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

The general solution to the differential equation t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) . The general solution to the differential equation t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) is z ( t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) _ . Theorem used: 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 Procedure used: Consider the differential equation a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 . The associated auxiliary equation is a r 2 + ( b − a ) r + c = 0 . Solving the auxiliary equation gives the roots r 1 and r 2 . (a) If the auxiliary equation has distinct real roots r 1 and r 2 , then y ( t ) = c 1 t r 1 + c 2 t r 2 is a general solution. (b) If the auxiliary equation has repeated root r , then y ( t ) = c 1 t r + c 2 t r ln t is a general solution. (c) If the auxiliary equation has complex root r = α ± i β , then y ( t ) = c 1 t α cos ( β ln t ) + c 2 t α sin ( β ln t ) is a general solution. Calculation: The given differential equation is, t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) . For easier calculation, assume the differential equation to be of the form t 2 y ″ − t y ′ + y = t ( 1 + 3 ln t ) . Compare t 2 y ″ − t y ′ + y = t ( 1 + 3 ln t ) with the general equation form a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 to obtain a = 1 , b = − 1 and c = 1 . The associated auxiliary equation is, a r 2 + ( b − a ) r + c = 0 becomes, 1 r 2 + ( − 1 − 1 ) r + ( 1 ) = 0 r 2 − 2 r + 1 = 0 Simplify r 2 − 2 r + 1 = 0 as, r 2 − r − r + 1 = 0 r ( r − 1 ) − 1 ( r − 1 ) = 0 ( r − 1 ) ( r − 1 ) = 0 r = 1 , 1 That is, the repeated roots are r 1 = 1 and r 2 = 1 . Thus, the given differential equation has solution of the form, y ( t ) = c 1 t + c 2 t ln t . That is, the two independent solutions to the homogenous equation t 2 y ″ − t y ′ + y = 0 are t and t ln t . Let y 1 ( t ) = t and y 2 ( t ) = t ln t . Then, y ′ 1 ( t ) = 1 and y ′ 2 ( t ) = ln t + 1 . 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 ) t ln t . For applying the variation of parameters, the equation is to be simplified in the form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) . Divide the equation t 2 y ″ − t y ′ + y = t ( 1 + 3 ln t ) by t 2 as, t 2 t 2 y ″ − t t 2 y ′ + 1 t 2 y = t ( 1 + 3 ln t ) t 2 y ″ − 1 t y ′ + 1 t 2 y = 1 t ( 1 + 3 ln t ) Compare y ″ − 1 t y ′ + 1 t 2 y = 1 t ( 1 + 3 ln t ) with the general equation form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) to obtain p ( t ) = − 1 t , q ( t ) = 1 t 2 and g ( t ) = 1 t ( 1 + 3 ln 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 t ln t 1 1 + ln t | = t ( 1 + ln t ) − 1 ( t ln t ) = t + t ln t − t ln t = t The equation v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 t ( 1 + 3 ln t ) , y 2 ( t ) = t ln t and W [ y 1 , y 2 ] ( t ) = t becomes, v 1 ( t ) = ∫ − 1 t ( 1 + 3 ln t ) t ln t t d t = − ∫ ln t t ( 1 + 3 ln t ) d t = − ∫ ( ln t t + 3 t ) d t = − ∫ ln t t d t − ∫ 3 t d t The equation v 2 ( t ) = ∫ g ( t ) y 1 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 t ( 1 + 3 ln t ) , y 1 ( t ) = t and W [ y 1 , y 2 ] ( t ) = t becomes, v 2 ( t ) = ∫ 1 t ( 1 + 3 ln t ) t t d t = ∫ 1 t ( 1 + 3 ln t ) d t = ∫ ( 1 t + 3 t ln t ) d t = ∫ 1 t d t + 3 ∫ 1 t ln t d t Let u = ln t . Then, d u = 1 t d t and d t = t d u . Thus, the equation v 1 ( t ) = − ∫ ln t t d t − ∫ 3 t d t becomes, v 1 ( t ) = − ∫ u t ( t d u ) − 3 ∫ 1 t d t = − ∫ u d u − 3 ∫ 1 t d t = − u 2 2 − 3 ln t + C 1 Thus, the equation v 2 ( t ) = ∫ 1 t d t + 3 ∫ 1 t ln t d t becomes, v 2 ( t ) = ∫ 1 t d t + 3 ∫ 1 t u ( t d u ) = ∫ 1 t d t + 3 ∫ 1 u d u = ln t + 3 ln u + C 2 Substitute u = ln t in the equations v 1 ( t ) = − u 2 2 − 3 ln t + C 1 and v 2 ( t ) = ln t + 3 ln u + C 2 as, v 1 ( t ) = − ( ln t ) 2 2 − 3 ln t + C 1 and v 2 ( t ) = ln t + 3 ln ( 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 ) = − ( ln t ) 2 2 − 3 ln t and v 2 ( t ) = ln t + 3 ln ( ln t ) . Thus, the particular solution y p ( t ) = v 1 ( t ) t + v 2 ( t ) t ln t becomes, y p ( t ) = v 1 ( t ) t + v 2 ( t ) t ln t = ( − ( ln t ) 2 2 − 3 ln t ) t + ( ln t + 3 ln ( ln t ) ) t ln t = − t ( ln t ) 2 2 − 3 t ln t + t ( ln t ) 2 + 3 t ln t ln ( ln t ) = t ( ln t ) 2 2 − 3 t ln t + 3 t ln t ln ( ln t ) Combine the general solution y ( t ) = c 1 t + c 2 t ln t of the corresponding homogenous equation and the particular solution y p ( t ) = t ( ln t ) 2 2 − 3 t ln t + 3 t ln t ln ( ln t ) to obtain as, y ( t ) = c 1 t + c 2 t ln t + t ( ln t ) 2 2 − 3 t ln t + 3 t ln t ln ( ln t ) = c 1 t + ( c 2 − 3 ) t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) ( ∵ c 2 − 3 is a constant ) That is, the solution for t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) will be, z ( t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) . Therefore, the general solution to the differential equation t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) is z ( t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) _ .

The general solution to the differential equation t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) . The general solution to the differential equation t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) is z ( t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) _ . Theorem used: 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 Procedure used: Consider the differential equation a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 . The associated auxiliary equation is a r 2 + ( b − a ) r + c = 0 . Solving the auxiliary equation gives the roots r 1 and r 2 . (a) If the auxiliary equation has distinct real roots r 1 and r 2 , then y ( t ) = c 1 t r 1 + c 2 t r 2 is a general solution. (b) If the auxiliary equation has repeated root r , then y ( t ) = c 1 t r + c 2 t r ln t is a general solution. (c) If the auxiliary equation has complex root r = α ± i β , then y ( t ) = c 1 t α cos ( β ln t ) + c 2 t α sin ( β ln t ) is a general solution. Calculation: The given differential equation is, t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) . For easier calculation, assume the differential equation to be of the form t 2 y ″ − t y ′ + y = t ( 1 + 3 ln t ) . Compare t 2 y ″ − t y ′ + y = t ( 1 + 3 ln t ) with the general equation form a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 to obtain a = 1 , b = − 1 and c = 1 . The associated auxiliary equation is, a r 2 + ( b − a ) r + c = 0 becomes, 1 r 2 + ( − 1 − 1 ) r + ( 1 ) = 0 r 2 − 2 r + 1 = 0 Simplify r 2 − 2 r + 1 = 0 as, r 2 − r − r + 1 = 0 r ( r − 1 ) − 1 ( r − 1 ) = 0 ( r − 1 ) ( r − 1 ) = 0 r = 1 , 1 That is, the repeated roots are r 1 = 1 and r 2 = 1 . Thus, the given differential equation has solution of the form, y ( t ) = c 1 t + c 2 t ln t . That is, the two independent solutions to the homogenous equation t 2 y ″ − t y ′ + y = 0 are t and t ln t . Let y 1 ( t ) = t and y 2 ( t ) = t ln t . Then, y ′ 1 ( t ) = 1 and y ′ 2 ( t ) = ln t + 1 . 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 ) t ln t . For applying the variation of parameters, the equation is to be simplified in the form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) . Divide the equation t 2 y ″ − t y ′ + y = t ( 1 + 3 ln t ) by t 2 as, t 2 t 2 y ″ − t t 2 y ′ + 1 t 2 y = t ( 1 + 3 ln t ) t 2 y ″ − 1 t y ′ + 1 t 2 y = 1 t ( 1 + 3 ln t ) Compare y ″ − 1 t y ′ + 1 t 2 y = 1 t ( 1 + 3 ln t ) with the general equation form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) to obtain p ( t ) = − 1 t , q ( t ) = 1 t 2 and g ( t ) = 1 t ( 1 + 3 ln 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 t ln t 1 1 + ln t | = t ( 1 + ln t ) − 1 ( t ln t ) = t + t ln t − t ln t = t The equation v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 t ( 1 + 3 ln t ) , y 2 ( t ) = t ln t and W [ y 1 , y 2 ] ( t ) = t becomes, v 1 ( t ) = ∫ − 1 t ( 1 + 3 ln t ) t ln t t d t = − ∫ ln t t ( 1 + 3 ln t ) d t = − ∫ ( ln t t + 3 t ) d t = − ∫ ln t t d t − ∫ 3 t d t The equation v 2 ( t ) = ∫ g ( t ) y 1 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = 1 t ( 1 + 3 ln t ) , y 1 ( t ) = t and W [ y 1 , y 2 ] ( t ) = t becomes, v 2 ( t ) = ∫ 1 t ( 1 + 3 ln t ) t t d t = ∫ 1 t ( 1 + 3 ln t ) d t = ∫ ( 1 t + 3 t ln t ) d t = ∫ 1 t d t + 3 ∫ 1 t ln t d t Let u = ln t . Then, d u = 1 t d t and d t = t d u . Thus, the equation v 1 ( t ) = − ∫ ln t t d t − ∫ 3 t d t becomes, v 1 ( t ) = − ∫ u t ( t d u ) − 3 ∫ 1 t d t = − ∫ u d u − 3 ∫ 1 t d t = − u 2 2 − 3 ln t + C 1 Thus, the equation v 2 ( t ) = ∫ 1 t d t + 3 ∫ 1 t ln t d t becomes, v 2 ( t ) = ∫ 1 t d t + 3 ∫ 1 t u ( t d u ) = ∫ 1 t d t + 3 ∫ 1 u d u = ln t + 3 ln u + C 2 Substitute u = ln t in the equations v 1 ( t ) = − u 2 2 − 3 ln t + C 1 and v 2 ( t ) = ln t + 3 ln u + C 2 as, v 1 ( t ) = − ( ln t ) 2 2 − 3 ln t + C 1 and v 2 ( t ) = ln t + 3 ln ( 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 ) = − ( ln t ) 2 2 − 3 ln t and v 2 ( t ) = ln t + 3 ln ( ln t ) . Thus, the particular solution y p ( t ) = v 1 ( t ) t + v 2 ( t ) t ln t becomes, y p ( t ) = v 1 ( t ) t + v 2 ( t ) t ln t = ( − ( ln t ) 2 2 − 3 ln t ) t + ( ln t + 3 ln ( ln t ) ) t ln t = − t ( ln t ) 2 2 − 3 t ln t + t ( ln t ) 2 + 3 t ln t ln ( ln t ) = t ( ln t ) 2 2 − 3 t ln t + 3 t ln t ln ( ln t ) Combine the general solution y ( t ) = c 1 t + c 2 t ln t of the corresponding homogenous equation and the particular solution y p ( t ) = t ( ln t ) 2 2 − 3 t ln t + 3 t ln t ln ( ln t ) to obtain as, y ( t ) = c 1 t + c 2 t ln t + t ( ln t ) 2 2 − 3 t ln t + 3 t ln t ln ( ln t ) = c 1 t + ( c 2 − 3 ) t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) ( ∵ c 2 − 3 is a constant ) That is, the solution for t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) will be, z ( t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) . Therefore, the general solution to the differential equation t 2 z ″ − t z ′ + z = t ( 1 + 3 ln t ) is z ( t ) = c 1 t + c 3 t ln t + t ( ln t ) 2 2 + 3 t ln t ln ( ln t ) _ .

Solution Summary: The author explains the general solution to the differential equation t2z′′-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 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. 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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