Explanation Given: The initial value problem: y ' ' + 10 y ' + k y = 0 y ( 0 ) = 1 y ' ( 0 ) = 0 Approach: Substitute value of k for each case and solve for the equation of motion. Calculation: Case 1: k = 20 The given initial value problem becomes y ' ' + + 10 y ' + 20 y = 0 y ( 0 ) = 1 y ' ( 0 ) = 0 (1) Auxiliary equation for equation (1) is r 2 + 10 r + 20 = 0 . Its roots are: r = − 10 ± 10 2 − 4 ∗ 20 2 = − 10 ± 20 2 = − 5 ± 5 General solution is: y = A e ( − 5 + 5 ) t + B e ( − 5 − 5 ) t Apply initial condition to get: A + B = 1 ( − 5 + 5 ) A + ( − 5 − 5 ) B = 0 − 5 ( A + B ) + 5 ( A − B ) = 0 A − B = 5 A = 1 + 5 2 B = 1 − 5 2 So the equation of motion is y = 1 + 5 2 e ( − 5 + 5 ) t + 1 − 5 2 B e ( − 5 − 5 ) t (2) Its graph is shown below: Figure(1) Case 2: k = 25 The given initial value problem becomes y ' ' + 10 y ' + 25 y = 0 y ( 0 ) = 1 y ' ( 0 ) = 0 (3) Auxiliary equation for equation (3) is r 2 + 10 r + 25 = 0

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 4.9, Problem 5E

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Equation of motion and sketch the graph for k=20,25,30.

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Chapter 4.9, Problem 4E

Chapter 4.9, Problem 6E

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Fundamentals of Differential Equations and Boundary Value Problems

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