UNIVESIDAD FERMIN TORO

FACULTAD DE INGENIERIA

CABUDARE – EDO. LARA

INTEGRANTES:

NARIÑO, KAREN 21.759.611

NAVAS, SADDAM 21.756.852

TAMPOA, ROBERTH 25.149.524

RESPUESTAS:

1era Parte:

L {F(t )}=∫0

+∞

F (t )e− stdt senh (wt )= ewt−e−wt

2

¿∫0

+∞

( 53−√7+5 senh (√7 t ))e− st dt

¿∫0

+∞53t e

−st

−√7 e−st+5 senh (√7 t )e−st dt

¿ 53∫0

+∞

t e−st dt−√7∫0

+∞

e− st+5∫0

+∞e√7 t−e−√7 t

2. e−st dt

¿ 53∫0

+∞

t e−st dt−√7∫0

+∞

e− st dt+ 52∫0

+∞

e(√7−5) tdt−¿ 52∫0

+∞

e−(√7−5) tdt ¿

¿ 53lima→+∞

∫0

a

t e−st dt−√7 lima→+∞

∫0

a

e−st dt+ 52lima→+∞

∫0

a

e (√7−5) tdt−¿ 52lima→+∞

∫0

a

e−(√7−5 )t dt ¿

¿ 53

lima→+∞

t e−st

−s+ e

−st

−s2|a0−√7lima→+∞

e− st

s |a0+ 52lima→+∞

e(√7−5) t

(√7−5 ) |a0−52lima→+∞

e−(√7+5 ) t

−(√7+5 ) |a0

¿ 53lima→+∞ ( ae−sa−s

+ e− sa

– s2−( ae

−s0

−s+ e

− s0

– s2 ))−√7 lima→+∞ ( e

− sa

s− e

−s0

s )+ 52 lima→+∞ ( e(√7−5)a

(√7−5 )− e (√7−5 )0

(√7−5 ) )−52 lima→+∞ ( e−(√7+5 )a

−(√7+5 )− e−(√7+5 )0

−(√7+5 ) )

¿ 53

lima→+∞

a e−sa

−s+ e

−sa

– s2− 1– s2

−√7 lima→+∞ ( e

−sa

s−1s )+52 lima→+∞ ( e(√7−5 )a

(√7−5 )− 1

(√7−5 ) )−52 lima→+∞ ( e−(√7+5)a

−(√7+5 )− 1

−(√7+5 ) )

¿ 53

lima→+∞

a

−sesa+ 1– s2 esa

− 1s2

−√7 lima→+∞ (−1s )+ 52 lima→+∞ ( −1

(√7−5 ) )−52 lima→+∞ ( −1(√7+5 ) )

¿53 (−1s2 )−¿

¿ 53 s2

−√7s

− 52 (√7−5 )

+ 52 (√7+5 )

ASI TENEMOS QUE:

L {53 t−√7+5 senh (√7 t )}= 5

3 s2−√7s

− 5

2 (√7−5 )+ 5

2 (√7+5 )

2da parte

a) F (t )=72e5 t (23 cos (2√5t )+2 senh (2√3 t )−4 t

¿¿)

F ( t )=73e5 t cos (2√5 t )+7 e5 t senh (2√3 t )−14 e5 t t

L {F(t )}=L {73 e5 t cos (2√5 t )+7e5 t senh (2√3 t )−14 e5 t t }¿ 73L {e5 t cos (2√5t ) }+7 L {e5 t senh (2√3 t ) }−14 L {e5 t t }

¿ 73

s−5(s−5)2+(2√5)2

+7 2√3(s−5)2−(2√3)2

−14 1(s−5)2

7 ( s−5 )

3((s−5)¿¿2+20)+ 14√3( s−5 )2−12

− 14(s−5 )2

¿

Asi tenemos que

L {F(t )}= 7 ( s−5 )3(s−5)2+60

+ 14√3(s−5 )2−12

− 14

( s−5 )2

b) F (t )=115t (7 senh (2 t )−5 cos (3 t)

t2 )¿ 775t senh (2 t )−11 cos (3 t)

t 2

No se puede calcular la transformada de Laplace de F(t) porque

∫0

+∞cos (3 t)t

e−st dt es una integral impropia mixta que es diferente.

c) F ( t )=74cos7 t−2e− st+ 3

5t3

¿−494sen7 t+6e−3 st+¿ 9

5t 2¿

¿ 3434cos7 t−18e−3 t+ 18

5t

F ' ' ' ( t )=24014sen7 t+54e−3 t+ 18

5

L {F' ' ' (t ) }=L {24014 sen7 t+54e−3 t+ 185 }

¿ 24014L {sen7 t }+54 L {e−3 t }+L{185 }

¿ 24014

7

s272+54 1

s−(−3)+ 185 s

¿ 16807

4 s2+196+ 54s+3

+ 185 s

Asi tenemos que

L {F' ' ' (t ) }= 16807

4 s2+196+ 54s+3

+ 185 s

3era parte:

A b c d

A) {11(s−34 )−√5

3(s−54 )2

−8+

11 (s−5 )+√79 (s2−10 s+125 )3

− 7 s−48 s2−18

+ 4 √5s2+3 }

a)54−34=12L−1( 11(s−

34 )−√5

3(s−54 )2

−8 )=L−1( 11(s−34 )−√5

3((s−54 )2

−83 ) )

¿ L−1( 11(s−54+ 54−34 )−√5

3((s−54 )2

−83 ) )

¿ L−1( 11(s−54 )+112 −√5

3 ((s−54 )2

−83 ) )

¿ L−1( 11(s−54 )3((s−54 )

2

−83 )

+

( 112 −√5)3

∗1

(s−54 )2

−83

)¿ L−1( 11

3∗s−5

4

(s−54 )2

−(√ 83 )2+

(112 −√5)3√ 83

∗√ 83(s−54 )

2

−83

)

¿ 113e54t

cosh (√ 83 t)+((112

−√5)3√ 83 )e 54 t cosh (√ 83 t )

b) ¿ L−1( 11 (s−5 )+√7

9 ( s−5 )6 )=119

∗(s−5 )

(s−5 )6+

√79

∗1

(s−5 )6=¿

¿ L−1( 1124∗9

∗24

( s−5 )5+

√7120 (9)

∗120

(s−5 )6 )¿ 11216

∗t 4 e5t+ √71080

t 4 e5 t

c ¿=L¿−1( 7 s−48 s2−18 )

¿ L−1 ¿

¿ L−1(78∗s−4

7

s2−188

)¿ L−1(

78∗s−4

7

s2−94

)¿ L−1(

78∗s

s2−94

−

47

s2−94

)

¿ L−1( 78∗ss2−94

−

4732

∗32

s2−94

)¿78L

−1( s

s2−94 )− 8

21L

−1(32

s2−94

)¿ 78cosh

32t− 821senh

32t

d) ¿4 √5s2+3

¿ L−1( 4√5s2+3 )¿ L−1(4 √5∗1

s2+3 )

¿ L−1( 4√5√3∗√3

s2+3 )¿ 4√5

√3sen√3 t

Asi Tenemos que:

L−1 {11(s−34 )−√5

3(s−54 )2

−8+11 ( s−5 )+√79 ( s2−10 s+125 )3

− 7 s−48 s2−18

+ 4√5s2+3 }=¿

¿ 113e54t

cosh (√ 83 t)+((112

−√5)3√ 83 )e 54 t cosh (√ 83 t )+ 11

216∗t 4 e5 t+ √7

1080t 4 e5 t−7

8cosh

32t− 821senh

32t+ 4 √5

√3sen √3 t

a b

B) L−1 { 4 s+7

s2+53s+ 174

− 6 s−4

s2−13s+9 }

L−1 { 4 s+7

(s+ 56)2

+ 329

− 6 s−4

(s+ 16)2

+ 263 }

a) L−1 { 4 s+7

(s+ 56)2

+ 329 }=¿

L−1 { 4 (s+74)

(s+ 56)2

+ 329

}L−1 {4 (s+ 56−56 +

74)

(s+ 56)2

+ 329

}

L−1 {4 (s+ 56 )−113(s+ 56)2

+ 329

}=L−1 { 4(s+

56)

(s+ 56)2

+ 329

−

113

(s+ 56)2

+ 329

}=

4 L−1{ s+56

(s+ 56)2

+ 329

−

113

(s+ 56)2

+ 329

}=4 L−1{( s+

56

(s+ 56 )2

+ 329

)− 3312√2 (

4√23

(s+ 56 )2

+ 329

)}=4 e

−56tcos( 4√23 )t− 33

12√2e56tsen( 4√23 ) t

b)L−1 { 6 s−4

(s−16 )2

+ 263 }=¿

L−1 { 6(s−23)

(s−16 )2

+ 263

}=¿

=L−1 {6(s−16+ 16−

23)

(s−16 )2

+ 263

}=¿

L−1 {6 (s−16 )−12¿¿¿(s−16 )

2

+ 263 }=¿

L−1 { 6(s−16)

(s−16 )2

+ 263

−

12

(s−16 )2

+ 263

}=¿

6 L−1 {( (s−16 )(s−16 )

2

+263

)+ 12 L−1(√26√3

(s−16 )2

+ 263

)}¿6e

16tcos (√26√3

t)−12 e16tse n(√26√3

t )

Asi tenemos que:

L−1 { 4 s+7

s2+53s+ 174

− 6 s−4

s2−13s+9 }=¿

4 e−56tcos( 4√23 )t− 33

12√2e56tsen( 4√23 ) t−6e

16tcos (√26√3

t )−12 e16tsen (√26√3

t)

C) L−1 ¿

L−1 { (s+1)2+7(s+1)2+1(s+1)2+4 }=¿

s2+2 s+8(s¿¿2+2 s+2)(s2+2 s+5)=¿¿

As+B

(s¿¿2+2 s+2)+Cs+D

(s2+2 s+5)=¿¿

s2+2 s+8= (As+B ) ( s2+2 s+5 )+(Cs+D )(s¿¿2+2 s+2)=¿¿

s2+2 s+8= (A s3+2 s2+5 As+Bs2+2Bs+5 B+C s3+2C s2+2Cs+Ds2+2Ds+2D )

s2+2 s+8= (A+C ) s3+(2 A+B+2C+D ) s3+(2 A+B+2C+D ) s2+(5 A+2B+2C+2D ) s+(5B+2D)

{0=A+C{1=2 A+B+2C+D{2=5 A+2B+2C+2D¿

0=A+CA=−C

1=−2C+B+2C+D

1= (B+D )∗2 ¿2B+2D

−B=D

2=5 A+2B+2C+2D2=−5 A+2B+2C+2D2=−3C+2B+2D2=−3C+2→0=C

1=B+B 1B

=D

8=5B+2D

−2=−2B−2D

8=5B+2D

6=3B

63=B→2=B

1=B+D

1=2+D

−2+1=D

−1=D

L−1 { 2

s2+2 s+2+ −1s2+2 s+5 }

L−1 { 2

(s+1)2+1+

−1(s+1)2+4 }=¿

L−1 {( 2(s+1)2+1 )−L−1( 1

(s+1)2+4 )}2 L−1( 1

(s+1)2+1 )− L−1

2 ( 2(s+1)2+4 )

¿2e−t se n(t )−12e−t sin(2 t)