ p ( t ) d t y 1 ( t ) 2 d t is a second, linearly independent solution. 2. Variation of parameters: If y 1 and y 2 are two linearly independent solutions to the homogenous equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = 0 on an interval I where p ( t ) , q ( t ) and g ( t ) are continuous, then a particular solution to y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) is y p = v 1 y 1 + v 2 y 2 , where v 1 and v 2 are determined up to a constant by the pairs of equations y 1 v ′ 1 + y 2 v ′ 2 = 0 and y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = g , which have the solution: v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t v 2 ( t ) = ∫ g ( t ) y 1 ( t ) W [ y 1 , y 2 ] ( t ) d t Calculation: The given differential equation is, ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 . Divide the equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 by 1 − t as, ( 1 − t ) 1 − t y ″ + t 1 − t y ′ − 1 1 − t y = ( 1 − t ) 2 1 − t y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t Compare y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t with the general equation form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) to obtain p ( t ) = t 1 − t , q ( t ) = − 1 1 − t and g ( t ) = 1 − t . A solution to the corresponding homogeneous equation of y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t is given as, f ( t ) = t . Let y 1 ( t ) = t . By reduction of order, the second linearly independent solution is y 2 ( t ) = y 1 ( t ) ∫ e − ∫ p ( t ) d t y 1 ( t ) 2 d t . Obtain the value of − ∫ p ( t ) d t as, − ∫ p ( t ) d t = − ∫ t 1 − t d t = ∫ t t − 1 d t = ∫ t − 1 + 1 t − 1 d t = ∫ t − 1 t − 1 d t + ∫ 1 t − 1 d t Simplify further as, − ∫ p ( t ) d t = ∫ d t + ∫ 1 t − 1 d t = t + ln ( t − 1 ) + C 1 Consider the constant to be zero. That is, − ∫ p ( t ) d t = t + ln ( t − 1 ) . The term e − ∫ p ( t ) d t y 1 ( t ) 2 for − ∫ p ( t ) d t = t + ln ( t − 1 ) and y 1 ( t ) = t becomes, e − ∫ p ( t ) d t y 1 ( t ) 2 = e t + ln ( t − 1 ) ( t ) 2 = e t e ln ( t − 1 ) t 2 = e t ( t − 1 ) t 2 ( ∵ e ln x = x ) Thus, y 2 ( t ) = y 1 ( t ) ∫ e − ∫ p ( t ) d t y 1 ( t ) 2 d t becomes, y 2 ( t ) = t ∫ e t ( t − 1 ) t 2 d t . Let u = e t ( t − 1 ) and v ′ = 1 t 2 . Obtain the value of ∫ e t ( t − 1 ) t 2 d t using integration by parts ∫ u v ′ = u v − ∫ u ′ v . The value of v will be, v = ∫ v ′ d t = ∫ 1 t 2 d t = − 1 t + C 1 Assuming the constant to be zero, v = − 1 t . The value of u ′ will be, u ′ = d d t [ e t ( t − 1 ) ] = e t d d t ( t − 1 ) + ( t − 1 ) d d t e t = e t ( 1 ) + ( t − 1 ) ( e t ) = e t + t e t − e t = t e t That is, u ′ = t e t . Thus, ∫ e t ( t − 1 ) t 2 d t is obtained as, ∫ e t ( t − 1 ) t 2 d t = e t ( t − 1 ) ( − 1 t ) − ∫ t e t ( − 1 t ) d t = e t ( 1 − t t ) + ∫ e t d t = e t ( 1 t − 1 ) + e t + C 2 = e t t − e t + e t + C 2 = e t t + C 2 Consider the constant to be zero. That is, ∫ e t ( t − 1 ) t 2 d t = e t t . The solution y 2 ( t ) = t ∫ e t ( t − 1 ) t 2 d t becomes, y 2 ( t ) = t ( e t t ) = e t That is, the linearly independent solutions to the corresponding homogeneous equation of y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t are e t and t . With y 1 ( t ) = t and y 2 ( t ) = e t , the first derivatives will be y ′ 1 ( t ) = 1 and y ′ 2 ( t ) = e 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 ) t + v 2 ( t ) e t . Obtain the value of W [ y 1 , y 2 ] ( t ) as, W [ y 1 , y 2 ] ( t ) = | y 1 ( t ) y 2 ( t ) y 1 ′ ( t ) y 2 ′ ( t ) | = | t e t 1 e t | = t ( e t ) − 1 ( e t ) = e t ( t − 1 ) The equation v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 − t , y 2 ( t ) = e t and W [ y 1 , y 2 ] ( t ) = e t ( t − 1 ) becomes, v 1 ( t ) = ∫ − ( 1 − t ) ( e t ) e t ( t − 1 ) d t = ∫ 1 d t = t + C 1 The equation v 2 ( t ) = ∫ g ( t ) y 1 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 − t , y 1 ( t ) = t and W [ y 1 , y 2 ] ( t ) = e t ( t − 1 ) becomes, v 2 ( t ) = ∫ ( 1 − t ) ( t ) e t ( t − 1 ) d t = − ∫ t e − t d t = − t ∫ e − t d t + ∫ − e − t d t = t e − t + e − t + C 2 = e − t ( t + 1 ) + 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 and v 2 ( t ) = e − t ( t + 1 ) . Thus, the particular solution y p ( t ) = v 1 ( t ) t + v 2 ( t ) e t becomes, y p ( t ) = ( t ) t + e − t ( t + 1 ) e t = t 2 + t + 1 Therefore, the particular solution to the differential equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 is y p ( t ) = t 2 + t + 1 _ .

The particular solution to the differential equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 , where f ( t ) = t is a solution to the corresponding homogeneous equation. The particular solution to the differential equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 is y p ( t ) = t 2 + t + 1 _ . Theorem used: 1. Reduction of order: If y 1 ( t ) is a solution, not identically zero, to the homogenous differential equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y ( t ) = 0 in an interval I , then y 2 ( t ) = y 1 ( t ) ∫ e − ∫ p ( t ) d t y 1 ( t ) 2 d t is a second, linearly independent solution. 2. Variation of parameters: If y 1 and y 2 are two linearly independent solutions to the homogenous equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = 0 on an interval I where p ( t ) , q ( t ) and g ( t ) are continuous, then a particular solution to y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) is y p = v 1 y 1 + v 2 y 2 , where v 1 and v 2 are determined up to a constant by the pairs of equations y 1 v ′ 1 + y 2 v ′ 2 = 0 and y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = g , which have the solution: v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t v 2 ( t ) = ∫ g ( t ) y 1 ( t ) W [ y 1 , y 2 ] ( t ) d t Calculation: The given differential equation is, ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 . Divide the equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 by 1 − t as, ( 1 − t ) 1 − t y ″ + t 1 − t y ′ − 1 1 − t y = ( 1 − t ) 2 1 − t y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t Compare y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t with the general equation form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) to obtain p ( t ) = t 1 − t , q ( t ) = − 1 1 − t and g ( t ) = 1 − t . A solution to the corresponding homogeneous equation of y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t is given as, f ( t ) = t . Let y 1 ( t ) = t . By reduction of order, the second linearly independent solution is y 2 ( t ) = y 1 ( t ) ∫ e − ∫ p ( t ) d t y 1 ( t ) 2 d t . Obtain the value of − ∫ p ( t ) d t as, − ∫ p ( t ) d t = − ∫ t 1 − t d t = ∫ t t − 1 d t = ∫ t − 1 + 1 t − 1 d t = ∫ t − 1 t − 1 d t + ∫ 1 t − 1 d t Simplify further as, − ∫ p ( t ) d t = ∫ d t + ∫ 1 t − 1 d t = t + ln ( t − 1 ) + C 1 Consider the constant to be zero. That is, − ∫ p ( t ) d t = t + ln ( t − 1 ) . The term e − ∫ p ( t ) d t y 1 ( t ) 2 for − ∫ p ( t ) d t = t + ln ( t − 1 ) and y 1 ( t ) = t becomes, e − ∫ p ( t ) d t y 1 ( t ) 2 = e t + ln ( t − 1 ) ( t ) 2 = e t e ln ( t − 1 ) t 2 = e t ( t − 1 ) t 2 ( ∵ e ln x = x ) Thus, y 2 ( t ) = y 1 ( t ) ∫ e − ∫ p ( t ) d t y 1 ( t ) 2 d t becomes, y 2 ( t ) = t ∫ e t ( t − 1 ) t 2 d t . Let u = e t ( t − 1 ) and v ′ = 1 t 2 . Obtain the value of ∫ e t ( t − 1 ) t 2 d t using integration by parts ∫ u v ′ = u v − ∫ u ′ v . The value of v will be, v = ∫ v ′ d t = ∫ 1 t 2 d t = − 1 t + C 1 Assuming the constant to be zero, v = − 1 t . The value of u ′ will be, u ′ = d d t [ e t ( t − 1 ) ] = e t d d t ( t − 1 ) + ( t − 1 ) d d t e t = e t ( 1 ) + ( t − 1 ) ( e t ) = e t + t e t − e t = t e t That is, u ′ = t e t . Thus, ∫ e t ( t − 1 ) t 2 d t is obtained as, ∫ e t ( t − 1 ) t 2 d t = e t ( t − 1 ) ( − 1 t ) − ∫ t e t ( − 1 t ) d t = e t ( 1 − t t ) + ∫ e t d t = e t ( 1 t − 1 ) + e t + C 2 = e t t − e t + e t + C 2 = e t t + C 2 Consider the constant to be zero. That is, ∫ e t ( t − 1 ) t 2 d t = e t t . The solution y 2 ( t ) = t ∫ e t ( t − 1 ) t 2 d t becomes, y 2 ( t ) = t ( e t t ) = e t That is, the linearly independent solutions to the corresponding homogeneous equation of y ″ + t 1 − t y ′ − 1 1 − t y = 1 − t are e t and t . With y 1 ( t ) = t and y 2 ( t ) = e t , the first derivatives will be y ′ 1 ( t ) = 1 and y ′ 2 ( t ) = e 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 ) t + v 2 ( t ) e t . Obtain the value of W [ y 1 , y 2 ] ( t ) as, W [ y 1 , y 2 ] ( t ) = | y 1 ( t ) y 2 ( t ) y 1 ′ ( t ) y 2 ′ ( t ) | = | t e t 1 e t | = t ( e t ) − 1 ( e t ) = e t ( t − 1 ) The equation v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 − t , y 2 ( t ) = e t and W [ y 1 , y 2 ] ( t ) = e t ( t − 1 ) becomes, v 1 ( t ) = ∫ − ( 1 − t ) ( e t ) e t ( t − 1 ) d t = ∫ 1 d t = t + C 1 The equation v 2 ( t ) = ∫ g ( t ) y 1 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 − t , y 1 ( t ) = t and W [ y 1 , y 2 ] ( t ) = e t ( t − 1 ) becomes, v 2 ( t ) = ∫ ( 1 − t ) ( t ) e t ( t − 1 ) d t = − ∫ t e − t d t = − t ∫ e − t d t + ∫ − e − t d t = t e − t + e − t + C 2 = e − t ( t + 1 ) + 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 and v 2 ( t ) = e − t ( t + 1 ) . Thus, the particular solution y p ( t ) = v 1 ( t ) t + v 2 ( t ) e t becomes, y p ( t ) = ( t ) t + e − t ( t + 1 ) e t = t 2 + t + 1 Therefore, the particular solution to the differential equation ( 1 − t ) y ″ + t y ′ − y = ( 1 − t ) 2 is y p ( t ) = t 2 + t + 1 _ .

Solution Summary: The author explains that the particular solution to the differential equation (1-t)y′′+t

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 4.7, Problem 45E

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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. The motion of a swinging door with...Ch. 4.3 - Prob. 36E Ch. 4.3 - Prob. 37E Ch. 4.3 - Prove the sum of angles formula for the sine...Ch. 4.4 - In Problems 18, decide whether or not the method...Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Prob. 5E Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 16E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 18E Ch. 4.4 - Prob. 19E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Prob. 28E Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - Prob. 31E Ch. 4.4 - Prob. 32E Ch. 4.4 - Prob. 33E Ch. 4.4 - Prob. 34E Ch. 4.4 - Prob. 35E Ch. 4.4 - In Problems 3336, use the method of undetermined...Ch. 4.5 - Given that y1(t) = cos t is a solution to y y + y...Ch. 4.5 - Prob. 2E Ch. 4.5 - Prob. 3E Ch. 4.5 - Prob. 4E Ch. 4.5 - Prob. 5E Ch. 4.5 - Prob. 6E Ch. 4.5 - Prob. 7E Ch. 4.5 - In Problems 38, a nonhomogeneous equation and a...Ch. 4.5 - Prob. 9E Ch. 4.5 - Prob. 10E Ch. 4.5 - Prob. 11E Ch. 4.5 - Prob. 12E Ch. 4.5 - Prob. 13E Ch. 4.5 - In Problems 916 decide whether the method of...Ch. 4.5 - Prob. 15E Ch. 4.5 - Prob. 16E Ch. 4.5 - Prob. 17E Ch. 4.5 - Prob. 18E Ch. 4.5 - Prob. 19E Ch. 4.5 - In Problems 1722, find a general solution to the...Ch. 4.5 - Prob. 21E Ch. 4.5 - Prob. 22E Ch. 4.5 - Prob. 23E Ch. 4.5 - Prob. 24E Ch. 4.5 - Prob. 25E Ch. 4.5 - Prob. 26E Ch. 4.5 - Prob. 27E Ch. 4.5 - Prob. 28E Ch. 4.5 - Prob. 29E Ch. 4.5 - Prob. 30E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 33E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 35E Ch. 4.5 - Prob. 36E Ch. 4.5 - Prob. 37E Ch. 4.5 - Prob. 38E Ch. 4.5 - In Problems 3740, find a particular solution to...Ch. 4.5 - Prob. 40E Ch. 4.5 - Prob. 41E Ch. 4.5 - Prob. 42E Ch. 4.5 - Prob. 43E Ch. 4.5 - Prob. 44E Ch. 4.5 - Prob. 46E Ch. 4.5 - Prob. 47E Ch. 4.5 - Prob. 48E Ch. 4.6 - Prob. 1E Ch. 4.6 - In Exercises 12, rectangles have been drawn to...Ch. 4.6 - Prob. 3E Ch. 4.6 - Prob. 4E Ch. 4.6 - Prob. 5E Ch. 4.6 - Prob. 6E Ch. 4.6 - Prob. 7E Ch. 4.6 - Prob. 8E Ch. 4.6 - Prob. 9E Ch. 4.6 - Prob. 10E Ch. 4.6 - Prob. 11E Ch. 4.6 - Prob. 12E Ch. 4.6 - Prob. 13E Ch. 4.6 - Prob. 14E Ch. 4.6 - Prob. 15E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - Prob. 17E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - For f(x) = 3x + 2, evaluate the Riemann sums: (a)...Ch. 4.6 - For g(x) = 4x 1, evaluate the Riemann sums: (a)...Ch. 4.6 - Prob. 21E Ch. 4.6 - Use Table 5.9 to evaluate the Riemann sums: Table...Ch. 4.6 - The graph of f(t) is in Figure 5.40. Which of the...Ch. 4.6 - Prob. 24E Ch. 4.6 - Prob. 25E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 3E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 5E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 7E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 9E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 11E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 13E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - Prob. 17E Ch. 4.7 - Prob. 18E Ch. 4.7 - Prob. 19E Ch. 4.7 - Prob. 20E Ch. 4.7 - Prob. 21E Ch. 4.7 - Prob. 22E Ch. 4.7 - Prob. 23E Ch. 4.7 - Solve the following CauchyEuler equations by using...Ch. 4.7 - Prob. 25E Ch. 4.7 - Prob. 26E Ch. 4.7 - Prob. 27E Ch. 4.7 - Prob. 28E Ch. 4.7 - Prove that if y1 and y2 are linearly independent...Ch. 4.7 - Prob. 30E Ch. 4.7 - Prob. 31E Ch. 4.7 - Prob. 32E Ch. 4.7 - Prob. 33E Ch. 4.7 - Prob. 34E Ch. 4.7 - Prob. 35E Ch. 4.7 - Prob. 36E Ch. 4.7 - Prob. 37E Ch. 4.7 - Prob. 38E Ch. 4.7 - Prob. 39E Ch. 4.7 - Prob. 40E Ch. 4.7 - Prob. 41E Ch. 4.7 - In Problems 41 through 44, a differential equation...Ch. 4.7 - Prob. 43E Ch. 4.7 - Prob. 44E Ch. 4.7 - Prob. 45E Ch. 4.7 - Prob. 46E Ch. 4.7 - Prob. 47E Ch. 4.7 - Prob. 48E Ch. 4.7 - Prob. 49E Ch. 4.7 - Prob. 50E Ch. 4.7 - Prob. 51E Ch. 4.7 - Prob. 52E Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Try to predict the qualitative features of the...Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Use reduction of order to derive the solution...Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.9 - Prob. 1E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 3E Ch. 4.9 - Prob. 4E Ch. 4.9 - Prob. 5E Ch. 4.9 - Prob. 6E Ch. 4.9 - Prob. 7E Ch. 4.9 - Prob. 8E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 10E Ch. 4.9 - Prob. 11E Ch. 4.9 - Prob. 12E Ch. 4.9 - Prob. 13E Ch. 4.9 - Prob. 14E Ch. 4.9 - Prob. 15E Ch. 4.9 - Prob. 16E Ch. 4.9 - Prob. 17E Ch. 4.9 - Prob. 18E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 2E Ch. 4.10 - Prob. 3E Ch. 4.10 - Prob. 4E Ch. 4.10 - Prob. 5E Ch. 4.10 - Prob. 6E Ch. 4.10 - Prob. 7E Ch. 4.10 - Prob. 8E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 10E Ch. 4.10 - Prob. 11E Ch. 4.10 - Prob. 12E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 14E Ch. 4.10 - Prob. 15E Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 11RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 15RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 25RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP Ch. 4 - Prob. 29RP Ch. 4 - Prob. 30RP Ch. 4 - Prob. 31RP Ch. 4 - Prob. 32RP Ch. 4 - Prob. 33RP Ch. 4 - Prob. 34RP Ch. 4 - Prob. 35RP Ch. 4 - Prob. 36RP Ch. 4 - Prob. 37RP Ch. 4 - Prob. 38RP Ch. 4 - A 32-lb weight is attached to a vertical spring,...Ch. 4 - Prob. 1TWE Ch. 4 - Prob. 2TWE Ch. 4 - Prob. 3TWE Ch. 4 - For students with a background in linear algebra:...

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