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

The general solution to the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ . The general solution to the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ is y ( θ ) = 1 2 θ e − θ sin θ + ( c 1 cos θ + c 2 sin θ ) e − θ _ . Procedure used: Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = C t m e α t cos β t , where m is a nonnegative integer, is of the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B m t m + ⋯ + B 1 t + B 0 ) e α t sin β t , with: (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: The given differential equation is, y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ . The corresponding auxiliary equation is r 2 + 2 r + 2 = 0 . Obtain the roots of r 2 + 2 r + 2 = 0 using the quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 2 ± 2 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = − 2 ± − 4 2 = − 2 ± 2 i 2 ( ∵ i = − 1 ) = − 1 ± i That is, the roots are r 1 = − 1 + i and r 2 = − 1 − i . Thus, the general solution for the homogenous equation is y ( θ ) = ( c 1 cos θ + c 2 sin θ ) e − θ . The RHS of the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ is e − θ cos θ . Comparing the term e − θ cos θ with the form C t m e α t cos β t , it is observed that m = 0 , α = − 1 and β = 1 . Since α + i β = − 1 + 1 i is a root of the associated auxiliary equation, s = 1 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B m t m + ⋯ + B 1 t + B 0 ) e α t sin β t for y as a function of θ with s = 1 , m = 0 , α = − 1 and β = 1 will be, y p ( θ ) = θ 1 ( A 0 ) e − θ cos θ + θ 1 ( B 0 ) e − θ sin θ = ( A 0 cos θ + B 0 sin θ ) θ e − θ Obtain the first and second order derivatives of y p ( θ ) = ( A 0 cos θ + B 0 sin θ ) θ e − θ as follows: Differentiate y p ( θ ) = ( A 0 cos θ + B 0 sin θ ) θ e − θ with respect to θ as, y ′ p ( θ ) = ( A 0 cos θ + B 0 sin θ ) ( − θ e − θ + e − θ ) + θ e − θ ( − A 0 sin θ + B 0 cos θ ) = − ( A 0 cos θ + B 0 sin θ ) θ e − θ + ( A 0 cos θ + B 0 sin θ ) e − θ + ( − A 0 sin θ + B 0 cos θ ) θ e − θ = [ ( − A 0 − B 0 ) sin θ + ( − A 0 + B 0 ) cos θ ] θ e − θ + ( A 0 cos θ + B 0 sin θ ) e − θ Differentiate y ′ p ( θ ) with respect to θ as, y ″ p ( θ ) = [ ( ( − A 0 − B 0 ) sin θ + ( − A 0 + B 0 ) cos θ ) ( − θ e − θ + e − θ ) + ( ( − A 0 − B 0 ) cos θ − ( − A 0 + B 0 ) sin θ ) ( θ e − θ ) + ( A 0 cos θ + B 0 sin θ ) ( − e − θ ) + ( − A 0 sin θ + B 0 cos θ ) e − θ ] = [ ( A 0 + B 0 + A 0 − B 0 ) θ e − θ sin θ + ( A 0 − B 0 − A 0 − B 0 ) θ e − θ cos θ + ( − A 0 − A 0 − B 0 − B 0 ) e − θ sin θ + ( − A 0 + B 0 − A 0 + B 0 ) e − θ cos θ ] Simplify further as, y ″ p ( θ ) = [ 2 A 0 θ e − θ sin θ − 2 B 0 θ e − θ cos θ + ( − 2 A 0 − 2 B 0 ) e − θ sin θ + ( − 2 A 0 + 2 B 0 ) e − θ cos θ ] = 2 ( A 0 θ − A 0 − B 0 ) e − θ sin θ + 2 ( − B 0 θ − A 0 + B 0 ) e − θ cos θ Thus, y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) for y p will be, y ″ p ( θ ) + 2 y ′ p ( θ ) + 2 y p ( θ ) = [ 2 ( A 0 θ − A 0 − B 0 ) e − θ sin θ + 2 ( − B 0 θ − A 0 + B 0 ) e − θ cos θ + 2 [ ( − A 0 − B 0 ) sin θ + ( − A 0 + B 0 ) cos θ ] θ e − θ + 2 ( A 0 cos θ + B 0 sin θ ) e − θ + 2 ( A 0 cos θ + B 0 sin θ ) θ e − θ ] Simpify further as, y ″ p ( θ ) + 2 y ′ p ( θ ) + 2 y p ( θ ) = [ ( 2 A 0 θ − 2 A 0 − 2 B 0 − 2 A 0 θ − 2 B 0 θ + 2 B 0 + 2 B 0 θ ) e − θ sin θ + ( − 2 B 0 θ − 2 A 0 + 2 B 0 − 2 A 0 θ + 2 B 0 θ + 2 A 0 + 2 A 0 θ ) e − θ cos θ ] = − 2 A 0 e − θ sin θ + 2 B 0 e − θ cos θ Setting − 2 A 0 e − θ sin θ + 2 B 0 e − θ cos θ = e − θ cos θ and comparing the coefficients result in − 2 A 0 = 0 and 2 B 0 = 1 . On simplification, A 0 = 0 and B 1 = 1 2 . Thus, the particular solution y p ( θ ) = ( A 0 cos θ + B 0 sin θ ) θ e − θ becomes y p ( θ ) = 1 2 θ e − θ sin θ . Combine the general solution y ( θ ) = ( c 1 cos θ + c 2 sin θ ) e − θ of the corresponding homogenous equation and the particular solution y p ( θ ) = 1 2 θ e − θ sin θ to obtain y ( θ ) = 1 2 θ e − θ sin θ + ( c 1 cos θ + c 2 sin θ ) e − θ . Therefore, the general solution to the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ is y ( θ ) = 1 2 θ e − θ sin θ + ( c 1 cos θ + c 2 sin θ ) e − θ _ .

The general solution to the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ . The general solution to the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ is y ( θ ) = 1 2 θ e − θ sin θ + ( c 1 cos θ + c 2 sin θ ) e − θ _ . Procedure used: Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = C t m e α t cos β t , where m is a nonnegative integer, is of the form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B m t m + ⋯ + B 1 t + B 0 ) e α t sin β t , with: (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: The given differential equation is, y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ . The corresponding auxiliary equation is r 2 + 2 r + 2 = 0 . Obtain the roots of r 2 + 2 r + 2 = 0 using the quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 2 ± 2 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = − 2 ± − 4 2 = − 2 ± 2 i 2 ( ∵ i = − 1 ) = − 1 ± i That is, the roots are r 1 = − 1 + i and r 2 = − 1 − i . Thus, the general solution for the homogenous equation is y ( θ ) = ( c 1 cos θ + c 2 sin θ ) e − θ . The RHS of the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ is e − θ cos θ . Comparing the term e − θ cos θ with the form C t m e α t cos β t , it is observed that m = 0 , α = − 1 and β = 1 . Since α + i β = − 1 + 1 i is a root of the associated auxiliary equation, s = 1 . Thus, the particular solution form y p ( t ) = t s ( A m t m + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B m t m + ⋯ + B 1 t + B 0 ) e α t sin β t for y as a function of θ with s = 1 , m = 0 , α = − 1 and β = 1 will be, y p ( θ ) = θ 1 ( A 0 ) e − θ cos θ + θ 1 ( B 0 ) e − θ sin θ = ( A 0 cos θ + B 0 sin θ ) θ e − θ Obtain the first and second order derivatives of y p ( θ ) = ( A 0 cos θ + B 0 sin θ ) θ e − θ as follows: Differentiate y p ( θ ) = ( A 0 cos θ + B 0 sin θ ) θ e − θ with respect to θ as, y ′ p ( θ ) = ( A 0 cos θ + B 0 sin θ ) ( − θ e − θ + e − θ ) + θ e − θ ( − A 0 sin θ + B 0 cos θ ) = − ( A 0 cos θ + B 0 sin θ ) θ e − θ + ( A 0 cos θ + B 0 sin θ ) e − θ + ( − A 0 sin θ + B 0 cos θ ) θ e − θ = [ ( − A 0 − B 0 ) sin θ + ( − A 0 + B 0 ) cos θ ] θ e − θ + ( A 0 cos θ + B 0 sin θ ) e − θ Differentiate y ′ p ( θ ) with respect to θ as, y ″ p ( θ ) = [ ( ( − A 0 − B 0 ) sin θ + ( − A 0 + B 0 ) cos θ ) ( − θ e − θ + e − θ ) + ( ( − A 0 − B 0 ) cos θ − ( − A 0 + B 0 ) sin θ ) ( θ e − θ ) + ( A 0 cos θ + B 0 sin θ ) ( − e − θ ) + ( − A 0 sin θ + B 0 cos θ ) e − θ ] = [ ( A 0 + B 0 + A 0 − B 0 ) θ e − θ sin θ + ( A 0 − B 0 − A 0 − B 0 ) θ e − θ cos θ + ( − A 0 − A 0 − B 0 − B 0 ) e − θ sin θ + ( − A 0 + B 0 − A 0 + B 0 ) e − θ cos θ ] Simplify further as, y ″ p ( θ ) = [ 2 A 0 θ e − θ sin θ − 2 B 0 θ e − θ cos θ + ( − 2 A 0 − 2 B 0 ) e − θ sin θ + ( − 2 A 0 + 2 B 0 ) e − θ cos θ ] = 2 ( A 0 θ − A 0 − B 0 ) e − θ sin θ + 2 ( − B 0 θ − A 0 + B 0 ) e − θ cos θ Thus, y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) for y p will be, y ″ p ( θ ) + 2 y ′ p ( θ ) + 2 y p ( θ ) = [ 2 ( A 0 θ − A 0 − B 0 ) e − θ sin θ + 2 ( − B 0 θ − A 0 + B 0 ) e − θ cos θ + 2 [ ( − A 0 − B 0 ) sin θ + ( − A 0 + B 0 ) cos θ ] θ e − θ + 2 ( A 0 cos θ + B 0 sin θ ) e − θ + 2 ( A 0 cos θ + B 0 sin θ ) θ e − θ ] Simpify further as, y ″ p ( θ ) + 2 y ′ p ( θ ) + 2 y p ( θ ) = [ ( 2 A 0 θ − 2 A 0 − 2 B 0 − 2 A 0 θ − 2 B 0 θ + 2 B 0 + 2 B 0 θ ) e − θ sin θ + ( − 2 B 0 θ − 2 A 0 + 2 B 0 − 2 A 0 θ + 2 B 0 θ + 2 A 0 + 2 A 0 θ ) e − θ cos θ ] = − 2 A 0 e − θ sin θ + 2 B 0 e − θ cos θ Setting − 2 A 0 e − θ sin θ + 2 B 0 e − θ cos θ = e − θ cos θ and comparing the coefficients result in − 2 A 0 = 0 and 2 B 0 = 1 . On simplification, A 0 = 0 and B 1 = 1 2 . Thus, the particular solution y p ( θ ) = ( A 0 cos θ + B 0 sin θ ) θ e − θ becomes y p ( θ ) = 1 2 θ e − θ sin θ . Combine the general solution y ( θ ) = ( c 1 cos θ + c 2 sin θ ) e − θ of the corresponding homogenous equation and the particular solution y p ( θ ) = 1 2 θ e − θ sin θ to obtain y ( θ ) = 1 2 θ e − θ sin θ + ( c 1 cos θ + c 2 sin θ ) e − θ . Therefore, the general solution to the differential equation y ″ ( θ ) + 2 y ′ ( θ ) + 2 y ( θ ) = e − θ cos θ is y ( θ ) = 1 2 θ e − θ sin θ + ( c 1 cos θ + c 2 sin θ ) e − θ _ .

Solution Summary: The author explains the general solution to the differential equation ay′′(theta )+2

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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