Explanation Formula used: The governing equation of Kirchhoff’s law is calculated as: L q ″ + R q ′ + 1 C q = E ( t ) . The quadratic equation a x 2 + b x + c = 0 is calculated as x = − b ± b 2 − 4 a c 2 a . Theorem used: “If r 1 = α + i β and r 2 = α − i β are the two complex roots of the auxiliary equation a r 2 + b r + c = 0 , then y = e α x ( c 1 cos β x + c 2 sin β x ) is the general solution of a y ″ + b y ′ + c y = 0 ”. Calculation: The voltage source is: E ( t ) = 20 cos t volt. In the circuit the voltage drops at the three points are numerically related as follows: Across the resistor R 4times the instantaneous change in charge it means R d q d t = 4 d q d t Across the capacitor, 10times the charge it means 1 C q = 10 q and across the inductor, 2times the instantaneous change in the current it means L d i d t = 2 d 2 q d t 2 . The initial value of the charge is q ( 0 ) = 2 coulombs and the current is q ′ ( 0 ) = 3 amperes Now, the governing equation of Kirchhoff’s law is: L q ″ + R q ′ + 1 C q = E ( t ) Where, q is the charge at a cross section of a conductor, measured in coulombs, I is the current or rate of change d q d t (flow of electrons) at a cross section of a conductor, measured in amperes, E is the electric (potential) source, measured in volts, and V is the difference in potential between two points along the conductor, measured in volts ( v ). Substitute the value of inductor L is 2 d 2 q d t 2 , the resistor R is 4 d q d t , and the capacitor C is 10 q in L q ″ + R q ′ + 1 C q = E ( t ) . 2 q ″ + 4 q ′ + 10 q = 20 cos t The characteristics equation is: 2 r 2 q + 4 r q + 10 q = 20 cos t ( r 2 + 2 r + 5 ) 2 q = 20 cos t r 2 + 2 r + 5 = 20 cos t The auxiliary equation is: r 2 + 2 r + 5 = 0 Comparing the above equation with a x 2 + b x + c = 0 . Here, a = 1 , b = 2 , c = 5 Substitute the value of a = 1 , b = 2 , c = 5 in the quadratic formula x = − b ± b 2 − 4 a c 2 a . r = − 2 ± 2 2 − 4 × 1 × 5 2 × 1 r = − 1 ± 2 i So, the general solution of differential equation is: q ( t ) = e α t ( c 1 cos β t + c 2 sin β t ) Substitute the value of α and β in q ( t ) = e α t ( c 1 cos β t + c 2 sin β t ) q c ( t ) = e − t ( c 1 cos 2 t + c 2 sin 2 t ) Solve the nonhomogeneous solution of the form q p ( t ) = A sin t + B cos t . Differentiate the equation q p = A sin t + B cos t is as follows: q ′ p ( t ) = A cos t − B sin t Again, differentiate the above equation q ′ p ( t ) = A cos t − B sin t is as follows: q ′ ′ p ( t ) = − A sin t − B cos t Substitute the derivative of q p ( t ) in the given equation 2 q ″ + 4 q ′ + 10 q = 20 cos t and find that A and B must satisfy the equation

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 17.3, Problem 26E

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Chapter 17.3, Problem 25E

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Chapter 17 Solutions

University Calculus: Early Transcendentals (4th Edition)

Ch. 17.1 - In Exercises 1−30, find the general solution of...Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E Ch. 17.1 - Prob. 7E Ch. 17.1 - Prob. 8E Ch. 17.1 - Prob. 9E Ch. 17.1 - Prob. 10E

Ch. 17.1 - Prob. 11E Ch. 17.1 - Prob. 12E Ch. 17.1 - Prob. 13E Ch. 17.1 - Prob. 14E Ch. 17.1 - In Exercises 1 – 30, find the general solution of...Ch. 17.1 - Prob. 16E Ch. 17.1 - Prob. 17E Ch. 17.1 - Prob. 18E Ch. 17.1 - Prob. 19E Ch. 17.1 - Prob. 20E Ch. 17.1 - Prob. 21E Ch. 17.1 - Prob. 22E Ch. 17.1 - Prob. 23E Ch. 17.1 - Prob. 24E Ch. 17.1 - Prob. 25E Ch. 17.1 - Prob. 26E Ch. 17.1 - Prob. 27E Ch. 17.1 - Prob. 28E Ch. 17.1 - Prob. 29E Ch. 17.1 - Prob. 30E Ch. 17.1 - Prob. 31E Ch. 17.1 - Prob. 32E Ch. 17.1 - Prob. 33E Ch. 17.1 - Prob. 34E Ch. 17.1 - Prob. 35E Ch. 17.1 - Prob. 36E Ch. 17.1 - Prob. 37E Ch. 17.1 - Prob. 38E Ch. 17.1 - Prob. 39E Ch. 17.1 - Prob. 40E Ch. 17.1 - Prob. 41E Ch. 17.1 - Prob. 42E Ch. 17.1 - Prob. 43E Ch. 17.1 - Prob. 44E Ch. 17.1 - Prob. 45E Ch. 17.1 - Prob. 46E Ch. 17.1 - In Exercises 41−55, find the general solution. 47....Ch. 17.1 - Prob. 48E Ch. 17.1 - Prob. 49E Ch. 17.1 - Prob. 50E Ch. 17.1 - Prob. 51E Ch. 17.1 - Prob. 52E Ch. 17.1 - Prob. 53E Ch. 17.1 - Prob. 54E Ch. 17.1 - Prob. 55E Ch. 17.1 - In Exercises 56−60, solve the initial value...Ch. 17.1 - Prob. 57E Ch. 17.1 - Prob. 58E Ch. 17.1 - Prob. 59E Ch. 17.1 - Prob. 60E Ch. 17.1 - Prob. 61E Ch. 17.1 - Prob. 62E Ch. 17.1 - Prob. 63E Ch. 17.1 - Prob. 64E Ch. 17.1 - Prob. 65E Ch. 17.1 - Show that if a, b, and c are positive constants,...Ch. 17.2 - Prob. 1E Ch. 17.2 - Prob. 2E Ch. 17.2 - Prob. 3E Ch. 17.2 - Prob. 4E Ch. 17.2 - Prob. 5E Ch. 17.2 - Prob. 6E Ch. 17.2 - Prob. 7E Ch. 17.2 - Prob. 8E Ch. 17.2 - Prob. 9E Ch. 17.2 - Prob. 10E Ch. 17.2 - Prob. 11E Ch. 17.2 - Prob. 12E Ch. 17.2 - Prob. 13E Ch. 17.2 - Prob. 14E Ch. 17.2 - Prob. 15E Ch. 17.2 - Prob. 16E Ch. 17.2 - Prob. 17E Ch. 17.2 - Prob. 18E Ch. 17.2 - Prob. 19E Ch. 17.2 - Prob. 20E Ch. 17.2 - Prob. 21E Ch. 17.2 - Prob. 22E Ch. 17.2 - Prob. 23E Ch. 17.2 - Prob. 24E Ch. 17.2 - Prob. 25E Ch. 17.2 - Prob. 26E Ch. 17.2 - Prob. 27E Ch. 17.2 - Prob. 28E Ch. 17.2 - Prob. 29E Ch. 17.2 - Prob. 30E Ch. 17.2 - Prob. 31E Ch. 17.2 - Prob. 32E Ch. 17.2 - Prob. 33E Ch. 17.2 - Prob. 34E Ch. 17.2 - Prob. 35E Ch. 17.2 - Prob. 36E Ch. 17.2 - Prob. 37E Ch. 17.2 - Prob. 38E Ch. 17.2 - Prob. 39E Ch. 17.2 - Prob. 40E Ch. 17.2 - Prob. 41E Ch. 17.2 - Prob. 42E Ch. 17.2 - Prob. 43E Ch. 17.2 - Prob. 44E Ch. 17.2 - Prob. 45E Ch. 17.2 - Prob. 46E Ch. 17.2 - Prob. 47E Ch. 17.2 - Prob. 48E Ch. 17.2 - Prob. 49E Ch. 17.2 - Prob. 50E Ch. 17.2 - Prob. 51E Ch. 17.2 - Prob. 52E Ch. 17.2 - Prob. 53E Ch. 17.2 - Prob. 54E Ch. 17.2 - Prob. 55E Ch. 17.2 - Prob. 56E Ch. 17.2 - Prob. 57E Ch. 17.2 - Prob. 58E Ch. 17.2 - Prob. 59E Ch. 17.2 - Prob. 60E Ch. 17.3 - Prob. 1E Ch. 17.3 - Prob. 2E Ch. 17.3 - Prob. 3E Ch. 17.3 - Prob. 4E Ch. 17.3 - Prob. 5E Ch. 17.3 - Prob. 6E Ch. 17.3 - Prob. 7E Ch. 17.3 - Prob. 8E Ch. 17.3 - Prob. 9E Ch. 17.3 - Prob. 10E Ch. 17.3 - Prob. 11E Ch. 17.3 - Prob. 12E Ch. 17.3 - Prob. 13E Ch. 17.3 - Prob. 14E Ch. 17.3 - Prob. 15E Ch. 17.3 - Prob. 16E Ch. 17.3 - Prob. 17E Ch. 17.3 - Prob. 18E Ch. 17.3 - Prob. 19E Ch. 17.3 - Prob. 20E Ch. 17.3 - Prob. 21E Ch. 17.3 - Prob. 22E Ch. 17.3 - Prob. 23E Ch. 17.3 - Prob. 24E Ch. 17.3 - Prob. 25E Ch. 17.3 - Prob. 26E Ch. 17.4 - In Exercises 1–24, find the general solution to...Ch. 17.4 - Prob. 2E Ch. 17.4 - Prob. 3E Ch. 17.4 - Prob. 4E Ch. 17.4 - Prob. 5E Ch. 17.4 - Prob. 6E Ch. 17.4 - Prob. 7E Ch. 17.4 - Prob. 8E Ch. 17.4 - Prob. 9E Ch. 17.4 - Prob. 10E Ch. 17.4 - Prob. 11E Ch. 17.4 - Prob. 12E Ch. 17.4 - Prob. 13E Ch. 17.4 - Prob. 14E Ch. 17.4 - Prob. 15E Ch. 17.4 - Prob. 16E Ch. 17.4 - Prob. 17E Ch. 17.4 - Prob. 18E Ch. 17.4 - Prob. 19E Ch. 17.4 - Prob. 20E Ch. 17.4 - Prob. 21E Ch. 17.4 - Prob. 22E Ch. 17.4 - In Exercises 1–24, find the general solution to...Ch. 17.4 - Prob. 24E Ch. 17.4 - Prob. 25E Ch. 17.4 - Prob. 26E Ch. 17.4 - Prob. 27E Ch. 17.4 - Prob. 28E Ch. 17.4 - Prob. 29E Ch. 17.4 - Prob. 30E Ch. 17.5 - Prob. 1E Ch. 17.5 - Prob. 2E Ch. 17.5 - Prob. 3E Ch. 17.5 - Prob. 4E Ch. 17.5 - Prob. 5E Ch. 17.5 - Prob. 6E Ch. 17.5 - Prob. 7E Ch. 17.5 - Prob. 8E Ch. 17.5 - Prob. 9E Ch. 17.5 - Prob. 10E Ch. 17.5 - Prob. 11E Ch. 17.5 - Prob. 12E Ch. 17.5 - Prob. 13E Ch. 17.5 - Prob. 14E Ch. 17.5 - Prob. 15E Ch. 17.5 - Prob. 16E Ch. 17.5 - Prob. 17E Ch. 17.5 - Prob. 18E

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