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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

The equation of motion if y ( 0 ) = − 1 and y ′ ( 0 ) = 5 . The equation of motion is y ( t ) = e − t sin 2 t − cos 3 t + sin 3 t _ . Procedure used: A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. Calculation: Given that, the mass of the spring is 1, the sping constant is 5, the damping coefficients is 2 and the sinusoidal external force is g ( t ) = 2 sin 3 t + 10 cos 3 t . That is, m = 1 , b = 2 and k = 5 . The general mass-spring system equation is m y ″ + b y ′ + k y = g ( t ) . Substitute m = 1 , b = 2 , k = 5 and g ( t ) = 2 sin 3 t + 10 cos 3 t in m y ″ + b y ′ + k y = g ( t ) to obtain y ″ + 2 y ′ + 5 y = 2 sin 3 t + 10 cos 3 t . The initial conditions are y ( 0 ) = − 1 and y ′ ( 0 ) = 5 . The corresponding auxiliary equation is r 2 + 2 r + 5 = 0 . Obtain the root using the quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 2 ± 2 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) = − 2 ± 4 − 20 2 = − 2 ± − 16 2 = − 2 ± 4 i 2 ( ∵ i = − 4 ) = − 1 ± 2 i That is, the roots are r 1 = − 1 + 2 i and r 2 = − 1 − 2 i . Here, α = − 1 and β = 2 . Thus, the general solution for the homogenous equation is, y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t . The RHS of the differential equation y ″ + 2 y ′ + 5 y = 2 sin 3 t + 10 cos 3 t is 2 sin 3 t + 10 cos 3 t . Comparing the term 2 sin 3 t + 10 cos 3 t with the form P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , it is observed that m = 0 , n = 0 , α = 0 and β = 3 . Since α + i β = 0 + 3 i is not a root of the associated auxiliary equation, s = 0 . Note that, k is the larger of m and n , which is 0. Thus, the particular solution form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t for s = 0 , k = 0 , α = 0 and β = 3 will be, y p ( t ) = t 0 ( A 0 ) e 0 cos 3 t + t 0 ( B 0 ) e 0 sin 3 t = A 0 cos 3 t + B 0 sin 3 t Differentiate y p ( t ) = A 0 cos 3 t + B 0 sin 3 t with respect to t to obtain, y ′ p ( t ) = − 3 A 0 sin 3 t + 3 B 0 cos 3 t . Differentiate y ′ p ( t ) = − 3 A 0 sin 3 t + 3 B 0 cos 3 t with respect to t to obtain, y ″ p ( t ) = − 9 A 0 cos 3 t − 9 B 0 sin 3 t . Thus, y ″ + 2 y ′ + 5 y for y p will be, y ″ p + 2 y ′ p + 5 y p = [ ( − 9 A 0 cos 3 t − 9 B 0 sin 3 t ) + 2 ( − 3 A 0 sin 3 t + 3 B 0 cos 3 t ) + 5 ( A 0 cos 3 t + B 0 sin 3 t ) ] = − 9 A 0 cos 3 t − 9 B 0 sin 3 t − 6 A 0 sin 3 t + 6 B 0 cos 3 t + 5 A 0 cos 3 t + 5 B 0 sin 3 t = ( − 9 A 0 + 6 B 0 + 5 A 0 ) cos 3 t + ( − 9 B 0 − 6 A 0 + 5 B 0 ) sin 3 t = ( 6 B 0 − 4 A 0 ) cos 3 t + ( − 4 B 0 − 6 A 0 ) sin 3 t Setting ( 6 B 0 − 4 A 0 ) cos 3 t + ( − 4 B 0 − 6 A 0 ) sin 3 t = 2 sin 3 t + 10 cos 3 t and comparing the coefficients result in 6 B 0 − 4 A 0 = 10 and − 4 B 0 − 6 A 0 = 2 . Simplify − 4 B 0 − 6 A 0 = 2 as, − 2 B 0 − 3 A 0 = 1 2 B 0 = − 1 − 3 A 0 Substitute 2 B 0 = − 1 − 3 A 0 in 6 B 0 − 4 A 0 = 10 , 3 ( 2 B 0 ) − 4 A 0 = 10 3 ( − 1 − 3 A 0 ) − 4 A 0 = 10 − 3 − 9 A 0 − 4 A 0 = 10 − 3 − 13 A 0 = 10 Simplify further as, − 13 A 0 = 13 A 0 = − 1 Substitute A 0 = − 1 in 2 B 0 = − 1 − 3 A 0 , 2 B 0 = − 1 − 3 ( − 1 ) 2 B 0 = − 1 + 3 2 B 0 = 2 B 0 = 1 That is, A 0 = − 1 and B 0 = 1 . Thus, the particular solution y p ( t ) = A 0 cos 3 t + B 0 sin 3 t becomes, y p ( t ) = − cos 3 t + sin 3 t . Combine the general solution y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t of the corresponding homogenous equation and the particular solution y p ( t ) = − cos 3 t + sin 3 t to obtain y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t . Obtain the value of the constants using the initial conditions y ( 0 ) = − 1 and y ′ ( 0 ) = 5 as follows: Substitute t = 0 and y ( 0 ) = − 1 in y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t , y ( 0 ) = c 1 e 0 cos ( 2 × 0 ) + c 2 e 0 sin ( 2 × 0 ) − cos ( 3 × 0 ) + sin ( 3 × 0 ) − 1 = c 1 cos ( 0 ) + c 2 sin ( 0 ) − cos ( 0 ) + sin ( 0 ) − 1 = c 1 ( 1 ) + c 2 ( 0 ) − 1 + 0 c 1 = − 1 + 1 c 1 = 0 Differentiate y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t with respect to t as, y ′ ( t ) = − 2 c 1 e − t sin 2 t − c 1 e − t cos 2 t + 2 c 2 e − t cos 2 t − c 2 e − t sin 2 t + 3 sin 3 t + 3 cos 3 t = ( − 2 c 1 e − t − c 2 e − t ) sin 2 t + ( − c 1 e − t + 2 c 2 e − t ) cos 2 t + 3 sin 3 t + 3 cos 3 t Substitute t = 0 and y ′ ( 0 ) = 5 in y ′ ( t ) = ( − 2 c 1 e − t − c 2 e − t ) sin 2 t + ( − c 1 e − t + 2 c 2 e − t ) cos 2 t + 3 sin 3 t + 3 cos 3 t , y ′ ( 0 ) = ( − 2 c 1 e 0 − c 2 e 0 ) sin ( 2 × 0 ) + ( − c 1 e 0 + 2 c 2 e 0 ) cos ( 2 × 0 ) + 3 sin ( 3 × 0 ) + 3 cos ( 3 × 0 ) 5 = ( − 2 c 1 − c 2 ) sin ( 0 ) + ( − c 1 + 2 c 2 ) cos ( 0 ) + 3 sin ( 0 ) + 3 cos ( 0 ) 5 = ( − 2 c 1 − c 2 ) ( 0 ) + ( − c 1 + 2 c 2 ) ( 1 ) + 0 + 3 5 = − c 1 + 2 c 2 + 3 Simplify further as, 5 = 0 + 2 c 2 + 3 ( ∵ c 1 = 0 ) 2 c 2 = 5 − 3 2 c 2 = 2 c 2 = 1 That is, c 1 = 0 and c 2 = 1 . Thus, the solution y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t becomes, y ( t ) = e − t sin 2 t − cos 3 t + sin 3 t . Therefore, the equation of motion is y ( t ) = e − t sin 2 t − cos 3 t + sin 3 t _ .

The equation of motion if y ( 0 ) = − 1 and y ′ ( 0 ) = 5 . The equation of motion is y ( t ) = e − t sin 2 t − cos 3 t + sin 3 t _ . Procedure used: A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. Calculation: Given that, the mass of the spring is 1, the sping constant is 5, the damping coefficients is 2 and the sinusoidal external force is g ( t ) = 2 sin 3 t + 10 cos 3 t . That is, m = 1 , b = 2 and k = 5 . The general mass-spring system equation is m y ″ + b y ′ + k y = g ( t ) . Substitute m = 1 , b = 2 , k = 5 and g ( t ) = 2 sin 3 t + 10 cos 3 t in m y ″ + b y ′ + k y = g ( t ) to obtain y ″ + 2 y ′ + 5 y = 2 sin 3 t + 10 cos 3 t . The initial conditions are y ( 0 ) = − 1 and y ′ ( 0 ) = 5 . The corresponding auxiliary equation is r 2 + 2 r + 5 = 0 . Obtain the root using the quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 2 ± 2 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) = − 2 ± 4 − 20 2 = − 2 ± − 16 2 = − 2 ± 4 i 2 ( ∵ i = − 4 ) = − 1 ± 2 i That is, the roots are r 1 = − 1 + 2 i and r 2 = − 1 − 2 i . Here, α = − 1 and β = 2 . Thus, the general solution for the homogenous equation is, y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t . The RHS of the differential equation y ″ + 2 y ′ + 5 y = 2 sin 3 t + 10 cos 3 t is 2 sin 3 t + 10 cos 3 t . Comparing the term 2 sin 3 t + 10 cos 3 t with the form P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , it is observed that m = 0 , n = 0 , α = 0 and β = 3 . Since α + i β = 0 + 3 i is not a root of the associated auxiliary equation, s = 0 . Note that, k is the larger of m and n , which is 0. Thus, the particular solution form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t for s = 0 , k = 0 , α = 0 and β = 3 will be, y p ( t ) = t 0 ( A 0 ) e 0 cos 3 t + t 0 ( B 0 ) e 0 sin 3 t = A 0 cos 3 t + B 0 sin 3 t Differentiate y p ( t ) = A 0 cos 3 t + B 0 sin 3 t with respect to t to obtain, y ′ p ( t ) = − 3 A 0 sin 3 t + 3 B 0 cos 3 t . Differentiate y ′ p ( t ) = − 3 A 0 sin 3 t + 3 B 0 cos 3 t with respect to t to obtain, y ″ p ( t ) = − 9 A 0 cos 3 t − 9 B 0 sin 3 t . Thus, y ″ + 2 y ′ + 5 y for y p will be, y ″ p + 2 y ′ p + 5 y p = [ ( − 9 A 0 cos 3 t − 9 B 0 sin 3 t ) + 2 ( − 3 A 0 sin 3 t + 3 B 0 cos 3 t ) + 5 ( A 0 cos 3 t + B 0 sin 3 t ) ] = − 9 A 0 cos 3 t − 9 B 0 sin 3 t − 6 A 0 sin 3 t + 6 B 0 cos 3 t + 5 A 0 cos 3 t + 5 B 0 sin 3 t = ( − 9 A 0 + 6 B 0 + 5 A 0 ) cos 3 t + ( − 9 B 0 − 6 A 0 + 5 B 0 ) sin 3 t = ( 6 B 0 − 4 A 0 ) cos 3 t + ( − 4 B 0 − 6 A 0 ) sin 3 t Setting ( 6 B 0 − 4 A 0 ) cos 3 t + ( − 4 B 0 − 6 A 0 ) sin 3 t = 2 sin 3 t + 10 cos 3 t and comparing the coefficients result in 6 B 0 − 4 A 0 = 10 and − 4 B 0 − 6 A 0 = 2 . Simplify − 4 B 0 − 6 A 0 = 2 as, − 2 B 0 − 3 A 0 = 1 2 B 0 = − 1 − 3 A 0 Substitute 2 B 0 = − 1 − 3 A 0 in 6 B 0 − 4 A 0 = 10 , 3 ( 2 B 0 ) − 4 A 0 = 10 3 ( − 1 − 3 A 0 ) − 4 A 0 = 10 − 3 − 9 A 0 − 4 A 0 = 10 − 3 − 13 A 0 = 10 Simplify further as, − 13 A 0 = 13 A 0 = − 1 Substitute A 0 = − 1 in 2 B 0 = − 1 − 3 A 0 , 2 B 0 = − 1 − 3 ( − 1 ) 2 B 0 = − 1 + 3 2 B 0 = 2 B 0 = 1 That is, A 0 = − 1 and B 0 = 1 . Thus, the particular solution y p ( t ) = A 0 cos 3 t + B 0 sin 3 t becomes, y p ( t ) = − cos 3 t + sin 3 t . Combine the general solution y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t of the corresponding homogenous equation and the particular solution y p ( t ) = − cos 3 t + sin 3 t to obtain y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t . Obtain the value of the constants using the initial conditions y ( 0 ) = − 1 and y ′ ( 0 ) = 5 as follows: Substitute t = 0 and y ( 0 ) = − 1 in y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t , y ( 0 ) = c 1 e 0 cos ( 2 × 0 ) + c 2 e 0 sin ( 2 × 0 ) − cos ( 3 × 0 ) + sin ( 3 × 0 ) − 1 = c 1 cos ( 0 ) + c 2 sin ( 0 ) − cos ( 0 ) + sin ( 0 ) − 1 = c 1 ( 1 ) + c 2 ( 0 ) − 1 + 0 c 1 = − 1 + 1 c 1 = 0 Differentiate y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t with respect to t as, y ′ ( t ) = − 2 c 1 e − t sin 2 t − c 1 e − t cos 2 t + 2 c 2 e − t cos 2 t − c 2 e − t sin 2 t + 3 sin 3 t + 3 cos 3 t = ( − 2 c 1 e − t − c 2 e − t ) sin 2 t + ( − c 1 e − t + 2 c 2 e − t ) cos 2 t + 3 sin 3 t + 3 cos 3 t Substitute t = 0 and y ′ ( 0 ) = 5 in y ′ ( t ) = ( − 2 c 1 e − t − c 2 e − t ) sin 2 t + ( − c 1 e − t + 2 c 2 e − t ) cos 2 t + 3 sin 3 t + 3 cos 3 t , y ′ ( 0 ) = ( − 2 c 1 e 0 − c 2 e 0 ) sin ( 2 × 0 ) + ( − c 1 e 0 + 2 c 2 e 0 ) cos ( 2 × 0 ) + 3 sin ( 3 × 0 ) + 3 cos ( 3 × 0 ) 5 = ( − 2 c 1 − c 2 ) sin ( 0 ) + ( − c 1 + 2 c 2 ) cos ( 0 ) + 3 sin ( 0 ) + 3 cos ( 0 ) 5 = ( − 2 c 1 − c 2 ) ( 0 ) + ( − c 1 + 2 c 2 ) ( 1 ) + 0 + 3 5 = − c 1 + 2 c 2 + 3 Simplify further as, 5 = 0 + 2 c 2 + 3 ( ∵ c 1 = 0 ) 2 c 2 = 5 − 3 2 c 2 = 2 c 2 = 1 That is, c 1 = 0 and c 2 = 1 . Thus, the solution y ( t ) = c 1 e − t cos 2 t + c 2 e − t sin 2 t − cos 3 t + sin 3 t becomes, y ( t ) = e − t sin 2 t − cos 3 t + sin 3 t . Therefore, the equation of motion is y ( t ) = e − t sin 2 t − cos 3 t + sin 3 t _ .

Solution Summary: The author explains how the equation of motion is underset_y(t)=e-t

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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Chapter 4.5, Problem 43E

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Chapter 4 Solutions

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

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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