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

(a) The equation of motion for the damped vibrating string with m = 10 kg , b = 60 kg/sec , k = 250 kg/sec 2 , y ( 0 ) = 0.3 m and y ′ ( 0 ) = − 0.1 m/sec . The equation of motion for the damped vibrating string is y ( t ) = 0.3 e − 3 t cos ( 4 t ) − 0.02 e − 3 t sin ( 4 t ) _ . The second order differential equation of the mass-spring oscillator is m y ″ + b y ′ + k y = 0 , which represents [ inertia ] y ″ + [ damping ] y ′ + [ stiffness ] y = 0 . Given that, m = 10 kg , b = 60 kg/sec , k = 250 kg/sec 2 , y ( 0 ) = 0.3 m and y ′ ( 0 ) = − 0.1 m/sec . Thus, the initial value problem will be 10 y ″ + 60 y ′ + 250 y = 0 , y ( 0 ) = 0.3 , y ′ ( 0 ) = − 0.1 . The associated auxiliary equation is 10 r 2 + 60 r + 250 = 0 , simplified as r 2 + 6 r + 25 = 0 . Factorize the polynomial r 2 + 6 r + 25 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as follows: r = − 6 ± ( 6 ) 2 − 4 ( 1 ) ( 25 ) 2 ( 1 ) = − 6 ± 36 − 100 2 = − 6 ± − 64 2 = − 6 ± 8 i 2 ( − 1 = i ) = − 3 ± 4 i That is, r = − 3 ± 4 i . Here, α = − 3 and β = 4 . Thus, the general solution will be, y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) . Apply the condition y ( 0 ) = 0.3 in y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) , y ( 0 ) = c 1 e − 3 × 0 cos ( 4 × 0 ) + c 2 e − 3 × 0 sin ( 4 × 0 ) 0.3 = c 1 cos ( 0 ) + c 2 sin ( 0 ) 0.3 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 0.3 Differentiate y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) with respect to t as, y ′ ( t ) = c 1 [ − 4 e − 3 t sin ( 4 t ) − 3 e − 3 t cos ( 4 t ) ] + c 2 [ 4 e − 3 t cos ( 4 t ) − 3 e − 3 t sin ( 4 t ) ] = − ( 4 c 1 + 3 c 2 ) e − 3 t sin ( 4 t ) + ( − 3 c 1 + 4 c 2 ) e − 3 t cos ( 4 t ) Apply the condition y ′ ( 0 ) = − 0.1 in y ′ ( t ) = − ( 4 c 1 + 3 c 2 ) e − 3 t sin ( 4 t ) + ( − 3 c 1 + 4 c 2 ) e − 3 t cos ( 4 t ) , y ′ ( 0 ) = − ( 4 c 1 + 3 c 2 ) e − 3 × 0 sin ( 4 × 0 ) + ( − 3 c 1 + 4 c 2 ) e − 3 × 0 cos ( 4 × 0 ) − 0.1 = − ( 4 c 1 + 3 c 2 ) sin ( 0 ) + ( − 3 c 1 + 4 c 2 ) cos ( 0 ) − 0.1 = − 3 c 1 + 4 c 2 4 c 2 = 3 c 1 − 0.1 c 2 = 3 c 1 − 0.1 4 Substitute c 1 = 0.3 in c 2 = 3 c 1 − 0.1 4 , c 2 = 3 ( 0.3 ) − 0.1 4 = 0.9 − 0.1 4 = 0.8 4 = 0.2 That is, c 1 = 0.3 and c 2 = 0.2 . The general solution y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) becomes, y ( t ) = 0.3 e − 3 t cos ( 4 t ) + 0.2 e − 3 t sin ( 4 t ) . Therefore, the equation of motion for the damped vibrating string is y ( t ) = 0.3 e − 3 t cos ( 4 t ) − 0.02 e − 3 t sin ( 4 t ) _ . (b) The time after which the mass in part (a) first cross the equilibrium point. The mass first cross the equilibrium point tends after [ π + tan − 1 ( − 1.5 ) 4 ] seconds _ . From part (a), the roots to the equation of motion for the damped vibrating string is y ( t ) = 0.3 e − 3 t cos ( 4 t ) + 0.2 e − 3 t sin ( 4 t ) . The aim is to find the time after which the mass first cross the equilibrium point. When the spring crosses the equilibrium point, y ( t ) = 0 . That is, solve for 0.3 e − 3 t cos ( 4 t ) + 0.2 e − 3 t sin ( 4 t ) = 0 . 0.2 e − 3 t sin ( 4 t ) = − 0.3 e − 3 t cos ( 4 t ) sin ( 4 t ) cos ( 4 t ) = − 0.3 0.2 tan ( 4 t ) = − 1.5 4 t = tan − 1 ( − 1.5 ) Since tan − 1 ( − 1.5 ) is negative, to make it positive add π , 4 t = π + tan − 1 ( − 1.5 ) t = π + tan − 1 ( − 1.5 ) 4 Therefore, the mass first cross the equilibrium point tends after [ π + tan − 1 ( − 1.5 ) 4 ] seconds _ . (c) The frequency of oscillation for the spring system. The frequency of oscillation for the spring system is f = 2 π _ . Formula used: The equation of the form y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t , the motion is said to be oscillatory with frequency β 2 π . Calculation: From part (a), the roots to the equation of a vibrating spring without damping for m = 10 kg , b = 60 kg/sec , k = 250 kg/sec 2 , y ( 0 ) = 0.3 m and y ′ ( 0 ) = − 0.1 m/sec is r = − 3 ± 4 i . That is, α = − 3 and β = 4 . Thus, the frequency β 2 π will be 4 2 π = 2 π . Therefore, the frequency of oscillation for the spring system is f = 2 π _ . (d) To compare: The results of Problems 32 and 33 to explain the effect of damping in the frequency of oscillation. In problem 32, the damping coefficient was 0 and the frequency of oscillation was f = 5 2 π ≈ 0.796 . In problem 33, the damping coefficient was 60 and the frequency of oscillation was f = 2 π ≈ 0.636 . That is, the damping will decrease the frequency of oscillation. The damping coefficient has led to the exponential term e − 3 t which will decay the solution to 0.

(a) The equation of motion for the damped vibrating string with m = 10 kg , b = 60 kg/sec , k = 250 kg/sec 2 , y ( 0 ) = 0.3 m and y ′ ( 0 ) = − 0.1 m/sec . The equation of motion for the damped vibrating string is y ( t ) = 0.3 e − 3 t cos ( 4 t ) − 0.02 e − 3 t sin ( 4 t ) _ . The second order differential equation of the mass-spring oscillator is m y ″ + b y ′ + k y = 0 , which represents [ inertia ] y ″ + [ damping ] y ′ + [ stiffness ] y = 0 . Given that, m = 10 kg , b = 60 kg/sec , k = 250 kg/sec 2 , y ( 0 ) = 0.3 m and y ′ ( 0 ) = − 0.1 m/sec . Thus, the initial value problem will be 10 y ″ + 60 y ′ + 250 y = 0 , y ( 0 ) = 0.3 , y ′ ( 0 ) = − 0.1 . The associated auxiliary equation is 10 r 2 + 60 r + 250 = 0 , simplified as r 2 + 6 r + 25 = 0 . Factorize the polynomial r 2 + 6 r + 25 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as follows: r = − 6 ± ( 6 ) 2 − 4 ( 1 ) ( 25 ) 2 ( 1 ) = − 6 ± 36 − 100 2 = − 6 ± − 64 2 = − 6 ± 8 i 2 ( − 1 = i ) = − 3 ± 4 i That is, r = − 3 ± 4 i . Here, α = − 3 and β = 4 . Thus, the general solution will be, y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) . Apply the condition y ( 0 ) = 0.3 in y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) , y ( 0 ) = c 1 e − 3 × 0 cos ( 4 × 0 ) + c 2 e − 3 × 0 sin ( 4 × 0 ) 0.3 = c 1 cos ( 0 ) + c 2 sin ( 0 ) 0.3 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 0.3 Differentiate y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) with respect to t as, y ′ ( t ) = c 1 [ − 4 e − 3 t sin ( 4 t ) − 3 e − 3 t cos ( 4 t ) ] + c 2 [ 4 e − 3 t cos ( 4 t ) − 3 e − 3 t sin ( 4 t ) ] = − ( 4 c 1 + 3 c 2 ) e − 3 t sin ( 4 t ) + ( − 3 c 1 + 4 c 2 ) e − 3 t cos ( 4 t ) Apply the condition y ′ ( 0 ) = − 0.1 in y ′ ( t ) = − ( 4 c 1 + 3 c 2 ) e − 3 t sin ( 4 t ) + ( − 3 c 1 + 4 c 2 ) e − 3 t cos ( 4 t ) , y ′ ( 0 ) = − ( 4 c 1 + 3 c 2 ) e − 3 × 0 sin ( 4 × 0 ) + ( − 3 c 1 + 4 c 2 ) e − 3 × 0 cos ( 4 × 0 ) − 0.1 = − ( 4 c 1 + 3 c 2 ) sin ( 0 ) + ( − 3 c 1 + 4 c 2 ) cos ( 0 ) − 0.1 = − 3 c 1 + 4 c 2 4 c 2 = 3 c 1 − 0.1 c 2 = 3 c 1 − 0.1 4 Substitute c 1 = 0.3 in c 2 = 3 c 1 − 0.1 4 , c 2 = 3 ( 0.3 ) − 0.1 4 = 0.9 − 0.1 4 = 0.8 4 = 0.2 That is, c 1 = 0.3 and c 2 = 0.2 . The general solution y ( t ) = c 1 e − 3 t cos ( 4 t ) + c 2 e − 3 t sin ( 4 t ) becomes, y ( t ) = 0.3 e − 3 t cos ( 4 t ) + 0.2 e − 3 t sin ( 4 t ) . Therefore, the equation of motion for the damped vibrating string is y ( t ) = 0.3 e − 3 t cos ( 4 t ) − 0.02 e − 3 t sin ( 4 t ) _ . (b) The time after which the mass in part (a) first cross the equilibrium point. The mass first cross the equilibrium point tends after [ π + tan − 1 ( − 1.5 ) 4 ] seconds _ . From part (a), the roots to the equation of motion for the damped vibrating string is y ( t ) = 0.3 e − 3 t cos ( 4 t ) + 0.2 e − 3 t sin ( 4 t ) . The aim is to find the time after which the mass first cross the equilibrium point. When the spring crosses the equilibrium point, y ( t ) = 0 . That is, solve for 0.3 e − 3 t cos ( 4 t ) + 0.2 e − 3 t sin ( 4 t ) = 0 . 0.2 e − 3 t sin ( 4 t ) = − 0.3 e − 3 t cos ( 4 t ) sin ( 4 t ) cos ( 4 t ) = − 0.3 0.2 tan ( 4 t ) = − 1.5 4 t = tan − 1 ( − 1.5 ) Since tan − 1 ( − 1.5 ) is negative, to make it positive add π , 4 t = π + tan − 1 ( − 1.5 ) t = π + tan − 1 ( − 1.5 ) 4 Therefore, the mass first cross the equilibrium point tends after [ π + tan − 1 ( − 1.5 ) 4 ] seconds _ . (c) The frequency of oscillation for the spring system. The frequency of oscillation for the spring system is f = 2 π _ . Formula used: The equation of the form y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t , the motion is said to be oscillatory with frequency β 2 π . Calculation: From part (a), the roots to the equation of a vibrating spring without damping for m = 10 kg , b = 60 kg/sec , k = 250 kg/sec 2 , y ( 0 ) = 0.3 m and y ′ ( 0 ) = − 0.1 m/sec is r = − 3 ± 4 i . That is, α = − 3 and β = 4 . Thus, the frequency β 2 π will be 4 2 π = 2 π . Therefore, the frequency of oscillation for the spring system is f = 2 π _ . (d) To compare: The results of Problems 32 and 33 to explain the effect of damping in the frequency of oscillation. In problem 32, the damping coefficient was 0 and the frequency of oscillation was f = 5 2 π ≈ 0.796 . In problem 33, the damping coefficient was 60 and the frequency of oscillation was f = 2 π ≈ 0.636 . That is, the damping will decrease the frequency of oscillation. The damping coefficient has led to the exponential term e − 3 t which will decay the solution to 0.

Solution Summary: The author explains the equation of motion for the damped vibrating string with m=10kg,b=60 kg/sec andyprime (0

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 4.3, Problem 32E

Chapter 4.3, Problem 34E

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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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Author:David Poole
Publisher:Cengage Learning

Text book image

Linear Algebra: A Modern Introduction

Algebra

ISBN:9781285463247

Author:David Poole

Publisher:Cengage Learning

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