Explanation Given: The linear differential equation is d 2 q d t 2 + q = 1 − 4 U ( t − π ) + 6 U ( t − 3 π ) . The initial values are q ( 0 ) = 0 and q ′ ( 0 ) = 0 . Definition used: (1)The Laplace transformation of a linear differential equation with constant coefficient is express as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] (2) If F ( s ) = L { f ( t ) } and a > 0 , then following relations holds. L { f ( t − a ) U ( t − a ) } = e − a s F ( s ) L { U ( t − a ) } = e − a s s (3) The linearity property of the Laplace transformation is as below: If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } . Calculation: Use the definition (1) to write the Laplace transform of the given linear differential equation. s 2 L { q } − s q ( 0 ) − q ′ ( 0 ) + L { q } = L { 1 − 4 U ( t − π ) + 6 U ( t − 3 π ) } … ( 1 ) Put the initial values q ( 0 ) = 0 and q ′ ( 0 ) = 0 in equation (1) and use the definition (2) to solve. ( s 2 + 1 ) L { q } = L { 1 } − 4 L { U ( t − π ) } + 6 { U ( t − 3 π ) } ( s 2 + 1 ) L { q } = 1 s − 4 e − π s s + 6 e − 3 π s s L { q } = 1 s ( s 2 + 1 ) − 4 e − π s s ( s 2 + 1 ) + 6 e − 3 π s s ( s 2 + 1 ) The obtained equation is as follows: L { q } = 1 s ( s 2 + 1 ) − 4 e − π s s ( s 2 + 1 ) + 6 e − 3 π s s ( s 2 + 1 ) … ( 2 ) In equation (2) the term 1 s ( s 2 + 1 ) is simplify as follows using partial fraction. 1 s ( s 2 + 1 ) = a 0 s + a 2 s + a 1 s 2 + 1 … ( 3 ) Multiply with s ( s 2 + 1 ) on both side of equation (3)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.4, Problem 72E

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The expression for the charge q(t) from the differential equation of LC series circuit using Laplace transform.

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Chapter 4.4, Problem 71E

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2. Consider the ODE u' = ƒ (u) = u² + r where r is a parameter that can take the values r = −1, −0.5, -0.1, 0.1. For each value of r: (a) Sketch ƒ(u) = u² + r and determine the equilibrium points. (b) Draw the phase line. (d) Determine the stability of the equilibrium points. (d) Plot the direction field and some sample solutions,i.e., u(t) (e) Describe how location of the equilibrium points and their stability change as you increase the parameter r. (f) Using the matlab program phaseline.m generate a solution for each value of r and the initial condition u(0) = 0.9. Print and turn in your result for r = −1. Do not forget to add a figure caption. (g) In the matlab program phaseline.m set the initial condition to u(0) = 1.1 and simulate the ode over the time interval t = [0, 10] for different values of r. What happens? Why? You do not need to turn in a plot for (g), just describe what happens.

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The expression for the charge q ( t ) from the differential equation of L C series circuit using Laplace transform.

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Chapter 4 Solutions