examples-ch4

examples-ch4 - MATH 152 Fall 2006-07 Applied Linear Algebra...

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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 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 ˛ ˛ 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 ˛ ˛ ˛ ˛ R + 1 2( a- s- ib ) e ( a- s- ib ) t ˛ ˛ ˛ ˛ 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 ˛ ˛ ˛ R- 1 a- s Z R e ( a- s ) t dt – ( s>a ) = lim R →∞ »- 1 ( a- s ) 2 e ( a- s ) t ˛ ˛ ˛ 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 ff 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....
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examples-ch4 - MATH 152 Fall 2006-07 Applied Linear Algebra...

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