= e 6t. = t 1 = t. 5 t 8L 1[ 1 = 3L 1 [ 1. L 1 [ π. = 3 π. = L 1 3s = L. = 3L 1 s t. = 3 cos(5t) sin(5t).

Μέγεθος: px

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Download "= e 6t. = t 1 = t. 5 t 8L 1[ 1 = 3L 1 [ 1. L 1 [ π. = 3 π. = L 1 3s = L. = 3L 1 s t. = 3 cos(5t) sin(5t)."

Μαθθαῖος Αλεξανδρίδης
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1 Worked Soluion 95 Chaper 25: The Invere Laplace Tranform 25 a From he able: L ] e c L 2 ] ] L! + 25 e L ] ] L in(5) 252 a L 6 + 2] L 6 ( 2)] 6L ( 2)] 6e c L 3 8 4] 3L ] 8L 4] 3L π π ] 3 π L π ] 8e 4 8e 4 3 8e 4 3 8e 4 π π 252 e L ] L ] ] L ] 3L ] 5 5 L co(5) + 5 in(5) 253 a From exercie 243 we have L in(ω)] 2ω ( 2 + ω 2) 2 y he definiion of he invere Laplace ranform lineariy i hen follow ha ] ] in(ω) L 2ω ( 2 + ω 2) 2ωL 2 ( 2 + ω 2) 2 Dividing hrough by 2ω hen yield L ] ( 2 + ω 2) 2 2ω in(ω)

2 96 The Invere Laplace Tranform 254 a L y + 9y ] L] L y ] + 9Ly] Y() y() ] + 9Y() }{{} 4 ( + 9)Y() 4 Y() Taking he invere: y() L Y()] L ] 4L ( 9)] 4e a Fir we mu find he parial fracion expanion: ( + 2)( ) A A( ) ( + 2)( ) + ( + 2) ( )( + 2) A( ) + ( + 2) ( + 2)( ) Cuing ou he middle muliplying hrough by ( + 2)( ) hen give A( ) + ( + 2) Solving for A i eaily done by repecively plugging 2 ino he la equaion olving : A( 2 ) + ( 2 + 2) 7( 2) + 5 A 7( 2) A( ) + ( + 2) Thu he parial fracion expanion of our funcion i ( + 2)( ) A Now we can ake he invere ranform: L L ( + 2)( )] ] 3L ] + 4L ] 3e 2 + 4e c 2 4 ( 2)( + 2) A A( + 2) ( 2)( + 2) + ( 2) ( + 2)( 2) A( + 2) + ( 2) ( 2)( + 2)

3 Worked Soluion 97 So A( + 2) + ( 2) Leing 2 yield A(2 + 2) + (2 2) A 4 while leing yield A( 2 + 2) + ( 2 2) 4 Thu L 2 4] 2 4 A 2 + L /4 2 + /4 + 2] 4 L /4 2 + /4 + 2 ] 4 L + 2 ] 4 e2 4 e e ( 4) A C 4 A( 4) + ( 4) + C2 2 ( 4) So A( 4) + ( 4) + C 2 Muliplying hi ou gahering like erm yield ( + C) 2 + (A 4) 4A giving u he yem which in urn mean ha + C A 4 4A A 4 A 4 6 C 6 So ] L A C 4 /4 2 /6 L /4 2 /6 + /6 4] ] 4 L 2 6 L + /6 4 ] + 6 L 4] e4 6 e4 4 6

4 98 The Invere Laplace Tranform 255 g ( + 6) ( ) A C ( ) A ( + C)( + 6) ( + 6) ( ) So ( ) A ( + C)( + 6) ( ) Muliplying hi ou gahering like erm yield (A + ) 2 + (6 + C) + (6A + 6C) giving u he yem A C 6 6A + 6C 4 ( ) The fir coefficien A i eaily found afer fir leing 6 in equaion ( ): ( ) A 6] ( 6] + C)( 6 + 6) 5 6] ] 4 A 5 6] ] 4 6] From hi yem ( ) we hen obain 5 A C So ( + 6) ( ) A C L ( + 6) ( )] L L + 6 2e 6 + 3L ] ] + 3L 2 2L + 6] 2 + 6] ] ] 4 4 L e co(4) 3 in(4) 255 i ( + ) ( ) ( + )( + 3)( + 7) A C + 7 A( + 3)( + 7) + ( + )( + 7) + C( + )( + 3) ( + )( + 3)( + 7)

5 Worked Soluion 99 So A( + 3)( + 7) + ( + )( + 7) + C( + )( + 3) ( ) Leing in equaion ( ): A( + 3)( + 7) + ( + 7) + C ( + 3) 6 ] ] + 92 A 6 ] ] + 92 ( + 3)( + 7) 3 Leing 3 in equaion ( ): A ( 3 + 7) + ( 3 + )( 3 + 7) + C( 3 + ) 6 3] ] ] ] + 92 ( 3 + )( 3 + 7) 5 Leing 7 in equaion ( ): A( 7 + 3) + ( 7 + ) + C( 7 + )( 7 + 3) 6 7] ] + 92 C 6 7] ] + 92 ( 7 + )( 7 + 3) 2 Thu ( + ) ( ) A C L 6 2 )] ( + ) ( L ] 3L + ] + 5L + 3 3e + 5e 3 2e 7 ] 2L + 7] 256 a Taking he ranform: L y 9y ] L] L y ] 9Ly]

6 2 The Invere Laplace Tranform 2 Y() y() y ] () 9Y() }{{}}{{} 4 9 ) ( 2 9 Y() 4 9 Y() Finding he parial fracion expanion for Y hen aking he invere ranform: ( 3)( + 3) A A( + 3) + ( 3) ( 3)( + 3) So A( + 3) + ( 3) Leing 3 3 repecively in hi la equaion: A(3 + 3) A A + ( 3 3) 4 3] ] Thu Y() A /2 3 + /2 + 3 y() L Y()] L 7/2 3 + /2 + 3] 7 ] 2 L + ] 3 2 L e3 + 2 e c Taking he ranform: L y + 8y + 7y ] L 65e 4] L y ] + 8L y ] + 7Ly] 65L e 4] 2 Y() y() y ] () }{{}}{{} 8 + 8Y() y() ] + 7Y() 65 }{{} 4 8 ( ) Y() Solving for Y : Y() 65 ( 4) ( ) (8 + 65)( 4) ( 4) ( ) ( 4)( + 7)( + )

7 Worked Soluion 2 Finding he parial fracion expanion of Y : Y() which require ha ( 4)( + 7)( + ) A C + A( + 7)( + ) + ( 4)( + ) + C( 4)( + 7) ( 4)( + 7)( + ) A( + 7)( + ) + ( 4)( + ) + C( 4)( + 7) ( ) Leing 4 in equaion ( ): A(4 + 7)(4 + ) + (4 + ) + C (4 + 7) A (4 + 7)(4 + ) Leing 7 in equaion ( ): 3 A ( 7 + ) + ( 7 4)( 7 + ) + C( 7 4) 8 7] ] ] ] 95 ( 7 4)( 7 + ) Leing in equaion ( ): A( + 7) + ( 4) + C( 4)( + 7) 8 ] ] 95 So C 8 ] ] 95 ( 4)( + 7) Y() A C y() L Y()] L ] 3L 4 ] + L 3e 4 + e 7 + 4e ( 7)] + 4L ( )] 257 a Uing he ranlaion ideniy we have ] L ( 7) 5 L F( 7)] e 7 f () ( )

8 22 The Invere Laplace Tranform where Replacing 7 wih X hi mean F( 7) ( 7) 5 Hence F(X) X 5 F() 5 f () L F()] L 5 ] 4! L 4! 4+ ] 24 4 Thu equaion ( ) become ] L ( 7) 5 e 7 f () e c Since he denominaor doe no facor we complee he quare of he denominaor: } 2 2 {{ } ( 3) ( 3) 2 So L ] L ( 3) ] L F( 3)] e 3 f () ( ) wih F( 3) ( 3) To ge F() fir le X 3 (equivalenly X + 3 ) in hi equaion: which mean Thu F(X) X + 3 X F() f () L F()] ] L ] L ] 6 6 L co(6) + 2 in(6) equaion ( ) coninue a L ] e 3 f () e 3 co(6) + 2 in(6) ]

9 Worked Soluion e Since he denominaor facor a ( + 4) 2 ( 4] ) 2 we can apply he ranlaion ideniy a follow: ] L 2 L ] ( ) 2 4] F( 4]) e 4 f () ( ) wih F( 4]) ( 4] ) 2 Leing X 4] hi become F(X) X 2 So F() 2 f () L F()] L 2 ] equaion ( ) become L ] e 4 f () e 4 e g Since we have ( + 6) L 2 L ] ( + 6) 2 + 4] F( + 6) F( 6]) e 6 f () ( ) wih Leing X + 6 hi become F( + 6) ( + 6) F(X) X Hence F() ] f () L F()] L ] 2 2 L in(2) equaion ( ) become L ] e 6 f () e 6 2 in(2) ] 2 in(2)e 6

10 24 The Invere Laplace Tranform 258 a L y 8y + 7y ] L] L y ] 8L y ] + 7Ly] 2 Y() y() y ] () }{{}}{{} 3 2 8Y() y() ] + 7Y() }{{} 3 ( ) Y() So wih Y() y() L 3 2 ( 4) 2 + ] F( 4) 3 2 ( 4) 2 + L F( 4)] e 4 f () ( ) 3 2 ( 4) 2 + Leing X 4 (equivalenly X + 4 ) we have F(X) 3X + 4] 2 X 2 + 3X X 2 + Thu F() f () L 3 2 3L + ] 2 + ] y() e 4 f () 3 co()e 4 3 co() 258 c L y + 6y + 3y ] L] L y ] + 6L y ] + 3Ly] 2 Y() y() y ] () }{{}}{{} Y() y() ] + 3Y() }{{} 2 ( ) Y() 2 2

11 Worked Soluion 25 So Y() ( + 3) ( ) 2 3] + 4 ] y() L ( ) 2 L 3] + 4 F( 3])] e 3 f () ( ) wih F( 3]) ( 3] ) Leing X 3] + 3 (equivalenly X 3 ) we have Thu F(X) F() 2X 3] ] f () L X ] ] 2L L X + 4 X co(2) + 7 in(2) y() e 3 f () e 3 2 co(2) + 7 in(2)] 259 a L y ] L e in() ] 2 Y() y() }{{} y () }{{} Le in() ] }{{} f () where 2 Y() F( ) So he la line conaining Y become F() L f ()] Lin()] 2 Y() F( }{{ } ) F(X) X 2 + X 2 + ( ) 2 + Solving for Y aring o find he correponding parial fracion expanion lead o Y() A 2 2( ( ) 2 + ) + + C + D ( ) 2 + A( ( ) 2 + ) + ( ( ) 2 + ) + (C + D) 2 2( ( ) 2 + )

12 26 The Invere Laplace Tranform which require ha A ( ( ) 2 + ) + ( ( ) 2 + ) + (C + D) 2 Muliplying hi ou gahering like erm hen yield 3 + A 2 + D] 2 + 2A + 2] + 2A which in urn require ha + C A 2 + D 2A + 2 2A From hi i follow ha A 2 A 2 D 2 A 2 C 2 hu he parial fracion expanion of Y i Y() A C + D ( ) 2 + ( ) + + ( ) ( 2 + ) ( ) 2 + Conequenly ( y() L ( ) 2 + )] ( L ] 2 2 +L ] ) L ( ) 2 + ] ) ( + L G( )] 2 ( ) + e g() 2 where Leing X hi become G( ) ( ) 2 + Hence G(X) X X 2 + g() L G()] L 2 + ] co() he la formula above for y become y() ( ) + e g() ( + e co() ) 2 2

13 Worked Soluion c L y 9y ] L 24e 3] L y ] L9y] 24L e 3] 2 Y() y() y ] 24 () 9Y() }{{}}{{} 3] 6 2 ) ( 2 9 Y() So Y() 24 ( + 3) ( 2 9 ) (6 + 2)( + 3) ( + 3) ( 2 9 ) ( + 3) 2 ( 3) Thi require ha A ( + 3) C 3 A( 3) + ( + 3)( 3) + C( + 3)2 ( + 3) 2 ( 3) A( 3) + ( + 3)( 3) + C( + 3) ( ) Muliplying hi ou gahering like erm yield + C] 2 + A + 6C] + 3A 9 + 9C] which in urn yield he yem + C 6 A + 6C 2 3A 9 + 9C 3 ( ) Leing 3 in ( ): A( 3 3) + ( 3 3) + C 2 6 3] ] + 3 A 6 3] ] Leing 3 in ( ): A + (3 + 3) + C(3 + 3) 2 63] ] + 3 C 23]2 + 23] + 3 (3 + 3) 2 4 Uing hee value wih he fir equaion in yem ( ): 6 C 6 4 2

14 28 The Invere Laplace Tranform Thu Y() A ( + 3) C 3 4 ( + 3) y() L 4 ( + 3) ] ] 4L ( + 3) 2 + 2L ] + 4L + 3 3] 4L F( 3]) ] + 2e 3 + 4e 3 4e 3 f () + 2e 3 + 4e 3 wih F( 3]) F( + 3) ( + 3) 2 F(X) X 2 f () L F()] L 2 ] y So y() 4e 3 f () + 2e 3 + 4e 3 4e 3 + 2e 3 + 4e 3 2e 3 4e 3 + 4e 3 25 a From heorem 245 on page 473 he definiion of he invere ranform we have L ] F() f (τ) dτ wih f (τ) L F()] τ In hi problem ( ) F() wih F() So )] L ( τ f (τ) L 2 + 9] L ] F() ] 3 L 3 τ f (τ) dτ 3 in(3τ) 3 in(3τ) dτ co(3)] 9 25 c In hi problem ( 3) 2 F() wih F() ( 3) 2 So ] τ f (τ) L ( 3) 2 LF( 3)] τ e 3 f () τe 3τ

15 Worked Soluion 29 ] L ( 3) 2 L ] F() f (τ) dτ τe 3τ dτ 3 τe3τ e 3τ dτ 3 3 e3 ] e (3 )e 3]

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