examples -ch2

examples -ch2 - MATH 152 Fall 2006-07 Applied Linear...

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Unformatted text preview: MATH 152 Fall 2006-07 Applied Linear Algebra & Differential Equations Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology September 1, 2006 ii Contents Table of Contents iii 1 Introduction 1 2 First-Order Differential Equations 5 3 Second-Order Linear Equations 39 4 Laplace Transform 83 5 Matrix 127 6 Systems of Linear Equations 135 7 Euclidean Vector 155 8 Eigenvalue and Eigenvector 177 9 Systems of Differential Equations 209 10 Orthogonality 231 iii Chapter 2 First-Order Differential Equations ¥ Example 2.1 (Integrating factor for linear equation) Solve the initial value problem ( ty + 2 y = t 2- t + 1 , y (1) = 1 / 2 . Solution For t 6 = 0, rewrite the equation in standard form y + 2 t y = t- 1 + 1 t . The integrating factor is e R ( 2 t ) dt = e 2 ln t = t 2 . Then consider ( t 2 y ) = t 2 y + 2 ty = t 2 „ y + 2 t y « = t 3- t 2 + t. Integrating both sides gives t 2 y = 1 4 t 4- 1 3 t 3 + 1 2 t 2 + C, or y ( t ) = 1 4 t 2- 1 3 t + 1 2 + C t 2 . The initial condition implies 1 2 = y (1) = 1 4- 1 3 + 1 2 + C = ⇒ C = 1 3- 1 4 = 1 12 . Hence, the solution is y ( t ) = 1 4 t 2- 1 3 t + 1 2 + 1 12 t 2 . 2 5 2. First-Order Differential Equations ¥ Example 2.2 (Integrating factor for linear equation) Solve the first-order linear equation 2 y + y = t- 1 . Solution Rewrite the equation in standard form y + 1 2 y = 1 2 ( t- 1) . The integrating factor is e R ( 1 2 ) dt = e 1 2 t . Then consider ( e 1 2 t y ) = e 1 2 t „ y + 1 2 y « = 1 2 e 1 2 t ( t- 1) . Integrating both sides gives e 1 2 t y = 1 2 Z e 1 2 t ( t- 1) dt = Z ( t- 1) d ( e 1 2 t ) = ( t- 1) e 1 2 t- Z e 1 2 t dt = ( t- 1) e 1 2 t- 2 e 1 2 t + C. Hence, the general solution is y ( t ) = t- 3 + Ce- 1 2 t . 2 ¥ Example 2.3 (Integrating factor for linear equation) Solve the first-order linear equation ty + 2 y = e t . Solution For t 6 = 0, y + 2 t y = e t t . The integrating factor is e R ( 2 t ) dt = e 2 ln t = t 2 . Then consider ( t 2 y ) = t 2 „ y + 2 t y « = te t . Integrating both sides gives t 2 y = Z te t dt = Z td ( e t ) = te t- Z e t dt = te t- e t + C. Hence, the general solution is y ( t ) = e t t- e t t 2 + C t 2 . 2 6 ¥ Example 2.4 (Integrating factor for linear equation) Solve the initial value problem ( (1- t 2 ) y- ty = t (1- t 2 ) , y (0) = 2 . Solution For | t | 6 = 1, y- t 1- t 2 y = t. The integrating factor is e R “- t 1- t 2 ” dt = e 1 2 R 1 1- t 2 d (1- t 2 ) = e 1 2 ln | 1- t 2 | = p 1- t 2 ,- 1 < t < 1 . Then consider ( p 1- t 2 y ) = p 1- t 2 „ y- t 1- t 2 y « = t p 1- t 2 . Integrating both sides gives p 1- t 2 y = Z t p 1- t 2 dt =- 1 2 Z p 1- t 2 d (1- t 2 ) =- 1 3 (1- t 2 ) 3 / 2 + C. At t = 0, y = 2, we have (1)(2) =- 1 3 (1) + C = ⇒ C = 7 3 ....
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This note was uploaded on 09/30/2010 for the course MATH MATH152 taught by Professor Kcc during the Spring '10 term at HKUST.

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examples -ch2 - MATH 152 Fall 2006-07 Applied Linear...

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