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The derivation of the system of differential equation shown in given figure if k 1 = 1 , k 2 = 1 , k 3 = 1 , m 1 = 1 , m 2 = 1 and x 1 ( 0 ) , x ′ 1 ( 0 ) = − 1 , x 2 ( 0 ) = 0 and x ′ 2 ( 0 ) = 1 using Laplace transformation.

The derivation of the system of differential equation shown in given figure if k 1 = 1 , k 2 = 1 , k 3 = 1 , m 1 = 1 , m 2 = 1 and x 1 ( 0 ) , x ′ 1 ( 0 ) = − 1 , x 2 ( 0 ) = 0 and x ′ 2 ( 0 ) = 1 using Laplace transformation.

Solution Summary: The author illustrates the derivation of the system of differential equation using Laplace transformation.

Advanced Engineering Mathematics

Advanced Engineering Mathematics

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.6, Problem 14E

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The derivation of the system of differential equation shown in given figure if k1=1,k2=1,k3=1, m1=1,m2=1 and x1(0),x′1(0)=−1,x2(0)=0 and x′2(0)=1 using Laplace transformation.

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Chapter 4.6, Problem 13E

Chapter 4.6, Problem 15E

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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.

True or False and why

10 5 Obtain by multiplying matrices the composite coordinate transformation of two transformations, first x' = (x + y√√2+2)/2 y' = z' (x√√2-2√2)/2 z = (-x+y√√2-2)/2 followed by x" = (x'√√2+z'√√2)/2 y" = (-x'y'√√2+2')/2 z" = (x'y'√√2-2')/2.

Chapter 4 Solutions

Advanced Engineering Mathematics

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