examples152-ch4

# Examples152-ch4 - MATH 152 Spring 2004-05 Applied Linear Algebra& Differential Equations Worked Examples Dr Tony Yee Department of Mathematics

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Unformatted text preview: MATH 152 Spring 2004-05 Applied Linear Algebra & Differential Equations Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology March 1, 2005 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 iii Chapter 4 Laplace Transform ¥ Example 4.1 (Laplace transform by definition) Find the Laplace transform of t 2 by definition. Solution L{ t 2 } def. = Z ∞ e- st ( t 2 ) dt =- 1 s lim R →∞ Z R t 2 d ( e- st ) =- 1 s lim R →∞ • R 2 e- sR- 0 + 2 s Z R td ( e- st ) ‚ ( s> 0) =- 2 s 2 lim R →∞ • te- st fl fl R- Z R e- st dt ‚ =- 2 s 2 lim R →∞ Re- sR + 2 s 2 lim R →∞ • e- st- s ‚ R = 0- 2 s 3 lim R →∞ ‡ e- sR- 1 · , L{ t 2 } = 2 s 3 , s > 0. 2 ¥ Example 4.2 (Laplace transform by definition) Find the Laplace transform of e at cos bt by definition. Solution L{ e at cos bt } def. = Z ∞ e- st ( e at cos bt ) dt = lim R →∞ Z R e ( a- s ) t cos btdt = lim R →∞ • 1 2 Z R e ( a- s + ib ) t dt + 1 2 Z R e ( a- s- ib ) t dt ‚ = lim R →∞ " 1 2( a- s + ib ) e ( a- s + ib ) t fl fl fl fl R + 1 2( a- s- ib ) e ( a- s- ib ) t fl fl fl fl R # ( s>a ) =- 1 2 • 1 a- s + ib + 1 a- s- ib ‚ =- 1 2 • 2( a- s ) ( a- s ) 2 + b 2 ‚ , L ' e at cos bt “ = s- a ( s- a ) 2 + b 2 , s > a . 2 83 4. Laplace Transform ¥ Example 4.3 (Laplace transform by definition) Find the Laplace transform of e- 3 t sin2 t by definition. Solution L ' e- 3 t sin2 t “ def. = Z ∞ e- st ( e- 3 t sin2 t ) dt = 1 2 i Z ∞ e (- s- 3+2 i ) t dt- 1 2 i Z ∞ e (- s- 3- 2 i ) t dt ( s>- 3) = 1 2 i • 1- s- 3- 2 i- 1- s- 3 + 2 i ‚ , L ' e- 3 t sin2 t “ = 2 ( s + 3) 2 + 4 , s >- 3. 2 ¥ Example 4.4 (Laplace transform by definition) Find the Laplace transform of te at by definition. Solution L{ te at } def. = Z ∞ e- st ( te at ) dt = lim R →∞ Z R te ( a- s ) t dt = lim R →∞ • 1 a- s Z R td ( e ( a- s ) t ) ‚ = lim R →∞ • 1 a- s te ( a- s ) t fl fl fl R- 1 a- s Z R e ( a- s ) t dt ‚ ( s>a ) = lim R →∞ •- 1 ( a- s ) 2 e ( a- s ) t fl fl fl R ‚ , L{ te at } = 1 ( s- a ) 2 , s > a . 2 ¥ Example 4.5 (Laplace transform by definition) Find the Laplace transform of cosh t by definition. Solution L{ cosh t } def. = L ‰ e t + e- t 2 def. = 1 2 Z ∞ e- st ( e t + e- t ) dt = 1 2 lim R →∞ • e- ( s- 1) t- ( s- 1) + e- ( s +1) t- ( s + 1) ‚ R = 1 2 1 s- 1 + 1 s + 1 ¶- 1 2 lim R →∞ • e- ( s- 1) R ( s- 1) + e- ( s +1) R s + 1 ‚ ( s> 1) = s s 2- 1- 1 2 [0 + 0] , L{ cosh t } = s s 2- 1 , s > 1. 2 84 ¥ Example 4.6 (Laplace transform by definition) Find by definition the Laplace transform of f ( t ) = ( 2 , < t < 3 , 3 , t > 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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Examples152-ch4 - MATH 152 Spring 2004-05 Applied Linear Algebra& Differential Equations Worked Examples Dr Tony Yee Department of Mathematics

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