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
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