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

The equation of motion of the mass and the resonance frequency for the system. The equation of motion of the mass is y ( t ) = 0.0469 e − 5 4 t cos ( t ) + 0.0016 e − 5 4 t sin ( t ) + 0.00312 cos t + 0.00016 sin t _ and the resonance frequency is 1.0779 cycles/sec _ . Given that, 2kg mass is attached to a spring cause stretching of 20cm upon coming to rest at equilibrium. Since 1 m = 100 cm , convert 20cm to meters as, 0.2m. Redefine y ( t ) to be measured from this equilibrium. Thus, m g k = 0.2 m . Simplify further as, k = m g 0.2 . Since m = 2 kg and g = 9.8 m/sec 2 , the value of the stiffness k will be, k = 2 × 9.8 0.2 = 98 The external force is, F ( t ) = 0.3 cos t N and the damping constant is 5 N-sec/m . That is, F = 0.3 cos t and b = 5 . Comparing the form F 0 cos γ t with F ( t ) = 0.3 cos t N , it is observed that F 0 = 0.3 and γ = 1 . Substitute m = 2 , k = 98 , b = 5 , F 0 = 0.3 and γ = 1 in m d 2 y d t 2 + b d y d t + k y = F 0 cos γ t to obtain as, 2 d 2 y d t 2 + 5 d y d t + 98 y = 0.3 cos t . The corresponding homogeneous equation is, 2 d 2 y d t 2 + 5 d y d t + 98 y = 0 . The associated auxiliary equation will be, 2 r 2 + 5 r + 98 = 0 . Factorize 2 r 2 + 5 r + 98 = 0 as, r = − 5 ± 5 2 − 4 ( 2 × 98 ) 2 ( 2 ) = − 5 ± 25 − 784 4 = − 5 ± − 759 4 = − 5 ± 27 − 30 4 Simplify further as, r = − 5 ± 27 30 i 4 ( ∵ i = − 1 ) = − 5 4 ± 27 4 30 i That is, the roots are r 1 = − 5 4 + 27 4 30 i and r 2 = − 5 4 − 27 4 30 i . Here, α = − 5 4 and β = 27 4 30 = 6.75 30 . It is known that, the auxiliary equation has complex conjugate roots α ± i β , then y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t is a general solution. Thus, the form y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t becomes y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) . The differential equation m d 2 y d t 2 + b d y d t + k y = F 0 cos γ t has particular solution of the form, y p ( t ) = F 0 ( k − m γ 2 ) 2 + b 2 γ 2 ( ( k − m γ 2 ) cos γ t + γ b sin γ t ) . Substitute m = 2 , k = 98 , b = 5 , F 0 = 0.3 and γ = 1 in y p ( t ) = F 0 ( k − m γ 2 ) 2 + b 2 γ 2 ( ( k − m γ 2 ) cos γ t + γ b sin γ t ) , y p ( t ) = 0.3 ( 98 − 2 × 1 2 ) 2 + ( 5 2 ) ( 1 2 ) ( ( 98 − 2 × 1 2 ) cos t + ( 5 × 1 ) sin t ) = 0.3 ( 98 − 2 ) 2 + 25 ( ( 98 − 2 ) cos t + 5 sin t ) = 0.3 ( 96 ) 2 + 25 ( 96 cos t + 5 sin t ) = 0.3 9241 ( 96 cos t + 5 sin t ) Thus, combining the particular solution y p ( t ) = 0.3 9241 ( 96 cos t + 5 sin t ) and the general solution y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) to obtain as, y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) . Given that, at time t = 0 , the mass is displaced 5cm blow the equilibrium and released. Since 1 m = 100 cm , convert 5cm to meters as, 0.05m. That is, y ( 0 ) = 0.5 and y ′ ( 0 ) = 0 . Substitute t = 0 and y ( 0 ) = 0.05 in y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) , y ( 0 ) = c 1 e − 5 4 × 0 cos ( 6.75 30 × 0 ) + c 2 e − 5 4 × 0 sin ( 6.75 30 × 0 ) + 0.3 9241 ( 96 cos ( 0 ) + 5 sin ( 0 ) ) 0.05 = c 1 e 0 cos ( 0 ) + c 2 e 0 sin ( 0 ) + 0.3 9241 ( 96 ( 1 ) + 5 ( 0 ) ) 0.05 = c 1 ( 1 ) + c 2 ( 0 ) + 0.3 9241 ( 96 ) 0.05 = c 1 + 28.8 9241 Simplify further as, c 1 = 0.05 − 28.8 9241 = 462.05 − 28.8 9241 = 433.25 9241 ≈ 0.0469 Differentiate y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) with respect to t as, y ′ ( t ) = [ − 6.75 30 c 1 e − 5 4 t sin ( 6.75 30 t ) − 5 4 c 1 e − 5 4 t cos ( 6.75 30 t ) + 6.75 30 c 2 e − 5 4 t cos ( 6.75 30 t ) − 5 4 c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( − 96 sin t + 5 cos t ) ] = [ ( − 6.75 30 c 1 − 5 4 c 2 ) e − 5 4 t sin ( 6.75 30 t ) + ( 6.75 30 c 2 − 5 4 c 1 ) e − 5 4 t cos ( 6.75 30 t ) + 0.3 9241 ( − 96 sin t + 5 cos t ) ] Substitute t = 0 and y ′ ( 0 ) = 0 in the above equation, y ′ ( 0 ) = [ ( − 6.75 30 c 1 − 5 4 c 2 ) e − 5 4 × 0 sin ( 6.75 30 × 0 ) + ( 6.75 30 c 2 − 5 4 c 1 ) e − 5 4 × 0 cos ( 6.75 30 × 0 ) + 0.3 9241 ( − 96 sin 0 + 5 cos 0 ) ] 0 = ( − 6.75 30 c 1 − 5 4 c 2 ) e 0 sin ( 0 ) + ( 6.75 30 c 2 − 5 4 c 1 ) e 0 cos ( 0 ) + 0.3 9241 ( − 96 ( 0 ) + 5 ( 1 ) ) 0 = ( − 6.75 30 c 1 − 5 4 c 2 ) ( 0 ) + ( 6.75 30 c 2 − 5 4 c 1 ) ( 1 ) + 0.3 9241 ( 5 ) Simplify as, 0 = 6.75 30 c 2 − 5 4 c 1 + 1.5 9241 6.75 30 c 2 = 5 4 c 1 − 1.5 9241 6.75 30 c 2 = 5 4 ( 433.25 9241 ) − 1.5 9241 ( ∵ c 1 = 433.25 9241 ) 6.75 30 c 2 = 2166.25 9241 × 4 − 1.5 × 4 9241 × 4 On further simplifications, 6.75 30 c 2 = 2166.25 − 6 9241 × 4 c 2 = 2160.25 9241 × 4 × 6.75 30 c 2 = 2160.25 1366606.122 c 2 ≈ 0.0016 That is, c 1 = 0.0469 and c 2 = 0.0016 . Thus, the solution y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) becomes, y ( t ) = 0.0469 e − 5 4 t cos ( 6.75 30 t ) + 0.0016 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) = 0.0469 e − 5 4 t cos ( 6.75 30 t ) + 0.0016 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 × 96 cos t + 0.3 9241 × 5 sin t = 0.0469 e − 5 4 t cos ( 6.75 30 t ) + 0.0016 e − 5 4 t sin ( 6.75 30 t ) + 0.00312 cos t + 0.00016 sin t Therefore, the equation of motion of the mass is y ( t ) = 0.0469 e − 5 4 t cos ( t ) + 0.0016 e − 5 4 t sin ( t ) + 0.00312 cos t + 0.00016 sin t _ . The response frequency is γ r 2 π , where γ r = k m − b 2 2 m 2 . Substitute m = 2 , k = 98 , b = 5 and γ = 1 in γ r = k m − b 2 2 m 2 , γ r = 98 2 − 5 2 2 ( 2 ) 2 = 49 − 25 8 = 367 8 = 45.875 ≈ 6.773 The response frequency is γ r 2 π , becomes, γ r 2 π = 6.773 2 π ≈ 1.0779 Therefore, the resonance frequency is 1.0779 cycles/sec _ .

The equation of motion of the mass and the resonance frequency for the system. The equation of motion of the mass is y ( t ) = 0.0469 e − 5 4 t cos ( t ) + 0.0016 e − 5 4 t sin ( t ) + 0.00312 cos t + 0.00016 sin t _ and the resonance frequency is 1.0779 cycles/sec _ . Given that, 2kg mass is attached to a spring cause stretching of 20cm upon coming to rest at equilibrium. Since 1 m = 100 cm , convert 20cm to meters as, 0.2m. Redefine y ( t ) to be measured from this equilibrium. Thus, m g k = 0.2 m . Simplify further as, k = m g 0.2 . Since m = 2 kg and g = 9.8 m/sec 2 , the value of the stiffness k will be, k = 2 × 9.8 0.2 = 98 The external force is, F ( t ) = 0.3 cos t N and the damping constant is 5 N-sec/m . That is, F = 0.3 cos t and b = 5 . Comparing the form F 0 cos γ t with F ( t ) = 0.3 cos t N , it is observed that F 0 = 0.3 and γ = 1 . Substitute m = 2 , k = 98 , b = 5 , F 0 = 0.3 and γ = 1 in m d 2 y d t 2 + b d y d t + k y = F 0 cos γ t to obtain as, 2 d 2 y d t 2 + 5 d y d t + 98 y = 0.3 cos t . The corresponding homogeneous equation is, 2 d 2 y d t 2 + 5 d y d t + 98 y = 0 . The associated auxiliary equation will be, 2 r 2 + 5 r + 98 = 0 . Factorize 2 r 2 + 5 r + 98 = 0 as, r = − 5 ± 5 2 − 4 ( 2 × 98 ) 2 ( 2 ) = − 5 ± 25 − 784 4 = − 5 ± − 759 4 = − 5 ± 27 − 30 4 Simplify further as, r = − 5 ± 27 30 i 4 ( ∵ i = − 1 ) = − 5 4 ± 27 4 30 i That is, the roots are r 1 = − 5 4 + 27 4 30 i and r 2 = − 5 4 − 27 4 30 i . Here, α = − 5 4 and β = 27 4 30 = 6.75 30 . It is known that, the auxiliary equation has complex conjugate roots α ± i β , then y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t is a general solution. Thus, the form y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t becomes y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) . The differential equation m d 2 y d t 2 + b d y d t + k y = F 0 cos γ t has particular solution of the form, y p ( t ) = F 0 ( k − m γ 2 ) 2 + b 2 γ 2 ( ( k − m γ 2 ) cos γ t + γ b sin γ t ) . Substitute m = 2 , k = 98 , b = 5 , F 0 = 0.3 and γ = 1 in y p ( t ) = F 0 ( k − m γ 2 ) 2 + b 2 γ 2 ( ( k − m γ 2 ) cos γ t + γ b sin γ t ) , y p ( t ) = 0.3 ( 98 − 2 × 1 2 ) 2 + ( 5 2 ) ( 1 2 ) ( ( 98 − 2 × 1 2 ) cos t + ( 5 × 1 ) sin t ) = 0.3 ( 98 − 2 ) 2 + 25 ( ( 98 − 2 ) cos t + 5 sin t ) = 0.3 ( 96 ) 2 + 25 ( 96 cos t + 5 sin t ) = 0.3 9241 ( 96 cos t + 5 sin t ) Thus, combining the particular solution y p ( t ) = 0.3 9241 ( 96 cos t + 5 sin t ) and the general solution y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) to obtain as, y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) . Given that, at time t = 0 , the mass is displaced 5cm blow the equilibrium and released. Since 1 m = 100 cm , convert 5cm to meters as, 0.05m. That is, y ( 0 ) = 0.5 and y ′ ( 0 ) = 0 . Substitute t = 0 and y ( 0 ) = 0.05 in y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) , y ( 0 ) = c 1 e − 5 4 × 0 cos ( 6.75 30 × 0 ) + c 2 e − 5 4 × 0 sin ( 6.75 30 × 0 ) + 0.3 9241 ( 96 cos ( 0 ) + 5 sin ( 0 ) ) 0.05 = c 1 e 0 cos ( 0 ) + c 2 e 0 sin ( 0 ) + 0.3 9241 ( 96 ( 1 ) + 5 ( 0 ) ) 0.05 = c 1 ( 1 ) + c 2 ( 0 ) + 0.3 9241 ( 96 ) 0.05 = c 1 + 28.8 9241 Simplify further as, c 1 = 0.05 − 28.8 9241 = 462.05 − 28.8 9241 = 433.25 9241 ≈ 0.0469 Differentiate y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) with respect to t as, y ′ ( t ) = [ − 6.75 30 c 1 e − 5 4 t sin ( 6.75 30 t ) − 5 4 c 1 e − 5 4 t cos ( 6.75 30 t ) + 6.75 30 c 2 e − 5 4 t cos ( 6.75 30 t ) − 5 4 c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( − 96 sin t + 5 cos t ) ] = [ ( − 6.75 30 c 1 − 5 4 c 2 ) e − 5 4 t sin ( 6.75 30 t ) + ( 6.75 30 c 2 − 5 4 c 1 ) e − 5 4 t cos ( 6.75 30 t ) + 0.3 9241 ( − 96 sin t + 5 cos t ) ] Substitute t = 0 and y ′ ( 0 ) = 0 in the above equation, y ′ ( 0 ) = [ ( − 6.75 30 c 1 − 5 4 c 2 ) e − 5 4 × 0 sin ( 6.75 30 × 0 ) + ( 6.75 30 c 2 − 5 4 c 1 ) e − 5 4 × 0 cos ( 6.75 30 × 0 ) + 0.3 9241 ( − 96 sin 0 + 5 cos 0 ) ] 0 = ( − 6.75 30 c 1 − 5 4 c 2 ) e 0 sin ( 0 ) + ( 6.75 30 c 2 − 5 4 c 1 ) e 0 cos ( 0 ) + 0.3 9241 ( − 96 ( 0 ) + 5 ( 1 ) ) 0 = ( − 6.75 30 c 1 − 5 4 c 2 ) ( 0 ) + ( 6.75 30 c 2 − 5 4 c 1 ) ( 1 ) + 0.3 9241 ( 5 ) Simplify as, 0 = 6.75 30 c 2 − 5 4 c 1 + 1.5 9241 6.75 30 c 2 = 5 4 c 1 − 1.5 9241 6.75 30 c 2 = 5 4 ( 433.25 9241 ) − 1.5 9241 ( ∵ c 1 = 433.25 9241 ) 6.75 30 c 2 = 2166.25 9241 × 4 − 1.5 × 4 9241 × 4 On further simplifications, 6.75 30 c 2 = 2166.25 − 6 9241 × 4 c 2 = 2160.25 9241 × 4 × 6.75 30 c 2 = 2160.25 1366606.122 c 2 ≈ 0.0016 That is, c 1 = 0.0469 and c 2 = 0.0016 . Thus, the solution y ( t ) = c 1 e − 5 4 t cos ( 6.75 30 t ) + c 2 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) becomes, y ( t ) = 0.0469 e − 5 4 t cos ( 6.75 30 t ) + 0.0016 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 ( 96 cos t + 5 sin t ) = 0.0469 e − 5 4 t cos ( 6.75 30 t ) + 0.0016 e − 5 4 t sin ( 6.75 30 t ) + 0.3 9241 × 96 cos t + 0.3 9241 × 5 sin t = 0.0469 e − 5 4 t cos ( 6.75 30 t ) + 0.0016 e − 5 4 t sin ( 6.75 30 t ) + 0.00312 cos t + 0.00016 sin t Therefore, the equation of motion of the mass is y ( t ) = 0.0469 e − 5 4 t cos ( t ) + 0.0016 e − 5 4 t sin ( t ) + 0.00312 cos t + 0.00016 sin t _ . The response frequency is γ r 2 π , where γ r = k m − b 2 2 m 2 . Substitute m = 2 , k = 98 , b = 5 and γ = 1 in γ r = k m − b 2 2 m 2 , γ r = 98 2 − 5 2 2 ( 2 ) 2 = 49 − 25 8 = 367 8 = 45.875 ≈ 6.773 The response frequency is γ r 2 π , becomes, γ r 2 π = 6.773 2 π ≈ 1.0779 Therefore, the resonance frequency is 1.0779 cycles/sec _ .

Solution Summary: The author explains the equation of motion of the mass and the resonance frequency for the system.

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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