Explanation Given: Given that E ( t ) = E 0 cos γ t V and at initial condition q ( 0 ) = q 0 C , i ( 0 ) = i 0 A . Calculation: The linear second order differential equation of LC series circuit is, L d 2 q d t 2 + 1 C q = E ( t ) Substitute the given value of E ( t ) in above equation, L d 2 q d t 2 + 1 C q = E 0 cos γ t d 2 q d t 2 + 1 L C q = E 0 L cos γ t So, the linear second order differential equation is, d 2 q d t 2 + 1 L C q = E 0 L cos γ t (1) The auxiliary equation of the equation (1) is, m 2 + 1 L C = 0 Solution of auxiliary equation is, m 2 = − 1 L C m = − 1 L C = ± i L C Roots of auxiliary equation are complex, m = ± i L C The complementary function is expressed as, q c ( t ) = c 1 cos t L C + c 2 sin t L C (2) The particular integral is expressed as, q p ( t ) = A cos γ t + B sin γ t (3) Differentiate both sides of equation (3) with respect to t , d q p d t = − γ A sin γ t + γ B cos γ t Differentiate both sides of above equation with respect to t , d 2 q p d t 2 = − γ 2 A cos γ t − γ 2 B sin γ t Substitute the values of q p ( t ) , d q p d t and d 2 q p d t 2 in equation (1), ( − γ 2 A cos γ t − γ 2 B sin γ t ) + 1 L C ( A cos γ t + B sin γ t ) = E 0 L cos γ t − γ 2 A cos γ t − γ 2 B sin γ t + 1 L C A cos γ t + 1 L C B sin γ t = E 0 L cos γ t ( − γ 2 + 1 L C ) A cos γ t + ( − γ 2 + 1 L C ) B sin γ t = E 0 L cos γ t Compare both sides of above equation, ( − γ 2 + 1 L C ) A = E 0 L and ( − γ 2 + 1 L C ) B = 0 Therefore, the value of A and B is, A = E 0 L ( − γ 2 + 1 L C ) = E 0 C ( 1 − γ 2 L C ) And, B = 0 Substitute the value of A and B in equation (3), q p ( t ) = ( E 0 C 1 − γ 2 L C ) cos γ t + ( 0 ) sin γ t So, the equation is, q p ( t ) = ( E 0 C 1 − γ 2 L C ) cos γ t (4) Thus, the general solution is, q ( t ) = q c ( t ) + q p ( t ) Substitute the values of q c ( t ) and q p ( t ) from equation (2) and (4) in above equation, q ( t ) = c 1 cos t L C + c 2 sin t L C + ( E 0 C 1 − γ 2 L C ) cos γ t (5) Differentiate both sides of equation (5) with respect to t , q ′ ( t ) = − 1 L C c 1 sin t L C + 1 L C c 2 cos t L C − E 0

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 5.1, Problem 61E

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Chapter 5.1, Problem 60E

Chapter 5.1, Problem 62E

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Chapter 5 Solutions

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

Ch. 5.1 - 5.1.1 Spring/Mass systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A force...Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - 5.1.1Spring/Mass Systems: Free Undamped Motion A...

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The charge on the capacitor and the current in LC series circuit.

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