Utilizar La Definición De Transformada De Laplace Y Resolver La Siguiente Función:

F (t )=53t 2−2√7+5cos 2√84 t

Por definición

L {F ( t ) }=∫0

∞

e− stF (t )dT

L {F ( t ) }=∫0

∞

e− st(53 t 2−2√7+5cos 2√84 t)dT

L {F (t ) }=∫0

∞

[e−st( 53 t 2)−e−st ( 2√7 )+e−st (5cos 2√84 t )]dTHaciendo las integrales cada una por separado tenemos:

⟹ 53∫0

∞

e−st (t 2 )dT=53 [−t2 e−stS

+ 2S∫0

∞

e−st (t )dT ]53 [−t 2e−stS

+ 2S (−t e−stS

+ 1S∫0

∞

e−st dT )]53 [−t 2e−stS

+ 2S (−t e−stS

−1S ( 1S e−st))]

53 [(−t2 e−stS

−2t e−st

S2−2e

−st

S3 )0

∞]=53 ( 2S3 )=103 ( 1S3 )

⟹−∫0

∞

e−st ( 2√7 )dT=− 2√7 [∫0

∞

e− st dT ]=−2√7[ e−stS ]0

∞

=−2√7S

⟹5[∫0

∞

e−st (cos 2√84 t )dT ]=5{[e−st( 2√84sin 2√84 t−Scos 2√84 t(−S )2+( 2√84 )2 )]0

∞

}= 5 SS2+84

∴F ( t )=53t2− 2√7+5cos 2√84 t=10

3 ( 1S3 )−2√7S

+ 5S

S2+84

Utilizar Propiedades Y Tabla Para Determinar La Transformada De Laplace, Enuncie Las Propiedades Antes De Resolver, Simplifique Los Resultados.

F ( t )=72e84 t ( 23 cos2 2√5t+2cosh 2 2√3 t−4 t 7)

Por linealidad

73L {e84 t (cos 2 2√5 t )}+7 L {e84 t (cosh 2 2√3 t )}−14 L {e84 t (t 7 )}

Haciendo las transformadas cada una por separada tenemos:

⟹ 73L {e84 t (cos2 2√5 t ) }

Por primer teorema de traslación

73L {e84 t (cos 2 2√5 t )}=7

3L {(cos2 2√5 t ) }S→S−84=73 ( S

S2+(2 2√5 )2 )S→S−8473 ( S−84

(S−84 )2+20 )= 7 S−5883S2−504 S+21220

⟹7 L {e84 t (cosh 2 2√3 t ) }

Por primer teorema de traslación

7 L {e84 t (cosh 2 2√3 t )}=7 L {(cosh 2 2√3 t )}S→S−84=7 ( S

S−(2 2√3 )2 )S →S−847( S−84S−84−12 )=7 S−588S−96

⟹−14 L {e84t ( t7 ) }

Por primer teorema de traslación

−14 L {e84 t (t 7 )}=−14 L {(t 7 )}S→S−84=−14 ( 7 !S8 )S→S−84

−14 ( 5040

(S−84 )8 )= 70560

(S−84 )8

∴F ( t )=72e84 t( 23 cos 2 2√5 t+2cosh 2 2√3 t−4 t7)=¿

¿ 7 S−5883S2−504 S+21220

+7 S−588S−96

+ 7 0560(S−84 )8

F ( t )=35t (84 sinh 2t−5 sin 3tt2 )

Por linealidad

2525L {t (sinh 2 t ) }−3 L{sin3 tt }

Haciendo las transformadas cada una por separada tenemos:

⟹ 2525L {t (sinh 2t ) }

Por multiplicación por t

2525L {t (sinh 2 t ) }=252

5[ (−1 ) (L {sinh 2 t } )´ ]=252

5 [−( 2

S2−(2 )2 )´ ]

¿ 2525 {−[−2S (2 )

(S2−4 )2 ]}=2525 [ 4SS4−8S2+16 ]= 1008S

5S4−40S2+80

⟹−3L {sin 3 tt }

Por división por t

−3 L{sin 3 tt }=−3(3∫S

∞1

u2+32du)=−3[( 32 tan−1 u3 )

S

∞]=−3( 3π4 −32tan−1 S

3 )

¿ −9π4

+ 92tan−1 S

3

∴F (t )=35t(84sinh 2 t−5 sin 3 tt 2 )= 1008S

5 S4−40S2+80−9π4

+92tan

−1 S3

F (t )=L {F (t )´ ´ };F ( t )=34cos84 t−2e−3 t+ 3

5t 5

F ( t )´=−34

(84 ) sin 84 t+2 (3 )e−3 t+3 t 4

F ( t )´ ´=−34

(84 ) (84 ) cos84 t−18e−3 t+12t 3

Por Derivación L {F (t )´ ´}=S2F ( t )−SF (0 )−F (0 )´

L {−34 (84 )2L {cos 84 t }−18 L {e−3 t }+12L {t 3 }}=S2F ( t )−SF (0 )−F (0 )´

−34

(84 )2( S

S2+ (84 )2 )−18( 1S+3 )+12( 6S4 )=S2 F ( t )−SF (0 )−F (0 )´

Haciendo las transformadas cada una por separada tenemos:

⟹−5292( S

S2+(84 )2 )=S2F (t )−SF (0 )−F (0 )´

−5292( S

S2+(84 )2 )=S2 F (t )−S ( 34 )

F ( t )= 1S2 [−5292( S

S2+(84 )2+ 34S)]

F ( t )= 1S2 [(−211685S+3S3+3 (84 )2S

4 (S2+(84 )2) )]= 1S2 ( 3S3

4 (S2+(84 )2) )=¿

F ( t )=34 ( S

S2+(84 )2 )

⟹−18( 1S+3 )=S2F ( t )−SF (0 )−F (0 )´

−18( 1S+3 )=S2F (t )−S (2 )−6

F ( t )= 1S2 (−18S+3

+6−2S)=−18+6S+18S+3

−2S=6S−2 S2−6S

S+3= −2S+3

⟹12( 6S4 )=S2F ( t )−SF (0 )−F (0 )´

12( 6S4 )=S2F ( t )

F ( t )=72S6

∴F (t )= 34 ( S

S2+ (84 )2 )− 2S+3

+72

S6

Utilizar El Teorema De Convolución Y Determine:

L−1 { 2 2√84S3 (S2+2 ) }

L−1 { 22√84

S3 (S2+2 ) }=2 2√84 L−1{ 1

S3 (S2+2 ) }

F ( s )=2√2S2+2

⟹F (t )=F (τ )= 12√2sin 2√2 τ

G (s )= 1

S3⟹G (t )=G ( t−τ )=1

2( t−τ )2

∫0

t

F (τ )G ( t−τ )dτ

1

22√2

∫0

t

{sin 2√2 τ [ (t−τ )2 ] }dτ

1

22√2

∫0

t

{t2 sin 2√2 τ−2 τt sin 2√2 τ+τ2 sin 2√2 τ }dτ

Haciendo las integrales cada una por separado tenemos:

⟹∫0

t

(t 2sin 2√2 τ )dτ=−t 22√2

(cos 2√2 τ )0t=−t2

2√2cos

2√2t+ t2

2√2

⟹−2t∫0

t

( τ sin 2√2 τ )dτ=( 2t2√2 τ cos 2√2 τ−t sin 2√2 τ )0t

=2 t2

2√2cos

2√2 t−t sin 2√2 t

⟹∫0

t

(τ2 sin 2√2 τ )dτ=(−2 τ2

2√2cos

2√2 τ+τ sin 2√2 τ+ 12√2cos

2√2 τ)0

t

=¿

¿−2 t2

2√2cos

2√2 t+t sin 2√2 t+ 12√2cos

2√2 t− 12√2

∴L−1 { 22√84

S3 (S2+2 ) }= t2

2√2+ 12√2

cos2√2t− 1

2√2

Aplicar tabla, simplificación y método correspondiente para determinar:

L−1 {f ( s) }=F ( t )

L−1 { 7(s−34 )−2√5

3(s−34 )2

−84+

5 ( s−5 )+ 2√79 ( s2−10 S+25 )3

− 7 s−48 s2−18

+ 42√5

s2+ 47

}L−1 { 7(s−34 )−2√5

3(s−34 )2

−84 }+L−1 {5 (s−5 )+ 2√79 (s−5 )3 }−L−1 { 7 s−48(s2−94 )}+L−1 { 4

2√5s2+ 4

7}

Haciendo las transformadas cada una por separada tenemos:

⟹ L−1{ 7 (s− 34 )−2√5

3(s−34 )2

−84 }=e 34 t (L−1{ 7 s−2√53 s2−84 })

¿e34t(L−1{ 7 s

3 s2−84 }−L−1 { 2√53 s2−84 })

¿e34t( 73 L

−1

{ Ss2−28 }−

2√53L

−1

{ 1s2−28 })

¿e34t( 73 cosh 2√28 t−

2√53 2√28

sinh2√28 t)

⟹ L−1{5 (s−5 )+ 2√79 (s−5 )3 }=e5 t(L−1 {5 s+ 2√7

9 s3 })¿e5 t(L−1 { 5 s9 s3 }+L−1{ 2√7

9 s3 })¿e5 t( 59 t+2√718t 2)

⟹−L−1{ 7 s−48(s2−94 ) }=−[ 78 L−1 { s

s2−94 }−48 L

−1

{ 1

s2−94 }]

¿−78cosh

32t+ 412sinh

32t

⟹ L−1{ 42√5

s2+ 47

}=4 2√5(L−1 { 1

s2+ 47 })=4

2√522√7

sin22√7t

∴L−1 { 7(s−34 )− 2√5

3(s−34 )2

−84+

5 (s−5 )+ 2√79 ( s2−10S+25 )3

− 7 s−48 s2−18

+ 42√5

s2+ 47

}=¿

¿e34t( 73 co sh 2√28t−

2√53 2√28

sinh2√28 t)+e5 t( 59 t+

2√718t 2)−78 cosh 32 t+ 412 sinh 32 t+ 4

2√522√7

sin22√7t

L−1 { s2+2 s+84( s2+2 s+2 ) (s2+2 s+5 ) }

s2+2 s+84(s2+2 s+2 ) ( s2+2 s+5 )

= As+Bs2+2 s+2

+ Cs+Ds2+2 s+5

A=0 ,B=823,C=0 ,D=−79

3

L−1 { 823

s2+2 s+2−

793

s2+2 s+5 }=823 L−1{ 1

(s+1 )2+1 }−793 L−1 { 1

( s+1 )2+4 }823e−t L−1 { 1

s2+1 }−793 e−t L−1{ 1

s2+22 }=823 e−t sin t−796 e−t sin 2 t

∴L−1 { s2+2 s+84( s2+2 s+2 ) ( s2+2 s+5 ) }=823 e−t sin t−796 e−t sin 2 t