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

The general solution to the differential equation y ″ + y = 3 sec t − t 2 + 1 . The general solution to the differential equation y ″ + y = 3 sec t − t 2 + 1 is y ( t ) = c 1 cos t + c 2 sin t + 3 cos t ln | cos t | + 3 t sin t − t 2 + 3 _ . 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 ″ + y = 3 sec t − t 2 + 1 . The corresponding auxiliary equation is r 2 + 1 = 0 . Simplify r 2 + 1 = 0 as, r 2 = − 1 r = ± − 1 r = ± i ( ∵ i = − 1 ) That is, the roots are r 1 = i and r 2 = − i . Here, α = 0 and β = 1 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos t + c 2 sin t . The differential equation y ″ + y = 3 sec t − t 2 + 1 is split into equations, y ″ + y = 3 sec t and y ″ + y = − t 2 + 1 . The solution to y ″ + y = 3 sec t is obtained using variation of parameters, the solution to y ″ + y = − t 2 + 1 is obtained using undetermined coefficients and the superposition principle is applied to combine both solutions. Obtain the solution to y ″ + y = − t 2 + 1 using the method of undetermined coefficients as follows: Since the RHS of the differential equation y ″ + y = − t 2 + 1 is − t 2 + 1 . The particular solution will be of the form y p ( t ) = A t 2 + B t + C . The second derivative of y p ( t ) = A t 2 + B t + C is y ″ p ( t ) = 2 A . Thus, y ″ + y for y p will be, 2 A + A t 2 + B t + C = A t 2 + B t + ( 2 A + C ) . Set A t 2 + B t + ( 2 A + C ) = − t 2 + 1 and comparing the coefficients will results in A = − 1 , B = 0 and 2 A + C = 1 . Substitute A = − 1 in 2 A + C = 1 to obtain C = 1 + 2 = 3 . Thus, the particular solution y p ( t ) = A t 2 + B t + C becomes y p ( t ) = − t 2 + 3 . That is, the solution to y ″ + y = − t 2 + 1 using the method of undetermined coefficients is y p ( t ) = − t 2 + 3 . Obtain the solution to y ″ + y = 3 sec t is obtained using variation of parameters as follows: Since the general solution for the homogenous equation y ″ + y = 0 is y ( t ) = c 1 cos t + c 2 sin t , the two independent solutions are cos t and sin t . That is, the two independent solutions to the homogenous equation y ″ + y = 0 are cos t and sin t . Let y 1 ( t ) = cos t and y 2 ( t ) = sin t . Then, y ′ 1 ( t ) = − sin t and y ′ 2 ( t ) = cos 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 t + v 2 ( t ) sin t . Compare y ″ + y = 3 sec t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 1 and f = 3 sec 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 t , y 2 ( t ) = sin t , y ′ 1 ( t ) = − sin t , y ′ 2 ( t ) = cos t and f = 3 sec t becomes, cos t ( v ′ 1 ) + sin t ( v ′ 2 ) = 0 … ( 1 ) − sin t ( v ′ 1 ) + cos t ( v ′ 2 ) = 3 sec t … ( 2 ) Multiply equation (1) by sin t and equation (2) by cos t ; and then add as follows: sin t cos t ( v ′ 1 ) + ( sin t ) 2 ( v ′ 2 ) = 0 + − sin t cos t ( v ′ 1 ) + ( cos t ) 2 ( v ′ 2 ) = 3 _ 0 + ( sin t ) 2 ( v ′ 2 ) + ( cos t ) 2 ( v ′ 2 ) = 3 Simplify ( sin t ) 2 ( v ′ 2 ) + ( cos t ) 2 ( v ′ 2 ) = 3 as, ( v ′ 2 ) [ ( sin t ) 2 + ( cos t ) 2 ] = 3 ( v ′ 2 ) [ 1 ] = 3 ( ∵ ( sin t ) 2 + ( cos t ) 2 = 1 ) v ′ 2 = 3 Integrating v ′ 2 ( t ) = 3 gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ 3 d t = 3 t + C 1 Substitute v ′ 2 ( t ) = 3 in equation (1), cos t ( v ′ 1 ) + sin t ( v ′ 2 ) = 0 cos t ( v ′ 1 ) + sin t ( 3 ) = 0 v ′ 1 = − 3 sin t cos t v ′ 1 = − 3 tan t Integrating v ′ 1 ( t ) = − 3 tan t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ − 3 tan t d t = 3 ln | cos 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 ) = 3 ln | cos t | and v 2 ( t ) = 3 t . Thus, the particular solution y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t becomes, y p ( t ) = 3 ln | cos t | cos t + 3 t sin t . By superposition principle, the particular solution to y ″ + y = 3 sec t − t 2 + 1 will be, y p ( t ) = 3 ln | cos t | cos t + 3 t sin t − t 2 + 3 . Combine the general solution y ( t ) = c 1 cos t + c 2 sin t of the corresponding homogenous equation and the particular solution y p ( t ) = 3 ln | cos t | cos t + 3 t sin t − t 2 − 1 to obtain y ( t ) = c 1 cos t + c 2 sin t + 3 cos t ln | cos t | + 3 t sin t − t 2 + 3 . Therefore, the general solution to the differential equation y ″ + y = 3 sec t − t 2 + 1 is y ( t ) = c 1 cos t + c 2 sin t + 3 cos t ln | cos t | + 3 t sin t − t 2 + 3 _ .

The general solution to the differential equation y ″ + y = 3 sec t − t 2 + 1 . The general solution to the differential equation y ″ + y = 3 sec t − t 2 + 1 is y ( t ) = c 1 cos t + c 2 sin t + 3 cos t ln | cos t | + 3 t sin t − t 2 + 3 _ . 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 ″ + y = 3 sec t − t 2 + 1 . The corresponding auxiliary equation is r 2 + 1 = 0 . Simplify r 2 + 1 = 0 as, r 2 = − 1 r = ± − 1 r = ± i ( ∵ i = − 1 ) That is, the roots are r 1 = i and r 2 = − i . Here, α = 0 and β = 1 . Thus, the general solution for the homogenous equation is y ( t ) = c 1 cos t + c 2 sin t . The differential equation y ″ + y = 3 sec t − t 2 + 1 is split into equations, y ″ + y = 3 sec t and y ″ + y = − t 2 + 1 . The solution to y ″ + y = 3 sec t is obtained using variation of parameters, the solution to y ″ + y = − t 2 + 1 is obtained using undetermined coefficients and the superposition principle is applied to combine both solutions. Obtain the solution to y ″ + y = − t 2 + 1 using the method of undetermined coefficients as follows: Since the RHS of the differential equation y ″ + y = − t 2 + 1 is − t 2 + 1 . The particular solution will be of the form y p ( t ) = A t 2 + B t + C . The second derivative of y p ( t ) = A t 2 + B t + C is y ″ p ( t ) = 2 A . Thus, y ″ + y for y p will be, 2 A + A t 2 + B t + C = A t 2 + B t + ( 2 A + C ) . Set A t 2 + B t + ( 2 A + C ) = − t 2 + 1 and comparing the coefficients will results in A = − 1 , B = 0 and 2 A + C = 1 . Substitute A = − 1 in 2 A + C = 1 to obtain C = 1 + 2 = 3 . Thus, the particular solution y p ( t ) = A t 2 + B t + C becomes y p ( t ) = − t 2 + 3 . That is, the solution to y ″ + y = − t 2 + 1 using the method of undetermined coefficients is y p ( t ) = − t 2 + 3 . Obtain the solution to y ″ + y = 3 sec t is obtained using variation of parameters as follows: Since the general solution for the homogenous equation y ″ + y = 0 is y ( t ) = c 1 cos t + c 2 sin t , the two independent solutions are cos t and sin t . That is, the two independent solutions to the homogenous equation y ″ + y = 0 are cos t and sin t . Let y 1 ( t ) = cos t and y 2 ( t ) = sin t . Then, y ′ 1 ( t ) = − sin t and y ′ 2 ( t ) = cos 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 t + v 2 ( t ) sin t . Compare y ″ + y = 3 sec t with the general equation form a y ″ + b y ′ + c y = f to obtain a = 1 , b = 0 , c = 1 and f = 3 sec 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 t , y 2 ( t ) = sin t , y ′ 1 ( t ) = − sin t , y ′ 2 ( t ) = cos t and f = 3 sec t becomes, cos t ( v ′ 1 ) + sin t ( v ′ 2 ) = 0 … ( 1 ) − sin t ( v ′ 1 ) + cos t ( v ′ 2 ) = 3 sec t … ( 2 ) Multiply equation (1) by sin t and equation (2) by cos t ; and then add as follows: sin t cos t ( v ′ 1 ) + ( sin t ) 2 ( v ′ 2 ) = 0 + − sin t cos t ( v ′ 1 ) + ( cos t ) 2 ( v ′ 2 ) = 3 _ 0 + ( sin t ) 2 ( v ′ 2 ) + ( cos t ) 2 ( v ′ 2 ) = 3 Simplify ( sin t ) 2 ( v ′ 2 ) + ( cos t ) 2 ( v ′ 2 ) = 3 as, ( v ′ 2 ) [ ( sin t ) 2 + ( cos t ) 2 ] = 3 ( v ′ 2 ) [ 1 ] = 3 ( ∵ ( sin t ) 2 + ( cos t ) 2 = 1 ) v ′ 2 = 3 Integrating v ′ 2 ( t ) = 3 gives, v 2 ( t ) = ∫ v ′ 2 ( t ) d t = ∫ 3 d t = 3 t + C 1 Substitute v ′ 2 ( t ) = 3 in equation (1), cos t ( v ′ 1 ) + sin t ( v ′ 2 ) = 0 cos t ( v ′ 1 ) + sin t ( 3 ) = 0 v ′ 1 = − 3 sin t cos t v ′ 1 = − 3 tan t Integrating v ′ 1 ( t ) = − 3 tan t gives, v 1 ( t ) = ∫ v ′ 1 ( t ) = ∫ − 3 tan t d t = 3 ln | cos 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 ) = 3 ln | cos t | and v 2 ( t ) = 3 t . Thus, the particular solution y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t becomes, y p ( t ) = 3 ln | cos t | cos t + 3 t sin t . By superposition principle, the particular solution to y ″ + y = 3 sec t − t 2 + 1 will be, y p ( t ) = 3 ln | cos t | cos t + 3 t sin t − t 2 + 3 . Combine the general solution y ( t ) = c 1 cos t + c 2 sin t of the corresponding homogenous equation and the particular solution y p ( t ) = 3 ln | cos t | cos t + 3 t sin t − t 2 − 1 to obtain y ( t ) = c 1 cos t + c 2 sin t + 3 cos t ln | cos t | + 3 t sin t − t 2 + 3 . Therefore, the general solution to the differential equation y ″ + y = 3 sec t − t 2 + 1 is y ( t ) = c 1 cos t + c 2 sin t + 3 cos t ln | cos t | + 3 t sin t − t 2 + 3 _ .

Solution Summary: The author explains the steps to obtain a particular solution to the differential equation y′′+y=3mathrmsect

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

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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