willian arroyo
TRANSCRIPT
7/25/2019 Willian Arroyo
http://slidepdf.com/reader/full/willian-arroyo 1/7
Willian Arroyo
CI: 26.461.435
1. UTILIZAR LA DEFINICION DE TRANSFORMADA DE LAPLACE Y RESOLVER LA
SIGUIENTE FUNCION
( ) t t F 2cosh5
8
3
5+=
Solución:
L {t }=∫0
+∞5
3 e
−st dt =
5
3 limh−∞∫0
h
e−st
dt
e−st
dt =¿−1
s e
−st
∫ ¿
¿5
3 limh→∞
∫0
h
e−st
dt =5
3 limh→∞ (
−1
s e
−sh+1
s )= 5
3 s
L{53}= 5
3s
L {cosh (2 t ) }=∫0
∞
cosh (2 t ) e−st
dt
Luego, haciendo el prier ca!io de "aria!le por par#e$ $e #iene
u=cosh (2 t ) , du=2sinh (2 t )dt ,dv=e−st
d t , v=−1
s e
−st
7/25/2019 Willian Arroyo
http://slidepdf.com/reader/full/willian-arroyo 2/7
∫cosh (2t ) e−st
dt =−1
s e
−st cosh (2t )+2
s∫ sinh (2 t ) e
−st dt
%l $egundo ca!io de "aria!le por par#e$ e$
u=sinh (2 t ) , du=2cosh (2 t )dt ,dv=e−st
d t , v=−1
s e−st
%n#once$,
∫cosh (2t ) e−st
dt =−1
s e
−st cosh (2t )+2
s (−1
s e
−st sinh (2 t )+ 2
s∫cosh (2 t ) e
−st dt )
∫cosh (2t ) e−st dt =
−1
s e
−st cosh (2t )− 2
s2e−st
sinh (2t )+ 4
s2∫ cosh (2 t )e−st
dt
∫cosh (2t ) e−st
dt −4
s2∫cosh (2 t ) e
−st dt =
−1
s e
−st cosh (2t )− 2
s2 e−st
sinh (2 t )
s2−4
s2 ∫cosh (2t ) e
−st dt =
−1
s e
−st cosh (2 t )− 2
s2 e−st
sinh (2 t )
&or #an#o
∫cosh (2t ) e−st
dt = s
2
s2−4 (−1
s e
−st cosh (2 t )− 2
s2 e−st
sinh (2 t ))%n con$ecuencia,
∫0
∞
cosh (2t ) e−st dt ¿ lim
h→∞ [ s2
s2−4 (−1
s e
−st cosh (2t )− 2
s2e−st
sinh (2t ))]h0
¿ s
2
s2−4
f ( s)= 5
3 s+8
5
s2
s2−4
7/25/2019 Willian Arroyo
http://slidepdf.com/reader/full/willian-arroyo 3/7
2. UTILIZAR PROPIEDADES Y TABLA PARA DETERMINAR LA TRANSFORMADA
DE LAPLACE. ENUNCIE LAS PROPIEDADES ANTES DE RESOLVER.
SIMPLIFIQUE LOS RESULTADOS.
( ) ( ){ } ( ) 42"
4
343cos
5
8t et t F sit F t F t
+−== −
L
Solución:
F ' ( t )=−24
5 sin (3t )+8e−2 t +3 t 3
F ' ' ( t )=−72
5 cos (3 t )−16 e
−2 t +9t 2
'#ili(areo$ la$ $iguien#e$ )orula$
L {cos(kt )}= s
s2
+k 2
, L {eat }= 1
s−a
, L {t n }= n !
sn+1
Ahora !ien,
F ( s )= L {−72
5 cos (3 t )−16e
−2 t +9 t 2}
¿−72
5 L {cos (3 t ) }−16 L {e−2t }+9 L {t
2 }
¿−72
5
s
s2+9−16
1
s+2+92 !
s3
7/25/2019 Willian Arroyo
http://slidepdf.com/reader/full/willian-arroyo 4/7
s
5(¿¿ 2+9)− 16
s+2+18
s3
F (s )=−72 s¿
3. Aplicar Tabla !i"pli#icaci$% & "'()*) c)rr+!p)%*i+%(+ para *+(+r"i%ar
( ){ } ( )t F s f L =−1
( )( )
+−++
−
+
+
+−
+−−−−
2
735
62
3
5
2
33
1688
742
2262
1
s s
s
s
s
s s
s L
Solución:
√ 2 ( s−4 )−7
8( s2−8s+16)3
=√ 2(s−4)
8 (s−4)6 −
7
8 (s−4 )6=√ 2
8
1
( s−4 )5−7
8
1
(s−4)6
L−1{ √ 2 (s−4 )−7
8 ( s2−8 s+16 )3 }=√ 2
8 L
−1 { 1
(s−4 )5 }+ 7
8 { 1
(s−4)6 }=√ 2
8e4 t t
4
4 !+7
8e4 t t
5
5 !( A)
* por o#ra par#e,
5s+3
s2−s+
7
2
= 5 s+3
(s−1
2 )2
+13
4
= 5 s
(s−1
2 )2
+13
4
+ 3
(s−1
2 )2
+13
4
¿5 s−
5
2+5
2
(s−1
2 )2
+13
4
+ 3
(s−1
2 )2
+13
4
=(5 s−
5
2 )(s−
1
2 )2
+13
4
+
5
2
(s−1
2 )2
+13
4
+ 3
(s−1
2 )2
+13
4
7/25/2019 Willian Arroyo
http://slidepdf.com/reader/full/willian-arroyo 5/7
¿5(s−1
2 )
(s−
1
2
)
2
+13
4
+
11
2
(s−
1
2
)
2
+13
4
=5
(s−1
2 )
(s−
1
2
)
2
+13
4
+11
2
1
(s−
1
2
)
2
+13
4
%n#once$
L−1
{ 5s+3
s2−s+
7
2 }= L
−1
{5
(s−1
2 )
(s−
1
2
)
2
+13
4
+11
2
1
(s−
1
2
)
2
+13
4
}¿5 L
−1 { (s−1
2 )(s−
1
2 )2
+13
4 }+ 11
2 L
−1
{ 1
(s−1
2 )2
+13
4 }
¿5
e
1
2t
cos
(√ 13
2 t
)+11
2
2
√ 13 e
1
2t
sin
(√ 13
2 t
)=5
e
1
2t
cos
(√ 13
2 t
)+ 11
√ 13 e
1
2t
sin
(√ 13
2 t
)(B)
√ 3(s+32 )+5
(s+ 3
2 )2
−6
=√ 3(s+ 3
2 )(s+ 3
2 )2
−6
+ 5
(s+ 3
2 )2
−6
L−1
{ √ 3(
s+3
2 )(s+
3
2 )2
−6 }=√ 3cosh ( √ 6 t )
7/25/2019 Willian Arroyo
http://slidepdf.com/reader/full/willian-arroyo 6/7
L−1
{ 5
(s+3
2 )2
−6 }= 5
√ 6sinh (√ 6 t )(C )
%n con$ecuencia co!inando +A , +- * +C #eneo$ lo $iguien#e
L−1{ √ 2 (s−4 )−7
8 ( s2−8s+16 )3+
√ 3(s+32 )+5
(s+ 3
2 )2
−6
− 5 s+3
s2−s+
7
2 }= L−1{ √ 2 ( s−4 )−7
8( s2−8 s+16)3 }− L−1{ √ 3(s+3
2 )(s+ 3
2 )2
−6 }− L−1
{ 5
(s+32
¿ √ 2
8 e
4 t t 4
4 !+7
8 e
4 t t 5
5!−√ 3cosh (√ 6 t )− 5
√ 6sinh (√ 6 t )+5e
1
2t
cos(√ 132 t )∓ 11
√ 13e
1
2t
sin (√ 132 t )
4. U(ili,ar +l (+)r+"a *+ C)%-)lci$% & *+(+r"i%+
( )
+−
16
52
231
s s
s L
Solución:
L−1{ 2√ 5
s3 ( s
2+16) }=2√ 5 L−1{1s3
1
s2+16}
¿2√ 5 L−1{1s3}∗ L
−1{ 1
s2+16}
¿2√ 5[ t 2∗14
sin (4 t ) ]
¿ √ 5
2 ∫
0
t
sin (4 τ )( t −τ )2dτ
7/25/2019 Willian Arroyo
http://slidepdf.com/reader/full/willian-arroyo 7/7
¿ √ 5
2 ∫
0
t
sin (4 τ ) (t 2−2 tτ +τ
2 ) dτ
¿ √ 5
2 ∫
0
t
[t 2sin (4 τ )−2 tτ sin (4 τ )+τ
2sin (4 τ ) ]dτ
¿ √ 5
2 [ t 2 (−cos (4 τ ) )4
−2t ( 14 sin (4 τ )− τ
4( cos (4 τ ) ))+(−τ
2
4(cos ( 4 τ ) )+ 2
4 ( 14 (cos ( 4 τ ) )+ τ
4sin ( 4 τ )))]
0
t
¿−√ 5
8 t
2cos (4 t )−√ 5
4 t sin ( 4 t )+√ 5
4 t
2cos (4 t )−√ 5
8 t
2cos ( 4 t )+√ 5
16 cos (4 t )+ √ 5
32 t sin (4 t )
(4 t )−√ 5
8 t
2cos (4 t )+√ 5
16 cos (4 t )=¿ √
5
16 cos ( 4 t )− 7
32t sin (4 t )
¿√ 5
8 t
2cos (4 t )−
7
32t sin¿
&or #an#o,
L−1{ 2√ 5
s3 ( s2+16) }=√ 5
16 cos ( 4 t )− 7
32 t sin (4 t )
5. Solución:
Coo la )unción sin (n w0
t ) e$ una )unción ipar para #odo n y la
)unción cos(nw0 t ) e$ una )unción par para #odo n, e$ de e$perar ue
a./ Si f (t ) e$ par, $u $erie de 0ourier no con#endr #rino$ $eno y
por lo #an#o bn=0 para #odo n.
!./ Sif (t )
e$ ipar, $u $erie de 0ourier no con#endr #rino$ co$eno
y por lo #an#o an=0 para #odo n.