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

The maximum displacement from the equilibrium that the mass will attain. The maximum displacement is 75 .67 cm _ . Formula used: The governing equation of a mass-spring system is: F e x t = [ i n e r t i a ] y ″ + [ d a m p i n g ] y ′ + [ s t i f f n e s s ] y = m y ″ + b y ′ + k y Consider F e x t = 0 . In the case of underdamped, b 2 − 4 m k < 0 and there will be two complex conjugates to the associated auxiliary equation. For the roots α ± i β , the general solution that determines the motion will be y ( t ) = e α t ( C 1 cos β t + C 2 sin β t ) . The solution form can also be expressed as y ( t ) = A e α t sin ( β t + ϕ ) , where A = C 1 2 + C 2 2 and tan ϕ = C 1 C 2 . The constant A is called the amplitude, A e α t is the damping constant, ϕ is the phase angle, the quasiperiod is P = 2 π β and the quasifrequency is 1 P . Calculation: Given that, mass 1 4 kg is attached to a spring with stiffness 8 N/m and the damping constant is 1 4 N-sec/m . That is, m = 1 4 , k = 8 and b = 1 4 . Thus, the governing equation of a mass-spring system F e x t = m y ″ + b y ′ + k y for 0 external force becomes, 1 4 y ″ + 1 4 y ′ + 8 y = 0 . Multiply by 4 to obtain y ″ + y ′ + 32 y = 0 . The associated auxiliary equation is, r 2 + r + 32 = 0 . Factorize r 2 + r + 32 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 1 ± ( 1 ) 2 − 4 ( 1 ) ( 32 ) 2 ( 1 ) = − 1 ± 1 − 128 2 = − 1 ± − 127 2 = − 1 ± 127 i 2 ( ∵ i = − 1 ) = − 0.5 ± 0.5 127 i That is, the roots are r 1 = − 0.5 + 0.5 127 i and r 2 = − 0.5 − 0.5 127 i . Here, α = − 0.5 and β = 0.5 127 . Thus, the solution form y ( t ) = e α t ( C 1 cos β t + C 2 sin β t ) becomes y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) . Given that, the mass is moved 1m to the left from the equilibrium and released. That is, y ( 0 ) = − 1 and y ′ ( 0 ) = 0 . The negative sign indicates the left direction. Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = − 1 in y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) , y ( 0 ) = e − 0.5 × 0 ( C 1 cos ( 0.5 127 × 0 ) + C 2 sin ( 0.5 127 × 0 ) ) − 1 = C 1 cos ( 0 ) + C 2 sin ( 0 ) C 1 ( 1 ) + C 2 ( 0 ) = − 1 C 1 = − 1 Differentiate y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) with respect to t as, y ′ ( t ) = [ e − 0.5 t ( − 0.5 127 C 1 sin ( 0.5 127 t ) + 0.5 127 C 2 cos ( 0.5 127 t ) ) − 0.5 e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) ] = [ e − 0.5 t ( − 0.5 127 C 1 sin ( 0.5 127 t ) + 0.5 127 C 2 cos ( 0.5 127 t ) ) + e − 0.5 t ( − 0.5 C 1 cos ( 0.5 127 t ) − 0.5 C 2 sin ( 0.5 127 t ) ) ] Simplify as, y ′ ( t ) = e − 0.5 t ( − 0.5 127 C 1 sin ( 0.5 127 t ) + 0.5 127 C 2 cos ( 0.5 127 t ) − 0.5 C 1 cos ( 0.5 127 t ) − 0.5 C 2 sin ( 0.5 127 t ) ) = e − 0.5 t ( ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0.5 127 t ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0.5 127 t ) ) Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = e − 0.5 t ( ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0.5 127 t ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0.5 127 t ) ) , y ′ ( 0 ) = e − 0.5 × 0 ( ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0.5 127 × 0 ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0.5 127 × 0 ) ) 0 = ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0 ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0 ) 0 = ( − 0.5 127 C 1 − 0.5 C 2 ) ( 0 ) + ( 0.5 127 C 2 − 0.5 C 1 ) ( 1 ) 0.5 127 C 2 − 0.5 C 1 = 0 Since C 1 = − 1 , the above equation becomes, 0.5 127 C 2 − 0.5 ( − 1 ) = 0 0.5 127 C 2 = − 0.5 C 2 = − 0.5 0.5 127 C 2 = − 1 127 That is, C 1 = − 1 and C 2 = − 1 127 . Thus, the solution y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) becomes, y ( t ) = e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) . The aim is to find the which maximum displacement. Note that, the maximum displacement will be obtained at y ′ = 0 . Differentiate y ( t ) = e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) with respect to t as, y ′ ( t ) = [ e − 0.5 t ( 0.5 127 sin ( 0.5 127 t ) − 0.5 127 127 cos ( 0.5 127 t ) ) − 0.5 e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) ] = [ e − 0.5 t ( 0.5 127 sin ( 0.5 127 t ) − 0.5 cos ( 0.5 127 t ) ) + e − 0.5 t ( 0.5 cos ( 0.5 127 t ) + 0.5 sin ( 0.5 127 t ) ) ] Simplify further as, y ′ ( t ) = e − 0.5 t ( 0.5 127 sin ( 0.5 127 t ) + 0.5 sin ( 0.5 127 t ) ) = 0.5 ( 127 + 1 ) e − 0.5 t sin ( 0.5 127 t ) Thus, y ′ = 0 implies, 0.5 ( 127 + 1 ) e − 0.5 t sin ( 0.5 127 t ) = 0 e − 0.5 t sin ( 0.5 127 t ) = 0 Note that, the value of the exponential is never zero. Thus, the only possibility is sin ( 0.5 127 t ) = 0 . This happenes only when 0.5 127 t = n π . Simplify further as, t = n π 0.5 127 . The first value is obtained when n = 1 . Thus, n = 1 implies t = π 0.5 127 . On simplification, t = 2 π 127 . That is, the maximum displacement occurs at t = 2 π 127 . Substitute t = 2 π 127 in y ( t ) = e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) , y ( 2 π 127 ) = e − 0.5 × 2 π 127 ( − cos ( 0.5 127 × 2 π 127 ) − 1 127 sin ( 0.5 127 × 2 π 127 ) ) = e − π 127 ( − cos ( π ) − 1 127 sin ( π ) ) = e − 0.2788 ( 1 − 1 127 ( 0 ) ) = 0.7567 ( 1 − 0 ) = 0.7567 That is, the maximum displacement is 0.7567m. In other words, 75.67cm. Therefore, the maximum displacement is 75 .67 cm _ .

The maximum displacement from the equilibrium that the mass will attain. The maximum displacement is 75 .67 cm _ . Formula used: The governing equation of a mass-spring system is: F e x t = [ i n e r t i a ] y ″ + [ d a m p i n g ] y ′ + [ s t i f f n e s s ] y = m y ″ + b y ′ + k y Consider F e x t = 0 . In the case of underdamped, b 2 − 4 m k < 0 and there will be two complex conjugates to the associated auxiliary equation. For the roots α ± i β , the general solution that determines the motion will be y ( t ) = e α t ( C 1 cos β t + C 2 sin β t ) . The solution form can also be expressed as y ( t ) = A e α t sin ( β t + ϕ ) , where A = C 1 2 + C 2 2 and tan ϕ = C 1 C 2 . The constant A is called the amplitude, A e α t is the damping constant, ϕ is the phase angle, the quasiperiod is P = 2 π β and the quasifrequency is 1 P . Calculation: Given that, mass 1 4 kg is attached to a spring with stiffness 8 N/m and the damping constant is 1 4 N-sec/m . That is, m = 1 4 , k = 8 and b = 1 4 . Thus, the governing equation of a mass-spring system F e x t = m y ″ + b y ′ + k y for 0 external force becomes, 1 4 y ″ + 1 4 y ′ + 8 y = 0 . Multiply by 4 to obtain y ″ + y ′ + 32 y = 0 . The associated auxiliary equation is, r 2 + r + 32 = 0 . Factorize r 2 + r + 32 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 1 ± ( 1 ) 2 − 4 ( 1 ) ( 32 ) 2 ( 1 ) = − 1 ± 1 − 128 2 = − 1 ± − 127 2 = − 1 ± 127 i 2 ( ∵ i = − 1 ) = − 0.5 ± 0.5 127 i That is, the roots are r 1 = − 0.5 + 0.5 127 i and r 2 = − 0.5 − 0.5 127 i . Here, α = − 0.5 and β = 0.5 127 . Thus, the solution form y ( t ) = e α t ( C 1 cos β t + C 2 sin β t ) becomes y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) . Given that, the mass is moved 1m to the left from the equilibrium and released. That is, y ( 0 ) = − 1 and y ′ ( 0 ) = 0 . The negative sign indicates the left direction. Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = − 1 in y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) , y ( 0 ) = e − 0.5 × 0 ( C 1 cos ( 0.5 127 × 0 ) + C 2 sin ( 0.5 127 × 0 ) ) − 1 = C 1 cos ( 0 ) + C 2 sin ( 0 ) C 1 ( 1 ) + C 2 ( 0 ) = − 1 C 1 = − 1 Differentiate y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) with respect to t as, y ′ ( t ) = [ e − 0.5 t ( − 0.5 127 C 1 sin ( 0.5 127 t ) + 0.5 127 C 2 cos ( 0.5 127 t ) ) − 0.5 e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) ] = [ e − 0.5 t ( − 0.5 127 C 1 sin ( 0.5 127 t ) + 0.5 127 C 2 cos ( 0.5 127 t ) ) + e − 0.5 t ( − 0.5 C 1 cos ( 0.5 127 t ) − 0.5 C 2 sin ( 0.5 127 t ) ) ] Simplify as, y ′ ( t ) = e − 0.5 t ( − 0.5 127 C 1 sin ( 0.5 127 t ) + 0.5 127 C 2 cos ( 0.5 127 t ) − 0.5 C 1 cos ( 0.5 127 t ) − 0.5 C 2 sin ( 0.5 127 t ) ) = e − 0.5 t ( ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0.5 127 t ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0.5 127 t ) ) Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = e − 0.5 t ( ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0.5 127 t ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0.5 127 t ) ) , y ′ ( 0 ) = e − 0.5 × 0 ( ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0.5 127 × 0 ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0.5 127 × 0 ) ) 0 = ( − 0.5 127 C 1 − 0.5 C 2 ) sin ( 0 ) + ( 0.5 127 C 2 − 0.5 C 1 ) cos ( 0 ) 0 = ( − 0.5 127 C 1 − 0.5 C 2 ) ( 0 ) + ( 0.5 127 C 2 − 0.5 C 1 ) ( 1 ) 0.5 127 C 2 − 0.5 C 1 = 0 Since C 1 = − 1 , the above equation becomes, 0.5 127 C 2 − 0.5 ( − 1 ) = 0 0.5 127 C 2 = − 0.5 C 2 = − 0.5 0.5 127 C 2 = − 1 127 That is, C 1 = − 1 and C 2 = − 1 127 . Thus, the solution y ( t ) = e − 0.5 t ( C 1 cos ( 0.5 127 t ) + C 2 sin ( 0.5 127 t ) ) becomes, y ( t ) = e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) . The aim is to find the which maximum displacement. Note that, the maximum displacement will be obtained at y ′ = 0 . Differentiate y ( t ) = e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) with respect to t as, y ′ ( t ) = [ e − 0.5 t ( 0.5 127 sin ( 0.5 127 t ) − 0.5 127 127 cos ( 0.5 127 t ) ) − 0.5 e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) ] = [ e − 0.5 t ( 0.5 127 sin ( 0.5 127 t ) − 0.5 cos ( 0.5 127 t ) ) + e − 0.5 t ( 0.5 cos ( 0.5 127 t ) + 0.5 sin ( 0.5 127 t ) ) ] Simplify further as, y ′ ( t ) = e − 0.5 t ( 0.5 127 sin ( 0.5 127 t ) + 0.5 sin ( 0.5 127 t ) ) = 0.5 ( 127 + 1 ) e − 0.5 t sin ( 0.5 127 t ) Thus, y ′ = 0 implies, 0.5 ( 127 + 1 ) e − 0.5 t sin ( 0.5 127 t ) = 0 e − 0.5 t sin ( 0.5 127 t ) = 0 Note that, the value of the exponential is never zero. Thus, the only possibility is sin ( 0.5 127 t ) = 0 . This happenes only when 0.5 127 t = n π . Simplify further as, t = n π 0.5 127 . The first value is obtained when n = 1 . Thus, n = 1 implies t = π 0.5 127 . On simplification, t = 2 π 127 . That is, the maximum displacement occurs at t = 2 π 127 . Substitute t = 2 π 127 in y ( t ) = e − 0.5 t ( − cos ( 0.5 127 t ) − 1 127 sin ( 0.5 127 t ) ) , y ( 2 π 127 ) = e − 0.5 × 2 π 127 ( − cos ( 0.5 127 × 2 π 127 ) − 1 127 sin ( 0.5 127 × 2 π 127 ) ) = e − π 127 ( − cos ( π ) − 1 127 sin ( π ) ) = e − 0.2788 ( 1 − 1 127 ( 0 ) ) = 0.7567 ( 1 − 0 ) = 0.7567 That is, the maximum displacement is 0.7567m. In other words, 75.67cm. Therefore, the maximum displacement is 75 .67 cm _ .

Solution Summary: The author explains the governing equation of a mass-spring system: cF_ext=left[inertiaright].

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 4.9, Problem 9E

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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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