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Métodos Matemáticos /2-ene#o-/
Ejercicios del cuadreno
SERIE COMPLEJA DE FOURIER
la seriede fourier complejade la funcion f , deinida en elintevalo abierto (0, L )
0
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Métodos Matemáticos /2-ene#o-/
Cn=1
2 [−2e−2 inπ
inπ +
e−2 inπ
n2
π 2 −
1
n2
π 2 ]
Cn=1
2 [−2inπ ( cos2nπ −isen2nπ )+ 1n2 π 2 (cos2nπ −isen 2nπ )− 1n2 π 2 ]
Cn=[−2inπ + 1n2 π 2− 1n2 π 2 ]
Cn=1
2 [−2inπ ]=−1inπ
Cn=(−1nπ ) (−i )= inπ si n=0
F ( x )=∑n=−∞
∞i
nπ e
inπxdonden≠0
Co=1
2∫ f ( x ) e
−2 i (0) π 2
x
dxCo=1
2∫0
2
x e0
dxCo=1
2∫0
2
xdx
f ( x )=Co+∑n=−∞
∞i
nπ e
inπx∴
f ( x )=1+∑n=−∞
∞i
nπ e
inπx
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Métodos Matemáticos /2-ene#o-/
2¿ f ( x )=1− x ; 0
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Métodos Matemáticos /2-ene#o-/
f ( x )= ∑n=−∞
∞−3 i
nπ e
inπ
3 x
donde n≠ o
f ( x )=Co+
∑n=−∞∞
−3 i
nπ e
inπ
3 x
∴
f ( x )=−2+∑n=−∞
∞−3inπ
einπ
3 x
definicion de serie compleja defourier : la seriecompleja de fourirer dela
funcion f ( x ) definidade un intervalo de− p
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Métodos Matemáticos /2-ene#o-/
TRANSFORMADA DE LAPLACE
∞∞eslafuncionF ( s )definida por lainte"ral [ o , ∞ ) dela funcion
F ( s)=∫0
∞
f (t )e−st dt
1¿ f ( t )=1
∫0
b
e−st
dt =¿
∫0
b
1 ∙ e−st
dt =¿ limb # ∞
¿
L {f (t )}= L {1 }=limb # ∞
¿
limb # ∞ [−1s e−st ]b0= limb # ∞ [−1s e−sb−(−1s e0)]
¿−1
s limb # ∞
1
esb+1
s∴
L {1}=1s
f ( t )=$
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Métodos Matemáticos /2-ene#o-/
L {$ }= $ s
2¿ L {eat }=limb #∞
∫0
b
eat
∙ e−st
dt
¿ limb #∞
∫0
b
eat −st
dt =limb # ∞ [ −1(s−a ) e−(s−a ) t ]bo
limb # ∞ [ −1s−a e−(s−a ) b−( −1s−a e−( s−a )0)]
¿− 1
s−a limb # ∞
1
e(s−a ) b
+ 1
s−a∴
L {eat }= 1
s−a
3¿ L {teat
}=limb # ∞∫0
b
t eat
∙ e−st
dt
¿ limb #∞
∫0
b
t e−(s−a ) t dt
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Métodos Matemáticos /2-ene#o-/
¿ limb #∞ [ −t s−a e−(s−a )t + 1s−a ( −1s−a e−(s−a)t )]b0
¿ limb #∞ [ −t s−a e−(s−a )t − 1(s−a)2 e−(s−a )t ]b0
¿ limb #∞ [ −bs−a e−(s−a ) b− 1(s−a )2 e−( s−a )b−( −0s−a e−( s−a)0− 1(s−a)2 e−(s−a)0)]
¿0+0+ 1
(s−a)2∴
L {teat }= 1(s−a)2
4 ¿ L {sen $t }=limb # ∞
∫0
b
sen$t e−st
u=sen$t dv=e−st dt
du=$cos$t v=−1
s
¿−1s
e−st
sen$t + $ s∫cos$t ∙e
−st dt
u=cos$t dv=e−st dt
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Métodos Matemáticos /2-ene#o-/
du=−$sen$t v=−1
s e
−st
¿−1
s e
−st sen$t +
$
s [−1s e−st cos$t −∫(−1s e−st ) (−$sen$t )]
∫0
∞
sen$te−st
dt =−1
s e
−st sen$t −
$
s2
e−st
cos$t −$
2
s2∫ sen$t e
−st dt
¿ limb #∞ [−1s e−st sen$t − $ s2 e−st cos$t ]
¿ $
s2+$ 2
%e desea encontrar la transformada de La &lacede la si"uiente funcion
1¿ L {15−3e−7 t +4 sen3 t }
L {15−3e−7 t +4 sen3 t }= L {15 }−3 L {e−7 t }+4 L {sen3 t }
¿15 ∙ 1
s−3
1
s− (−7 )+4
3
s2+9
¿15
s −
3
s+7+ 12
s2+9
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Métodos Matemáticos /2-ene#o-/
2 f ( t )=cos5 t −t 5+t 3 e2 t
t 5
L {f (t )}= L {cos5 t }− L {¿ }+ L {t 3 e2 t }
¿ 5
s2+52
−5 '
s6+
3 '
(s−2)2
$#ans8o#mada de na &e#i"ada
S9 L {f (t ) = F ( s ) entonces
/ L {f ( ( t ) }=% L { f ( t ) }−f (0 )=%F (s )−f (0)
2 L {f ( ( (t ) }=%2 L { f (t ) }−%f (0 )−f ( (s )=%2 F (s )−sf (0 )−f ( (0)
3 L {f ( ( ( (t ) }=%3 L { f (t ) }−%2 f (0 )−sf ( (0 )−f ( ( (0)
L {f )* (t ) }=%4 L { f (t ) }−%3 f (0 )−%2 f ( (0 )−%f ( ( (0 )−f ( ( ( (0)
Reso"e# as si;ientes ecaciones di8e#enciaes
1) + ( +4 +=e−4 t
; + (0 )=2
d+dt +4 +=e−4 t
L { + ( +4 + }= L {e−4 t }
L { + }+4 L { + (t ) }= L {e−4 t }
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Métodos Matemáticos /2-ene#o-/
( % ( s)− + (0 ) )+4 ( s)= 1
s+4
% ( s)−2+4 ( s )= 1
s+4
(s ) ( %+4 )= 1
s+4+2
¿1+2(s+4)
s+4
¿1+2s+8
s+4
¿2 s+9s+4
(s ) ( %+4 )=2 s+9
s+4
(s )=2 s+9
(s+4 )2
L { ( s) }= L−1{ 2 s+9( s+4 )2 } + (t )= L−1 { 2 s+9(s+4 )2 }2 s+9s+4 =
-( s+4 ) +
.( s+4 )2
¿ - ( s+4 )+.
(s+4 )2
2 s+9= - ( s+4 )+.
2 s+9= -s+4 -+.
2= -
9=4 -+. # .=9−4 -⇒.=9−4 (2 )=1
L−1{ 2 s+9( s+4 )2 }= L−1{ 2( s+4 ) }+ L−1 { 1( s+4 )2 }
¿2 L−1 { 1(s+4 ) }+ L−1{ 1(s+4 )2 }
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Métodos Matemáticos /2-ene#o-/
¿2e−4 t +t e−4 t
+ (t )=2 e−4 t + t e−4 t
2) + ( + +=sent ; + (0 )=0
L { + ( + + }= L {sent }
L { + ( }+ L { + ( t ) }= L {sent }
( % ( s)− (0 ) )+ (s )= 1
s2+1
% ( s)−0+ (s )= 1
s2+1
(s ) ( s+1 )= 1
s2
+1
s
(¿¿2+1)(s+1)
(s )=1
¿
s
(¿¿ 2+1)( s+1)1
¿ L
−1 { + (s ) }= L−1¿
s
s(¿¿ 2+1)( s+1)
(¿¿ 2+1)(s+1)=( s+1 ) ( -s+. )+C (s2+1 )
¿1
¿
1=( s+1 ) ( -s+. )+c ( s2+1 )
1= - s2
+.s+ -s+.+C s2
+C 1=( -+C ) s2+ ( -+. ) s+( .+C )
-+C =0
-+.=0
.+C =1
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Métodos Matemáticos /2-ene#o-/
(1 0 11 1 00 1 1
|001)−1 f 1+ f 2# (
1 0 0
0 1 −10 1 1
|001)−f 2+f 3# (
1 0 0
0 1 −10 0 2
|001) 12 f 3
# (1 0 00 1 −10 0 1 |
0
0
1
2) f 3+ f 2# −f 3+ f 1# (100
-=−12
.=−12
C =1
2
+ (t )= L−1
{ -s+.
s2+1 }+ L−1
{ C
s+1 }¿ L−1{−12 s+12
s2+1
}+ L−1{ 12s+1 }¿ L−1{ ss2+1 }+
1
2 L
−1{ 1s2+1 }+1
2 L
−1 { 1s+1 }t +¿
1
2 sent +
1
2 e
−t
+ (t )=−1
2 cos¿
Ci#cito %RC
/ (t )=c
L d
2
/d t
2 +i (t ) 0+1
c / (t )= E (t )
i (t )= -
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Métodos Matemáticos /2-ene#o-/
i (t )=d/
dt
i (t )=/ ( (t )
1) Un ind!"#$ d% 3 &%n$i#' %'"( %n '%$i% !#n n( $%'i'"%n!i( d%
30 O*' + n( ,%$-( %.%!"$#*#"$i- d% 150 /#."i#' '#ni%nd#% %n " 0 .( !#$$i%n"% %' !%$# (..($ .( !#$$i%n"% %n!(.i%$ "i%*# "0
L d
2/
d t 2 +i (t ) 0= E (t )
3d
2/
d t 2 +30 i (t )=150 ; i (0 )=0
L {3/ ( ( (t )+30 i (t ) }= L {150 }
3 L {/ ( ( (t ) }+30 L {i (t ) }= L {150 }
3 [s21 (s )−%/ (0 )−/ ( (0 ) ]+30 )s= L {150 }
3 s2
1 (s )−3 %/ (0 )−3/ ( (0 )+30 )s=150
s
3 s2
1 (s )−3 %/ (0 )−3/ ( (0 )+30 )s=150
s
3 s2
1 (s )−3 %/ (0 )+30 )s=150
s
/ ( (t )=i (t )
d2
/ (t )dt
= d
dt [ d/dt ]
¿ d
dt [ / ( (t ) ]
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Métodos Matemáticos /2-ene#o-/
50
s ( s+10 )=
- (s+10 )+. (s )s (s+10 )
50= - ( s+10 )+. (s )
50= -s+10 -+. (s )
10 -=50# -=5
-+.=0# .=−5
L−1{5s − 5s+10 }
L−1
{5s }− L
−1
{ 5
s+10 }5 L
−1 {1s }−5 L−1 { 1s+10 }
¿5 (1 )−5e−10 t
i (t )=5−5 e−10 t
2)dx
dt = x−2 + x (0 )=−1
d+
dt =5 x− + + (0 )=2
L { x }= L { x }−2 L { + }
L { + }=5 L { x }− L { + }
%2 ( s)− x (0 )= 2 (s )−2 ( s )
%, ( s)− + (0 )=5 2 (s )−, (s )
%2 ( s)+1= 2 (s )−2 ( s)
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Métodos Matemáticos /2-ene#o-/
%, ( s)−2=5 2 (s )−, (s )
%2 ( s)− 2 (s )=−2 ( s)−1
, (s ) ( s−1 )=−2, ( s )−1
2 (s )=−2 (s )−1
(s−1 )
5 2 (s )=5 , (s )−2+,s
2 (s )=% ( s)−2+s
5
2 (s )=% ( s)−2+s
5
2 (s )= ( s)
−2 ( s )−1s−1
=% ( s )−2+ (s)
5
5 (−2 , (s )−1 )=(s−1 ) (%, ( s)−2+, ( s ))
−10 (s )−5=( %2 ( s )−2 (s )+% ( s )−% ( s )+2+ ( s) )
−5−2+2 s=%2 (s )− (s )+10 ( s)
−7+2 s=s2
, ( s )+9 , (s )
2 s−7= ( s ) ( s2+9 )
(s )=2 s−7
(s2+9)
L−1 { (s ) }= L−1 {2 s−7s2+9 }
+ (t )= L−1
{ 2 s
s2+9−
7
s2+9 } + (t )=2 L−1{ 2 ss2+9 }−7 L−1 {
1
s2+9 }#
3
3(s2+32)
+ (t )=2cos3 t −7
3 sen3 t
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Métodos Matemáticos /2-ene#o-/
&e#i"ada de na $#ans8o#mada
Si L {f ( t ) }= F (s) < entonces ⟹ L {t n
f ( t ) }=(−1 )n d(n )
d%
(n ) F (s)
(se a &e#i"ada de na $#ans8o#mada a#a encont#a#
1) L {t e3t }= 1
(s−3)2
Aicando a #oiedad de a &e#i"ada
L {t e3t }=(−1 )n d(n )
d %(n )
F ( s)
¿ (−1 )1 d
d% F (s )
f ( t )=e3 t
L {f ( t ) }= L {e3 t }
F ( s)= 1
s−3
L {t e3t }=(−1 ) dds [ 1s−3 ]
¿−( s−3 ) (0 )−1 (1 )
( s−3 )2
¿−( −1(s−3 )2 )
L {t e3t }= 1( s−3 )2
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Métodos Matemáticos /2-ene#o-/
2) L {t 3 cos2 t }= 3 ' s
s4(s2+4 )
= 3 ' s
s6+4 s4
L {t 3 cos2 t }=(−1 )n dn
d sn L {f ( t ) }
¿−13 d
3
d %3
L {cos2 t }
¿− d
3
d %3 [ ss2+4 ]
F ( s )=[ ss2+4 ]
F ( (% )=( s2+4) (1 )−% (2 s )
(s2+4 )2
= s2+4−2 s2
( s2+4 )2 =
4−s2
(s2+4 )2
2 ( s2+4 )(2 s)¿
( s2+4 )2(2 s)−(4−s2)¿
F ( ( ( s)=¿
¿−2 s ( s2+4 )2−4 s (4−s2)( s2+4 )
( s2+4 )4
−2 s¿¿
( s2+4 )¿¿¿
¿−2 s3−8 s−16 s+4 s3
(s2+4 )3
¿2 s
3−24 s
( s2+4 )3
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Métodos Matemáticos /2-ene#o-/
F ( ( ( (s )=( s2+4 )
3
(6 s2−24 )−(2 s3−24 s ) (3 ( s2+4) (2 s ) )( s2+4 )
6
( s2+4 ) (6 s2−24 )−(2 s3−24 s)(6 s)¿
( s2
+4
)
2
¿¿¿
¿6 s
4−24 s2+24 s2−96−12 s4+144 s2
( s2+4 )4
¿−6 s4+144 s2−96
( s2+4)4
¿ d3
d s3 [
s
s2+4 ]=−
6 s4+144 s2−96(s2+4 )
4
3) L {t 2
cosh5 t }=−12 d2
d s2 L {cosh5 t }
¿ d
2
d s2
[ s
s2
−2 s ] F ( (s )=
( s2−25 ) (1 )−(s)(2s )(s2−25)2
¿ s
2−25−2 s2
(s2−25)2
¿−s2
−25( s2−25)2
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Métodos Matemáticos /2-ene#o-/
2( s2−25 )2 s¿
(( s2−25)2−25)−25−s2¿ F ( ( ( s)=¿
¿
( s2−25 )[−25 (s2−25 )−(−25−s2) 4 s ](s2−25 )
4
¿−2 s3+50 s+100 s+4 s3
( s2−25 )3
F ( ( (s)=2 s
3+150 s
( s2−25 )3
/e#. $eo#ema de $#asaci=n
S9 L {f ( t ) }= F ( s ) + a∈ 0 < entonces L{eat
f ( t ) }= F (s−a)
L{eat f ( t ) }= L {f ( t )}s # s−a
¿ F (s )|s # s−a
¿ F (s−a)
1) L {e5 t
t 3 }= L { f (t ) }s− s−5a=5
¿ L {t 3 }s−s−5
¿ 3 '
s4 |
s # s−5
¿ 3 '
(s−5)4=
6
(s−5)4
2) L {e−2 t
cos4 t }= L {cos 4 t }s # s+2a=−2
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Métodos Matemáticos /2-ene#o-/
( >t-s? 7 t 3
/ tB< 3
M ( >t-a? M ( a >t ? o si t a
/ si t B< a
M --
$
3 e(2−3 )
>t-2
e−(2−3 )
>t-2
? 2 8 >t? e(−3 )
e(−23) e(−3 )
e(−23)
>/Ds/
? e(−23)
D>s/
F#$*( %. (."%$n# d% 2 "%#$%*( d% "$('.(!in 8 >t >t-a ? e
(2−3 )
8 >ta
1) , "3)
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Métodos Matemáticos /2-ene#o-/
? e(2−3 )
8 >ta
Ft ?t
F>t3 ?>t3
? e(2−3 )
8 >t 3
? e(2−3 )
>/Ds 3Ds ? e(2−3 )
3s e(2−3 )
D>s2
? s Ds /1?s2D >s2 /1
3 t - e−t
3 e−4 t
H cos t
Cos t - e−t
cos t 3 e−4 t
cos t
cos t J e−t
cos t 3
F>K ?d7D2 L An cos > nD2 N7 sen nD H
cos > nD2 x= en 4 /2
+¿en 4/2
x2
sen nD H?e
n 4 /2+¿en 4/2 x2
F>$
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Métodos Matemáticos /2-ene#o-/
7
/
2
3
1
>t
a Encont#a# 8>t en 8o#ma sa
N Encont#a# 8>t en té#minos de a 8nci=n aia
c encont#a# 8t
() 8>t? / si 7< t/
3 si 7< t
:) 8 >t ? / >t-7 -/ >t-/ 3>t-/ ?3 t-/ 2>t-
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Métodos Matemáticos /2-ene#o-/
/-/>t- 3>t-/ 2 >t-
F>t ? / 2 >t-/ J >t-
!) / 2 >t-/ J>t-
/ 2 t-/ J >t-
/Ds /Ds -/Ds
SE;UNDO TEOREMA DE TRASLACION
Si 8>t ? 8 >s 0 a 7 entonces
> t-a >t-a ? en 4/2
8 >t
/ Encont#a# a t#ans8o#mada de aace a#a >-t >t-/ 8 > ? t a ? /8 >t-a ?t-a8 >t-/ ?t-/
>-t >t-/ ? es
8 >t
? e−s
D s
2 >t-/ e−t
>t-/
>t-/e−t
>t-/ ? e
−t
>t-/
? e−t
8>t
? e−t
>sD>s-/ ? 1e D>s-/
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Métodos Matemáticos /2-ene#o-/
TRANSFORMADA DE UNA DERIt ? 8 >s entonces
/ 8>t ? s 8>- 8Q ? 8> J 8>e2 8>t ? s2 8>- s 8Q ? 8> J s2 8>e J 8>E 3 8>t ? s3 8>- s 8Q ? 8> J s3 8>e J 8>E 8>t ? s 8>- s 8Q ? 8> J s 8>e J 8>E -8>
%a t#ans8o#mada de aace con na 8nction nita#ia conaB7 es.
>t-a ?/Ds e
aa# a t#ans8o#mada de aace de
F>t ? -s >t-/ 1 >t-7
-s >t-/ 1 t-7
-s >t-/ 1 t-7
? -s >t-/ 1 t-7
? -Ds e 1D
8>t ? 1-eDs
4s-/Ds-4s-1aa#
F>t t>2 si 7< t 2
3t2 si 2< t
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Métodos Matemáticos /2-ene#o-/
si tB
F>t ? t >t-7 Jt2 a>t-2-3t 2>t- ( >t-
F>t? t2 >t2 Jt >t2 >t-
? t>2>-t>23t -2>t-2->3t-3 >t-
F>t
>t
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Métodos Matemáticos /2-ene#o-/
I!S$I$($O PO%I$TC!ICO
!ACIO!A%(!I&A& PROFESIO!A% I!$ER&ISCIP%I!ARIA
&E I!GE!IER'A U CIE!CIAS SOCIA%ES U
A&MI!IS$RA$IVAS
PROBLEMAS TERCER PARCIAL
Métodos Matemáticos
I!GE!IERIA I!&S$RIA%
ELA=ORO> CRU? CORT@S DANIEL ANTONIO
PROFESOR> ;ON?ALES NA
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Métodos Matemáticos /2-ene#o-/
• Se#ie Comea de Fo#ie#
/ Encont#a# se#ie comea de a 8nci=n::
F ( x )= x ; 0
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Métodos Matemáticos /2-ene#o-/
2nπ
cos2nπ −i sin ¿+ 1
n2
π 2 (cos2nπ −isen2nπ )−
1
n2
π 2
−2inπ
¿
Cn=1
2 ¿
Cn=1
2 [ −2inπ + 1n2 π 2− 1n2 π 2 ]
Cn=1
2 [−2inπ ]=−1inπ
Cn=
(−1
inπ
)(−i )= i
nπ
F ( x )=∑n=−∞
∞ x
nπ e
inπ donde n≠ 0
Ee#cicio 2
f ( x )=1− x ;
0! x<
6
Cn=1
6∫0
6
x e
−2 inπ 6 dx
¿ 1
6∫0
6
(1− x)e−lnπ 3
x
dx
¿ 1
6∫0
6
e
−lnπ 3
x
dx−1
6∫0
6
x e
−lnπ 3
x
dx
Cn=1
6 [−3lnπ e−lnπ 3 6
0]−16 [−3 xlnπ e−lnπ 3
x
− 9
x2
n2
π 2 e
−lnπ 3
x ]60
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Métodos Matemáticos /2-ene#o-/
−3lnπ
e−lnπ 3 x dx
(¿)
−3 xlnπ
e
−lnπ 3
x
−∫¿
¿−1
6
¿
−3 (0 )lnπ
e
−lnπ 3
(0)+ 9n2
π 2 e
−lnπ 3
(0 )
(−3 (6 )lnπ e−lnπ 3
(6 )
+ 9
n2
π 2 e
−lnπ 3
(6 ))−¿cn=
−12 lnπ
[ e−lnπ 3
(6 )−e
−lnπ 3
(0 )]−16 ¿
Cn= −1
2lnπ [e−2lnπ −1 ]−1
6 [−18
lnπ e
−2 lnπ + 9
n2
π 2
e−2 lnπ −
9
n2
π 2 ]
¿− 1
2 lnπ e
−2 lnπ + 1
2lnπ + 3
lnπ e
−2lnπ − 3
2n2
π 2
e−2 lnπ +
3
2n2
π 2
2nπ −isen2nπ cos¿
¿− 1
2 lnπ (cos2nπ −isen 2nπ )+
1
2 lnπ + 3
lnπ (cos 2nπ 5 isen2nπ )−
3
2n2 π ¿
¿− 1
2 lnπ +
1
2lnπ + 3
lnπ −
3
2n2
π 2
%a se#ie se#9a:
f ( x )=∑n=−∞
∞−3 i
nπ e
lnπ
3 x
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Métodos Matemáticos /2-ene#o-/
• $#as8o#mada de a ace
/ F (t )=1
L {f (t )}= L {1 }=limb # ∞∫0
b
1−e−st dt
¿ limb #∞
∫0
b
e−st
dt
¿ limb #∞ [−1s e−st ]b0
¿ limb #∞ [−1s e−sb−(−1s e0)]
¿limb # ∞
−1
s e
−sb+limb #∞
1
s
¿−1
s
1
esb+1
s
L {1}=1s
2 L {eat }=lim
b # ∞∫0
b
eat
e−st
dt
¿ limb #∞
∫0
b
e( at −st ) dt
¿ limb #∞
∫0
b
e−( s−a )t dt
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Métodos Matemáticos /2-ene#o-/
¿ limb #∞ [ −1(s−a) e−(s−a ) t ]b0−1
s−ae−(s−a) 0
−1
s−a e
−( s−a )b−¿
¿ limb # ∞
¿
¿limb # ∞
−1
s−a e
−( s−a) b+limb# ∞
1
s−a (1 )
¿− 1
s−a limb # ∞
e−( s−a) b+ 1
s−a
L {eat }= 1s−a
3 L {t eat }= lim
b# ∞∫0
b
t eat −st
dt =limb # ∞
∫0
b
t e−( s−a) t dt
Inte;#aci=n o# a#tes:
(?t ∫du=∫e−(s−a ) t dt
d?dt u= −t s−a
e−( s−a) t
¿ limb #∞ [ −t s−a e−(s−a )t + 1s−a∫ e−( s−a )t dt ]
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Métodos Matemáticos /2-ene#o-/
2t ¿
t 3e¿
¿ L {cos 5t }− L {t 5 }+ L¿
¿ s
s2
+52−
5 '
s6+
3 '
(s−2)4
In/%$'( d% .( .(!%
/ L−1 { f ( s ) }=f (t )
L−1 { $f (s )}=$ L−1 {f (s)}
L−1 { $f (s )}+∝6 (s )= L−1 {$f (s)}+ L−1 {∝6(s) }
¿ 7 L−1 {f (s) }+∝ L−1 {6(s)}
¿ 7f ( t )+∝" (t )
2 f ( s)= s
s2+25
−5 's6+ 3 '(s−2)4
s−2¿¿¿3 '¿
¿ L−1{ ss2+25 }−{5 's6 }+ L−1 ¿
¿ L−1{ s
s2+52 }− L
−1
{ 5 '
s5+1 }+ L
−1
{ 3'
(s−2)3+1 }f ( t )=cos5 t −t 5+t 5 e2 t
T$(n',#$*(d( d% n( d%$i/(d(
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Métodos Matemáticos /2-ene#o-/
/ +8 +4 +=e−4 t ; + (0 )=2
d+
dx+4 +=e−4 t
L { + 8 +4 + }= L {e−4 t }
L { + 8 (t )+4 L { + (t )} }= L {e−4 t }
%+ (s )−2+4 + (s )= 1
s+4
+ (s ) ( s+4 )= 1s+4
+2
¿1+2 s+s
s+4
2 s8 +9
(s+4 )2=
- (s+4 )+.
(s+4)2
1 '(s+4 )2
= 1'(s−(−4 ))
2 +8 + +=sent ; + (0 )=0
L { + 8 }+ L { + }= L {sent }
(s+1 ) + ( s)= 1
s2+1
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Métodos Matemáticos /2-ene#o-/
L−1 { + (s)}= L−1 { 1( s2+1 )(s+1)}
1
( s2+1 ) (s+1)=
( s+1 ) ( -s+. ) c (s2+1)
( s2+1 ) (s+1)
1=( -+C ) s2+ ( -+. ) %+(.+C )
-+C =0, -+.=0, .+C =1
Resoci=n o# mat#ices
Sstitci=n
+ (t )=−12
L−1{ ss2+1 }+
1
2 { 1s2+1 }+1
2 L
−1 { 1s+1 }
+ (t )=−12 cos t +
1
2 sent +
1
2 e
−t
A.i!(!i#n%'
L d2
d t 2 =i (t ) 0+
1
c / (t )= E(t )
L {3/8 8 (t )+30i(t )}= L {150 }
3 [s21 (s )−s/ (0 )−/ 8 (0) ]+30 ) ( s )= L {150 }
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Métodos Matemáticos /2-ene#o-/
3 s2
1 (s )−2 %/ (0 )−3/ (0 )+30 ) (s )=150
s
3 s) (s )+30 ) (s )=150
5
) ( s)= 50s (s+10 )
50
s (s+10)=
- (s+10 )+.ss(s+10)
=0=% -s+10 -+.%
i (t ) 0 s−s e−10 t
INSTITUTO POLITÉCNICO NACIONAL
UNIDAD PROFESIONALINTERDISCIPLINARIA DE INGENIERÍA Y
CIENCIAS SOCIALES Y ADMINISTRATIVAS
Métodos Matemáticos a!a "a I#$e#ie!%a
E&e!cicios a!a te!ce! a!cia"
'IV(A
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Métodos Matemáticos /2-ene#o-/
¿−1
s e
−st sen$t +
$
s∫cos$t e−st dt
u=cos$t dv=∫ e−st
dt
du=−$ sen$t dt v=−1
s e−st
−1s
e−st
cos $t −∫ 1s
$sen$t e−st
dt
¿−1
s e
−st sen$t +
$
s ¿
¿−1
s e
−st sen$t −
$
s2 e−st
cos $t −$
2
s2∫ sen$t e
−st dt
∫0
∞
sen $t e−st
dt =−1
s e
−st sen $t −
$
s2 e
−st cos $t −
$ 2
s2∫sen $t e
−st dt
sen$te−st
dt =¿−1
s e
−st sen$t −
$
s2
e−st
cos$t
$ 2
s2∫
0
∞
sen$t e−st
dt +∫0
∞
¿
( $ 2
s2+1) limb #∞∫0
b
sen $t e−st
dt =limb # ∞
[−1
s e
−st sen$t −
$
s2
e−st
cos$t ]b0
¿ limb #∞
[(−1s e−st sen $t − $ s2 e−st
cos $t )−(−1s e−s (0) sen $ (0 )− $ s2 e−s (0 )cos $ (0 ))]
( $ 2
s2+1) limb #∞∫0
b
sen $t e−st
dt =$
s2
$ 2+s2
s2 lim
b # ∞∫0
b
sen$te−st
dt = $
s2
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Métodos Matemáticos /2-ene#o-/
¿ s
2($ )
s2($ 2+s2)
% {sen $t }= ($ )
($ 2+s2)
Transformada !n"ersa de Laplae
Si % { f (t ) } =F(S) entonces %-1 { F ( % ) }=f ( t ) , se llama transformada inversa de
F(S) y también es lineal.
E#emplos$ Determinar la transformada inversa de:
1.- %-1 { 1
%4
}=¿ %-1
{ 1
%3+1
}=¿ %-1
{ 3 ' (1 )
3 ' %3+1
}= 1
3 ' %-1 { 3 '
%3+1
}= 1
3' t 3
∴ f ( t )=1
6 t 3
2.- %-1 { 3%6 }=¿ %-1 { 3
%5+1 }=¿ %-1 { (3 )5 '5 ' %5+1 }= 35 ' %-1 {
5 '
%5+1 }= 35' t 5∴ f ( t )= 3120 t 5
.- %-1 { 4 %4 %2+1 }=¿ %-1 { 4 %
4 (%4+ 14 ) }=¿ %-1 { %
%2+
1
4 }=¿ %-1
{ %
%2+( 12 )
2 }=cos 1
2t ∴ f ( t )=cos
1
2t
4.- %-1 { 14 %+1 }=¿ %-1 { 1
4 (%+ 14 ) }=1
4 %-1 { 1%+ 1
4 }=14 %-1
{ 1
%−(−14 )}=
1
4 e
−1
4
t
∴ f (t )=1
4 e
−1
4
t
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Métodos Matemáticos /2-ene#o-/
!.- %-1 { 1%2−16 }=¿ %-1 { 1
%2−(4 )2 }=¿ %-1 { (4 ) (1 )4 (%4−(4 )2 )}=14 %-1
{ 4%2−(4 )2 }∴ f ( t )=14 sin 94 t
".- %-1 {3%+5%2+7 }=¿ %-1 { 3 %
%2+7
+ 5
%2+7 }=¿ %-1 {
3 %
%2+7 }+¿ %-1 {
5
%2+7 }=¿ %-1
{ %%2+7 }+¿ ! %-1 { 1
%2+7 }
= %-1 { %%2+(√ 7 )2 }+¿ ! %-1 { 1
%2+(√ 7 )
2 }=¿ %-1 { %%2+(√ 7 )2 }+¿ ! %-1
{ (√ 7 ) (1 )(√ 7 ) (%2+(√ 7)2) }∴ f ( t )=3cos√ 7 t −
5
√ 7sin √ 7 t
#.- %-1
{ 5
%−6
− 6%
%2
+9
+ 3
2%2
+8%+10 }=¿
%-1
{ 5
%−6 }−¿
%-1
{ 6 %
%2
+9 }+¿
%-1
{ 32%2+8%+10 }
=! %-1 { 1%−6 }−¿ " %-1 { %%2+32 }+¿ %-1 { 3
2 (%2+4 %+5) }=5 e6 t +6cos93 t +3
2 %-1
{ 1
%2
+4 %+5 }Donde:
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Métodos Matemáticos /2-ene#o-/
3
2 %-1 { 1%2+4 %+5}=3
2 %-1 { 1%2+4 %+(2 )2−(2 )2+5 }=32 %-1 { 1( %+2 )2+1 }=32 %-1
{ 1( %−(−2 ) )2+1 }
=3
2e−2t
sin t ∴ f (t )=5e6 t +6cos 93t +3
2e−2 t
sin t
$.- %-1 { %−12%2+%+6 }=¿ %-1 { %−1
2(%2+ 12 %+3) }=¿ %-1
{ %−1
2(%2+ 1
2 %+(
14 )
2
−( 14 )
2
+3)}=¿
%-1
{ %−1
2[(%+ 14 )
2
+ 4716 ] }
1
2 %-1 { %−1
(%+ 14 )2
+( √ 474 )2 }=¿ 12 %-1 { %+
1
4−
1
4−1
(%+ 14 )2
+( √ 474 )2 }=12 %-1
{ (%+ 1
4 )(%+ 14 )
2
+( √ 474 )2−
5
4 ( 1
(%+ 14 )2
+( √ 474 )2 )}
1
2 %-1
{
%+1
4
(%+ 14 )2
+( √ 474 )2
}−( 54 )( 12 ) %-1 {
1
(%+ 14 )2
+( √ 474 )2 }
f ( t )=1
2e−14
t
cos ( √ 474 t )−58 %-1 {( 4√ 47 )√ 474
(%+ 14 )2
+( √ 474 )2 }
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Métodos Matemáticos /2-ene#o-/
f ( t )=1
2e−14
t
cos ( √ 474 t )−(58 )( 4√ 47 )e−14
t
sin( √ 474 t )
∴ f ( t )=1
2 e
−14
t
cos ( √ 474 t )− 12√ 47 e−14
t
sin( √ 474 t )
%.- %-1 {−8 %2−5%+9
%3−2%2−%+2 }
Donde: %3−2%2−%+2≠0= (%−1 ) ( %2−%−2 )=( %−1 ) ( %−2 ) ( %+1 )
∴ %-1 { −8 %2−5 %+9
( %−1 ) ( %−2 ) ( %+1 ) }=¿ %-1
{ -%−1+ .%−2 + C %+1 }= (%−2 ) (%+1 ) -+ (%−1 ) (%+1 ) .+( %−1 ) ( %−2 )C (%−1 ) (%−2 ) (%+1 )∴−8 %2−5 %+9=( %−2 ) ( %+1 ) -+( %−1 ) ( %+1 ) .+ (%−1 ) (%−2 ) C
Si S=2
−8 (2 )2−5 (2 )+9=(2−2 ) (2+1 ) -+(2−1 ) (2+1 ) .+(2−1 ) (2−2 )C
−32−10+9=(1 ) (3 ) .
−33=3. #∴.=−11
Si S=1
−8 (1 )2−5 (1 )+9=(1−2 ) (1+1 ) -+ (1−1 ) (1+1 ) .+(1−1 ) (1−2 )C
−8−5+9=(−1 ) (2 ) -
−4=−2 - #∴ -=2
Si S=-1
−8 (−1 )2−5 (−1 )+9=(−1−2 ) (−1+1 ) -+(−1−1 ) (−1+1 ) .+(−1−1 ) (−1−2 ) C
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Métodos Matemáticos /2-ene#o-/
−5+5+9= (−2 ) (−3 )C
6=6C #∴C =1
∴ %-1 { 2%−1+ −11%−2 + 1%+1 }=¿ 2%-1 { 1%−1 }−11 %-1 { 1%−2 }+¿ %-1
{ 1%− (−1 ) }
f ( t )=2e t −11e2 t +e−t
1&.- %-1 { 2 %
2+10 %
( %2−2 %+5 )( %+1 ) }=¿ %-1 {
-%+.
( %2−2%+5 )+ C
%+1 }=¿ %-1
{( %+1 ) ( -%+. )+( %2−2%+5 ) C
( %2−2%+5 ) ( %+1 ) }∴2 %
2+10%=( %+1 ) ( -%+. )+ (%2−2%+5 ) C
'ara S= -1
2 (−1 )2+10 (−1 )=(−1+1 ) ( - (−1 )+. )+( (−1 )2−2 (−1 )+5 )C
2−10=(1+2+5 )C #−8=8C #−1=C
2%2+10 %= -%2+.%+ -%+.+C %2−2C%+5C
2 %2+10 %=( -+C ) %2+( .+ -−2C ) %+( .+5 C )
Donde:
-+C =2# -=2−C # -=2−(−1 )∴ -=3
.+5C =0# .=−5 C # .=−5 (−1 )∴.=5
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Métodos Matemáticos /2-ene#o-/
%-1 { 2%2+10 %
( %2−2%+5 ) ( %+1 ) }=¿ %-1 { 3 %+5( %2−2 %+5 ) }−¿ %-1 { 1( %+1 ) }Donde:
%2−2%+5=%2−2%+12−12+5#=(%−1 )2+4
∴ %-1 { 3 %+5( %2−2 %+5 ) }=¿ %-1 { 3%+5( %−1 )2+4 }=¿ %-1 {3%−3+3+5(%−1 )2+4 }=¿ %-13 (%−1 )+8
( %−1 )2+4
∴
%-1
{ 3 %+5
( %2
−2 %+5 ) }−¿
%-1
{ 1
( %+1 ) }=¿
%-1
{ %−1
( %−1 )2
+22
}+¿
%-1
{ (4 ) (2 )( %−1 )2+22 }−¿ %-1 { 1( %+1 ) }
%-1 { %−1( %−1 )2+22 }+¿ 4 %-1 { (2 )( %−1 )2+22 −¿ %-1 { 1( %+1 ) }
f ( t )=3 et
cos2 t +4 et
sin 2t −e−t
11.- %-1 { 5%2+34 %+53
%3+7 %2+15%+9 }
Donde:
%
3
+7
%
2
+15
%+9
=(%+
1 ) (%
2
+6
%+9 )=
(%+
1 ) (%+
3 ) (%+
3 )=
(%+
1 ) (%+
3 )2
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Métodos Matemáticos /2-ene#o-/
%-1 {5%2+34 %+53
( %+1 ) (%+3 )2 }=¿ %-1
{ -%+1+ .%+3 + C (%+3 )2 }# 5%2+34 %+53
( %+1 ) (%+3 )2 =
(%+3 )2 -+ (%+1 ) ( %+3 ) .+(%+1 )C
( %+1 ) ( %+3 )2
∴5 %2+34 %+53=(%+1 )2 -+ (%+1 ) ( %+3 ) .+(%+1 ) C
'ara S= -1
5 (−1 )2+34 (−1 )+53=(−1+3 )2 -+(−1+1 ) (−1+3 ) .+(−1+1 )C
5−34+53=4C #24=4 C #∴ -=6
'ara S=-
5 (−3 )2+34 (−3 )+53=(−3+3 )2 -+ (−3+1 ) (−3+3 ) .+(−3+1 ) C
45−102+53=−2C #−4=−2C #∴C =2
5%2+34%+53=( %+1 )2 -+( %+1 ) (%+3 ) .+( %+1 )C
5%2+34%+53=( %2+6%+9 ) -+( %2+4%+3) .+(%+1 )C
5%2+34%+53=( -+. ) %2+(6 -+4.+C ) %+(9 -+3 .+C )
Donde : -+.=5# .=5− - ; .=5−6 ; .=−1
%-1 { 6%+1 }+¿ %-1 { −1%+3 }+¿ %-1 { 2( %+3 )2 }=¿ "%-1 { 1%+1 }−¿ %-1 { 1%+3 }+¿ 2 %-
1
{ 1'
( %+3 )1+1 }∴ f ( t )=6 e−1−e−3 t +2te−3 t
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Métodos Matemáticos /2-ene#o-/
Teorema de %ea"!s!de
Sean '(S) y (S) olinomios en los c*ales '(S) es de +rado menor *e (S) y
(S) tiene n races diferentes /, con /=1, 2, , 0, n. ntonces:
%-1 { & (%)1(%)}=∑$ =1n
&(:$ )
1 ) (:$ )
e:$ t
1)
%-1 { 2 s+4s3+2 s2−5 s−6 } & (% )=2s+4
1 (% )=s3+2 s2−5 s−6
-1 1 2 -! -"-1 -1 "
1 1 -" &
1 (% )=(s+1)( s2+5−6)
1 (% )=(s+1)( s+3)( s−2)
1 ) ( % )=3 s2+4 s−5
: 1=−1
: 2=−3
: 3=2
%-1 { & (%)1(%)}=∑$ =1n
&(:$ )
1 ) (:$ )
e:$ t =
&(: 1)
1 ) (: 1)
e: 1 t +
&(: 2)
1 ) (: 2)
e: 2 t +
&(: 3)
1 ) (: 3)
e: 3 t
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Métodos Matemáticos /2-ene#o-/
¿ &(−1)
1 ) (−1)
e−t +
&(−3)
1 ) (−3)
e−3 t +
&(2)
1 ) (2)
e2 t
& (−1 )=2 (−1 )+4=2
& (−3
)=2
(−3
)+4
=−2
& (2 )=2 (2 )+4=8
1 ) (−1 )=3 (−1 )2+4 (−1 )−5=−6
1 ) (−3 )=3 (−3 )2+4 (−3 )−5=10
1 ) (2 )=3 (2 )2+4 (2 )−5=15
f ( t )=−26
e−t −
2
10e−3 t +
8
15e2 t
%-1 { 2 s+4s3+2 s2−5 s−6 }
%-1 { 2 s+4(s+1)( s+3)( s−2) }2 s+4
(s+1)(s+3)(s−2)=
-
s+1+
.
s+3+
C
s−2
2 s+4(s+1)(s+3)(s−2)
= - ( s+3 ) ( s−2 )+. ( s+1 ) (s−2 )+C (s+1)( s+3)
(s+1)( s+3)( s−2)
2 s+4= - (s+3 ) (s−2 )+. (s+1 ) ( s−2 )+C (s+1 ) (s+3 )
si s=−3
2(−3)+4= - (−3+3 ) (−3−2 )+. (−3+1 ) (−3−2 )+C (−3+1 ) (−3+3 )
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Métodos Matemáticos /2-ene#o-/
−6+4=.(−2)(−5)
−2=10. # .=−210
si s=2
2(2)+4= - (2+3 ) (2−2 )+. (2+1 ) (2−2 )+C (2+1 ) (2+3 )
8=15C #C = 8
15
si s=−1
2(−
1)+
4= - (−
1+3
) (−1−2
)+. (−1+1
) (−1−2
)+C (∓1
) (−1+3
)
2=−6 - #C =−13
%-1 { 2 s+4(s+1)( s+3)( s−2) } = %-1 { -s+1+ .s+3 + C s−2 }
?%-1 {−13
s+1 } %-1 {−15
s+3 } %-1 { 8
15
s−2 }=
−13 %-1 { 1s+1 }−15 %-1 { 1s+3 }+ 815 %-1 { 1s−2 }
¿−1
3 e
−t −1
5e−3 t +
8
15e2 t
Der!"ada de &na 'ransformada
Si % { f (t ) = F (%) entonces
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Métodos Matemáticos /2-ene#o-/
% {t n
f (t ) }=(−1 )n d(n )
ds (n) F (%)
3se la derivada de *na transformada ara encontrar
1)
% {te3 t }= 1
(s−3)2
% {teat }= 1
(s−a)2
f ( t )=e3 t
f ( s)= 1s−3
% {te3 t }=(−1) d
ds [ 1s−3 ]
% {te3 t }=−( s−3 ) (0 )−1(1)
(s−3)2
% {te3 t }= −−1
( s−3 )2=
1
(s−3)2
2)
% {t 3
cos2 t }=(−1 )n d(n )
ds (n ) F (%)
% {t 3cos2 t }=(−1)3 d
3
ds3 % {cos2 t }
¿− d
3
ds3 [ ss2+4 ]
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Métodos Matemáticos /2-ene#o-/
)
% {t 2cosh 95 t }=(−1 )n d
( n)
ds( n) F (%)
¿ (−1 )2 d
2
ds(2 ) F (%) L {t 2 cosh95 t }
¿− d
2
ds2 [ ss2−25 ]
F ( % )= s
s2−25
F ) (% )=(s
2
−25) (1 )
−(25 ) (
s)
( s2−25)2 = s
2
−25
−2
s
2
(s2−25 )2 =−
25−s
2
( s2−25 )2
F )) (% )=
(s2−25 )2
(−2 s )−(−25−s2)2 ( s2−25 )2 s
(s2−25 )4
F )) (% )=
(s2−25 )[−2 s (s2−25 )−(25−s2 )4 s]( s2−25 )
4
F )) (% )=
−2 s3+50 s+100 s+4 s3
( s2−25 )4
=2 s
3+150 s
( s2−25 )3
¿− d
2
ds2 [ ss2−25 ]
¿− d
2
ds2 [ F (%) ]
¿ F )) (%)
¿2 s
3+150 s
(s2−25 )3
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Métodos Matemáticos /2-ene#o-/
PRIMER TEOREMA DE TRASLACI(N
Si % { f (t ) } =F(S) y a∈ 0 entonces % {ea% f ( t ) }= F (%−a) % {e
a% f ( t ) }=¿
% { f (t )}% # %−a
ncontrar la transformada de 5alace, *sando el rimer teorema de traslacin.
1.- % {e5 t
t 3 } a=5 ; f ( t )=t 3
% {t 3 }%# %−5=
3 '
%4
% # %−5
= 3'
( %−5 )4=
6
(%−5 )4∴ F (% )=
6
(%−5 )4
2.- % {e−2 t
cos4 t } a=−2; f (t )=cos4 t
% {cos 4 t }% # %+2=
%
%2+42
= %+2
(%+2 )2+16∴ F (% )=
%+2
(%+2 )2+16
.- % {t (e t +e2 t )2 }=¿ % {t ( e
2 t +2e t e2 t +e4 t )}=¿ % {te2 t +2e3t +te4 t }
=% {t e2 t }+¿ 2% {t e
3 t }+¿ % {te4 t }=¿ % {t }% # %−2 , 2% {t }% # %−3 , % {t }% # %−4
¿ 1
%2
% # %−2
, 1
%2
% # %−3
, 1
%2
% # %−4
∴ F (% )= 1
(%−2 )2+
2
( %−3 )2+
1
( %−4 )2
Si ∫ {f (t )} ? F >s 0 a ∈ 0
∫ { eat
f (t ) } ? F >s-a
∫eat f ( t ) ? ∫ {f ( t ) } s ¿ s-a
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Métodos Matemáticos /2-ene#o-/
? F >s
? F >s-a
/ ∫ { est
t 3
} ?3 '
( s−5 )4 ?6
( s−5 )4
∫ { e5 t t 3 } ? ∫ {f (t )} s ¿s−5
a=5 ? ∫ {t 3 } s B s -
?3 '
s4 s B s J
?3 '
( s−5 )4 ?6
( s−5 )4
2 ∫ { e−2 t
cos 4 t } ? ∫ {f (t )} s B s-a
? ∫ {cos 4 t } s B s J >-2
?5
s2+16 sB s2
?s+2
(s+2)2+16
3 ∫ { [1−et +3e−4 t ]cos5 t }
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Métodos Matemáticos /2-ene#o-/
∫ {cos5t −et cos5 t +3e−4 t cos5 t }
∫ {cos5 t }−∫ {et cos5 t }+3∫ {e−4 t cos5 t }
s
s2+25 - ∫ {cos5 t } s B s J / 3 ∫ {cos5 t } s>s+4
s
s2+25
− s
s2+25 ⋮ s B s J / 3
s
s2+25 ⋮ s B s
?s
s2+25
− (s−1)
( s−1 )2+25
3(s+4 )
(s+4 )2+25
Fnci=n esca=n nita#io
%a 8nci=n esca=n nita#io ( >t- a ? (a>t se de@ne como
( >t Ja ? (a >t ? {0 sit t J 3 ? {0 ; t t- /7 ? {0 ; t t Ja ? M (a >t ? { 0 si t
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Métodos Matemáticos /2-ene#o-/
4 ( >t J /7 ? {0 ; t t ? - ( > t -/ 1 ( >t J 7
∫ {f (t )}=∫ {−5= (t −1 )+6 = (t −0) }
? ∫ {−5= (t −1 ) }+∫ {6= (t −0 ) }
? −5∫ {= (t −1 ) }+6∫ {= ( t −0)}
? -1
s e
−s+61
s e
0=−5e−5
s +
6
5
? 1 e−5
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Métodos Matemáticos /2-ene#o-/
Se)&ndo 'eorema de 'rasla!*n
Si % { f (t ) } =F(S) y a6& entonces % { f (t −a )u (t −a ) }=e−a%
% { f (t ) }
Forma alternativa de S.7.7 % { " (t ) u (t −a ) }=e−a%
% { " ( t −a ) }
E#emplo$ 8li*e el se+*ndo teorema de traslacin ara encontrar
1.- % { (t −2 ) u (t −2 ) }a=2 ; f ( t )=t
¿e−2% % {t }=e−2%
> 1
%2=
e−2%
%2
2.- % {(t −1 )3
e( t −1 ) u (t −1 ) } a=1 ; f (t )=t 3 et
¿e−% % {t 3
et }=e−% 3'
( %−1 )4=e−%
6
(%−1 )4=
6e−%
( %−1 )4
.- % {tu (t −2 ) } a=2 ; " (t )=t ; " (t +a )=t +a∴ " ( t +2 )=t +2
% {tu (t −2 ) }=e−2%
% { (t +2) }=e−2%
9% {t }+¿ % {2} = e−2%[ 1%2 + 2% ]
4.- % {t 2
u (t −3 ) } a=3 ; " (t )=t 2∴" (t +2 )=( t +3 )2=t 2+6 t +9
¿e−3% % {t 2+6t +9 }=e−3% % {t
2 }+¿ % {6 t }+¿ % {9 }=e−3 % [ 2 '%3 + 6%2+ 9% ]
Transformada de la !n'e)ral
Si % { f (t ) } =F(S), entonces % {∫0t
f ( ? )d? }=1
% % { f (t ) } .
ncontrar la transformada de la inte+ral dada or:
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Métodos Matemáticos /2-ene#o-/
1.- % {∫0
t
e?
d? }; f ( t )=e t ∴=1% % { f (t ) }=1% [ 1%−1 ]
2.- % {∫0
t
cos? d? }; f (t )=cos t ∴=1% % { f (t ) }=1% [ %%2+1 ]= 1%2+1
.- % {∫0
t
e−?
cos ? d? }; f ( t )=e−t cos t ∴=1% % { f (t ) }=1% [ %− (−1 )( %+1 )2+1 ]=1% [ %+1(%+1 )2+1 ]
%-1 { F (%)}=¿ %-1 {1% }+3 %-1 { 1%2 }+¿ %-1 { 2
%3 }
∴ f ( t )=1+3t +t 2
Con"ol&!*n
Si las f*nciones f(t) y +(t) son contin*as or tramos en el intervalo ¿ , la
convol*cin de f(t) y +(t) se reresenta or f;+ y se define como:
f ∗"=∫0
t
f ( ? ) " (t −? ) d?
Teorema de la on"ol&!*n
Si f(t) y +(t) son contin*as or tramos en el intervalo ¿ y de orden
e
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Métodos Matemáticos /2-ene#o-/
¿[ 1%−1 ][ 1%2+1 ]= 1( %−1 ) ( %2+1)
2.- % {∫0
t
?e (t −? ) d? }; f (t )=t ; " (t )=e t ∴=¿ % {t } % {e t }=[ 1%2 ] [ 1%−1 ]= 1%2 (%−1 )>es*elva con resecto a f(t) la ec*acin inte+ro-diferencial dada or:
1.- f ( t )=2 t +1+∫0
t
f ( t −? )e−? d? ;f ( t )=f ( t ); " (t )=e−t
% { f (t ) =¿ 2% {t } % {1 } % {∫o
t
f (t −? )e−? d? }
F ( % )= 2
%2+1
%+¿ % { f (t )} % {e
−t } # F (% )= 2%
2+1
%+ F (%)[ 1%+1 ]
F ( % )− F (% )[ 1%+1 ]= 2%2+ 1
%# F ( % )[1− 1%+1 ]=2+%%2 # F (% )[ %+
1−1%+1 ]=2+%%2
F ( % )[ %%+1 ]=2+%%2 # F ( % )=( %+2 ) ( %+1 )
%3
# F ( % )=%
2+3 %+2%
3
F ( % )=%2
%3+3 %
%3 +
2
%3 # F ( % )=
1
%+ 3
%2+ 2
%3
INSTITUTO POLITECNICO NACIONAL
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Métodos Matemáticos /2-ene#o-/
(!I&A& PROFESIO!A% I!$ER&ISCIP%I!ARIA &E I!GE!IERIA
U CIE!CIAS SOCIA%ES U A&MI!IS$RA$IVAS
T($%( T%$!%$ P($!i(.
Nom+res$
8me?c*a @ernAnde? Besar@ermida on?Ale? Cara Fernanda'ére? 8lvarado BarolinaSAnce? van+elista 8t?iri
Se&en!a$
2E8
Ma'er!a$
Cétodos CatemAticos De 5a En+eniera
Carrera$
En+eniera End*strial
Profesor$
on?Ale? Gavarrete Barlos
T$(n',#$*(d( d% L(.(!%
F (t )=1
L {f (t )}= L {1 }=limb # ∞
∫0
b
1−e−st dt =limb # ∞
∫0
b
e−st
dt =limb# ∞ [−1s e−st ]b0
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Métodos Matemáticos /2-ene#o-/
¿ limb #∞ [−1s e−sb−(−1s e0)]=
limb # ∞
−1
s e
−sb+limb# ∞
1
s =
−1s
1
esb+1
s L {1 }=1
s
F (t )= 7
L {f (t )}= L {$ }= limb # ∞
∫0
b
$ e−st
dt = limb# ∞
$ ∫0
b
e−st
dt = limb# ∞
$ [−1
s e
−st
]b0
¿ limb #∞
$ [−1s e−sb−(−1s e0)]=lim
b # ∞
−$
s e
−sb+limb # ∞
$
s =
−$ s lim
b# ∞
e−sb+
$
s=
$
s
L {eat
}=limb # ∞∫0
b
eat
e−st
dt
limb # ∞
∫0
b
e(at −st ) dt = lim
b #∞∫0
b
e−( s−a )t dt = lim
b #∞ [ −1(s−a) e−( s−a )t ]b0−1
s−ae−( s−a) 0
−1
s−a e
−( s−a )b−¿=limb # ∞
−1
s−a e
−(s−a ) b+limb # ∞
1
s−a (1 )
¿ limb # ∞
¿
¿− 1
s−a limb # ∞
e−( s−a) b+ 1
s−a L {eat }= 1
s−a
L {t eat }
limb # ∞
∫0
b
t eat −st
dt = limb # ∞
∫0
b
t e−( s−a )t dt = =t ……∫ du=∫e−( s−a) t dt
du=d t … … u= −t s−a
e−( s−a) t ¿ limb #∞ [ −t s−a e−(s−a )t + 1s−a∫ e−( s−a )t dt ]
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Métodos Matemáticos /2-ene#o-/
¿ limb #∞
[
−t s−a
e−(s−a )t −
1
(s−a )
2e−(s−a ) t
]b0limb # ∞
[
−bs−a
e−(s−a ) b−
1
(s−a)
2e−(s−a )t
]b0
¿ limb #∞ [( −bs−a e−( s−a) b− 1( s−a )2 e−( s−a) b)−( −0s−a e−( s−a) 0− 1( s−a )2 e−( s−a) 0)]
¿ 1
(s−a)
limb # ∞
b
e ( s−a) b−
1
(s−a)2
limb # ∞
1
e( s−a) b+
limb # ∞
1
(s−a)2=
1
(s−a)2
% {sen $t }
limb # ∞
∫0
b
sen$te−st
dt ..u=sen$t … ..dv=∫ e−st dt…du=$ cos $tdt ..v=−1s
e−st
¿−1
s e
−st sen$t +
$
s∫cos$t e−st dt … u=cos $t …du=−$ sen$t dt
dv=∫ e−st
dt … v=−1
s e−st
…
−1s
e−st
cos $t −∫ 1s
$sen$t e−st
dt
¿−1
s e
−st sen$t +
$
s ¿
¿−1
s e
−st sen$t −
$
s2 e−st
cos $t −$
2
s2∫ sen$t e
−st dt
? ∫0
∞
sen $t e−st
dt =−1
s e
−st sen $t −
$
s2 e
−st cos $t −
$ 2
s2∫sen $t e
−st dt
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Métodos Matemáticos /2-ene#o-/
sen$te−st
dt =¿−1
s e
−st sen$t −
$
s2
e−st
cos$t
¿ $
2
s2∫
0
∞
sen$te−st
dt +∫0
∞
¿
¿($ 2
s2 +1)
limb # ∞∫0
b
sen$te−st
dt =limb # ∞ [−1
s e−st
sen$t − $
s2 e
−st
cos $t ]b0
¿ limb #∞
[(−1s e−st sen $t − $ s2 e−st
cos $t )−(−1s e−s (0) sen $ (0 )− $ s2 e−s (0 )cos $ (0 ))]
¿( $ 2
s2+1) limb # ∞∫0
b
sen$te−st
dt = $
s2 ..
$ 2+s2
s2 lim
b# ∞∫0
b
sen$t e−st
dt =$
s2
¿ s
2($ )
s2($ 2+s2)
= ($ )
($ 2+s2)
L {15−3e−7 t +4 sen3 t }
¿ L {15 }− L {3 e−7 t
}+ L {4 sen3 t }=15 L {1 }−3 L {e−7 t
}+4 L {sen3 t }
¿151
s−3
1
s−(−7 )+4
3
s2+9
=15
s −
3
s+7+ 12
s2+9
L (t )=cos 5t −t 5+t 3 e2 t
¿ L {f (t ) }= L {cos5 t −t 5+t 3 e2 t }= L {cos5 t }− L {t 5 }+ L {t 3 e2 t }
¿ s
s2+52
−5 '
s6+
3 '
(s−2)4
T$(n',#$*(d( in/%$'( d% L(.(!%
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Métodos Matemáticos /2-ene#o-/
L−1{ ss2+25−5 's6 + 3 '(s−2)4 }
s−2¿¿¿3 '¿
¿ L−1{ ss2+25 }−{5 's6 }+ L−1 ¿
¿ L−1{ ss2+52 }− L−1{ 5 '
s5+1 }+ L−1{ 3'(s−2)3+1 }=cos5 t −t 5+t 5 e2 t
L
−1
{ 1
%4
}¿
L−1{ 1%3+1 }=¿ L−1{ 3 ' (1 )3 ' %3+1 }= 13' L−1{
3 '
%3+1 }= 13 ' t 3=16 t 3
L−1{ 5%−6− 6 %%2+9 +
3
2 %2+8%+10 }
¿ L−1
{ 5
%−6 }−¿ L−1
{ 6 %%2+9 }+ L
−1
{ 3
2 %2+8%+10 }
? L−1{ 1%−6 }−¿ 1 L−1{ %%2+32 }+¿ L−1{
3
2 ( %2+4%+5 ) }=5e6 t +6cos93 t +3
2
L−1{ 1%2+4%+5 }
¿3
2 L
−1{ 1%2+4%+5 }=3
2 L
−1{ 1%2+4 %+( 2 )2−(2 )2+5 }=32 L−1{ 1( %+2 )2+1 }=32 L
−1{ 1( %−(−2 ) )2+1
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Métodos Matemáticos /2-ene#o-/
?3
2e−2t
sin t =5e6 t +6cos 93t +3
2e−2 t
sin t
L−1{ %−12 %2+%+6 }
¿ L
−1
{ %−1
2(%2+ 12 %+3) }=¿ L
−1
{ %−1
2(%2+ 12 %+( 14 )2
−( 14 )2
+3) }=¿
L−1
{ %−1
2 [(%+ 14 )2
+47
16 ]}
?
1
2 L
−1
{ %−1
(%+ 14 )2
+( √ 474 )2 }=¿ 12 L−1 { %+
1
4−
1
4−1
(%+ 14 )2
+( √ 474 )2 }=12
L−1
{ (%+ 14 )
(%+
1
4
)
2
+
( √ 47
4
)
2−
5
4
(
1
(%+
1
4
)
2
+
( √ 47
4
)
2
)}?
1
2 L
−1 { %+1
4
(%+14 )2
+( √ 474 )2 }−( 54 )( 12 ) L−1{ 1(%+ 14 )2+( √ 474 )2 }
¿ 12
e
−14 t cos ( √
474
t )−58
L−1
{( 4√ 47 )
√ 47
4
(%+ 14 )2
+( √ 474 )2
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Métodos Matemáticos /2-ene#o-/
¿−2
6 e
−t − 2
10e−3 t +
8
15e2 t
L−1{ 2 s+4s3+2 s2−5 s−6 }… L−1{
2 s+4(s+1)(s+3)(s−2)}
2 s+4(s+1)(s+3)(s−2)
= -
s+1+
.
s+3+
C
s−2
2 s+4(s+1)(s+3)(s−2)
= - ( s+3 ) ( s−2 )+. ( s+1 ) (s−2 )+C (s+1)( s+3)
(s+1)( s+3)( s−2)
2 s+4= - (s+3 ) (s−2 )+. (s+1 ) ( s−2 )+C (s+1 ) (s+3 )
si s=−3
2(−3)+4= - (−3+3 ) (−3−2 )+. (−3+1 ) (−3−2 )+C (−3+1 ) (−3+3 )
−6+4=.(−2)(−5)
−2=10. … .=−210
si s=2
2(2)+4= - (2+3 ) (2−2 )+. (2+1 ) (2−2 )+C (2+1 ) (2+3 )
8=15C … C = 8
15
si s=−1
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Métodos Matemáticos /2-ene#o-/
2(−1)+4= - (−1+3 ) (−1−2 )+. (−1+1 ) (−1−2 )+C (∓1 ) (−1+3 )
2=−6 - … C =−13
¿ L−1
{ 2 s+4
(s+1)( s+3)( s−2)}? L
−1
{ -
s+1 + .
s+3+ C
s−2 }
? L−1{ −13s+1 } L−1{
−15
s+3 } L−1{ 8
15
s−2 }?
−13
L−1{ 1s+1 }−15 L−1{ 1s+3 }+ 815 L−1{ 1s−2 }=−13 e−t −15 e−3 t + 815 e2 t
T$(n',#$*(d( d% n( d%$i/(d(
+8 +4 +=e−4 t ; + (0 )=2
d+
dx+4 +=e−4 t … L { + 8 +4 + }= L {e−4 t }
L { + 8 (t )+4 L { + (t )} }= L {e−4 t }… %+ ( s)−2+4 + ( s)= 1s+4
+ (s ) ( s+4 )= 1
s+4+2…=
1+2 s+ss+4
(% )= 2 s8 +9
(s+4)2
D%$i/(d( d% n( "$(n',#$*(d(
L {te3t }= 1(s−3)2
L {teat }= 1(s−a)2
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Métodos Matemáticos /2-ene#o-/
L {te3t }=(−1 ) dds [ 1s−3 ]
L {te3t }=−(s−3 ) (0 )−1(1)(s−3)2
L {te3t }= −−1(s−3 )2
= 1
(s−3)2
L {t 3 cos2 t }=(−1 )n d(n)
ds(n)
F (% )
L {t 3
cos2 t }=(−1)3 d
3
ds3 L {cos2t } …=
−d3
ds3
[ s
s2+4 ]
F ) (%)=
(s2+4 ) (1 )−s(2 s )( s2+4)
2 =
s2+4−2 s2
( s2+4)2 =
4−s2
(s2+4 )2
F )) (%)=
( s2+4 )2 (−2 s )−(4−s2)(2 (s2+4 ) (2 s ))
(s2+4 )4
F )) (%)=
−2 s ( s2+4 )2−4 s(4−s2) ( s2+4 )
(s2+4 )4
F )) (% )=
(s2+4 ) [−2s ( s2+4 )−4 s (4−s2 ) ]( s2+4 )
4
F )) (% )=−
2 s3−8 s−16 s+4 s3
( s2
+4 )
3 =
2 s3−24 s
( s2
+4 )
3
F ))) (% )=
( s2+4 )3
(6 s2−24 )− (2s3−24 s)(3 ( s2+4 )2(25 ))
(s2+4 )6
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Métodos Matemáticos /2-ene#o-/
s
( s2+4 )2 [(¿¿ 2+4 )(6 s−24 )−(2 s3−24 s)(6 s)]
( s2+4 )6
F ))) (% )=¿
F )))
(% )=6 s
4−24 s2+24 s2−96−12 s4+144 s2
( s2+4 )4
F ))) (% )=
−6 s4+144 s2−96
(s2+4 )4
F ))) (% )=
−d3
ds3 [ ss2+4 ]=−[−6 s
4+144 s2−96
( s2+4 )4 ]
F ))) (% )=−
6 s4+144 s2−96
(s2+4 )4
L {t 2 cosh95 t }=(−1 )n d(n )
ds(n ) F (%)
¿ (−1 )2 d
2
ds(2 )
F (% ) L {t 2 cosh95 t }…=−d2
ds2 [ ss2−25 ]
F ( % )= s
s2−25
F ) (% )=
(s2−25) (1 )−(25 ) ( s)
( s2−25)2
=s2−25−2 s2
(s2−25 )2 =
−25−s2
( s2−25 )2
F )) (% )=
(s2−25 )2 (−2 s )−(−25−s2)2 ( s2−25 )2 s
(s2−25 )4
F )) (% )=
(s2−25 )[−2 s (s2−25 )−(25−s2 )4 s]( s2−25 )
4
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Métodos Matemáticos /2-ene#o-/
F )) (% )=
−2 s3+50 s+100 s+4 s3
( s2−25 )4
=2 s
3+150 s
( s2−25 )3
¿− d
2
ds2 [ ss2−25 ]
¿− d
2
ds2 [ F (%) ]
¿ F )) (%)
¿2 s
3+150 s
(s2−25 )3
P$i*%$ "%#$%*( d% "$('.(!in
L {e5 t t 3 }
a=5… f ( t )=t 3
L {t 3 }% # %−5=3'
%4
% # %−5
= 3 '
(%−5 )4=
6
( %−5 )4=
6
( %−5 )4
L {e−2 t cos4 t }
a=−2… f (t )=cos4 t
L {cos4 t }% # %+2= %
%2+42
= %+2
( %+2 )2+16=
%+2
(%+2 )2+16
L {[1−et +3e−4 t ]cos5 t }
L {cos5 t −et cos5 t +3e−4 t cos5 t }
L {cos5t }− L {e t cos5 t }+3 L {e−4 t cos5 t }
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Métodos Matemáticos /2-ene#o-/
s
s2+25 -
L {cos5t } s B s J / 3 L {cos5t } s>s+4
s
s2+25
− s
s2+25
⋮ s B s J / 3s
s2+25
⋮ s B s ?
s
s2+25
− (s−1)( s−1 )
2+25
3(s+4 )(s+4 )2+25
Fn!in %'!(.n ni"($i#
F >t ? - ( >t -/ 1 ( >t J 7
L {f (t )}= L {−5= (t −1 )+6= (t −0)}
? L {−5= (t −1 ) }+ L {6= (t −0 ) }
−5 L {= (t −1 ) }+6 L {= (t −0)}
? -1
s e
−s+61
s e
0=−5e−5
s +
6
5…=
6−5e−5
s
T$(n',#$*(d( d% .( in"%$(.
L
{∫0t
e?
d?
}f ( t )=e t …=
1
% L {f ( t ) }=1
% [ 1%−1 ]
L{∫0
t
cos ? d? }
f ( t )=cos t …=1
% L {f ( t ) }=1
% [ %%2+1 ]= 1%2+1
L{∫0
t
e−? cos ? d? }
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Métodos Matemáticos /2-ene#o-/
f ( t )=e−t cos t …=1
% L {f ( t ) }=1
% [ %−(−1 )(%+1 )2+1 ]=1% [ %+1( %+1 )2+1 ]C#n/#.!in
T%#$%*( d% .( !#n/#.!in
L{∫0
t
e? sin (t −? ) d? }
f ( t )=e t ; " (t )=sin t …=¿ L {et } L {sin t }
¿[ 1%−1 ][ 1%2+1 ]= 1( %−1 ) ( %2+1)
INSTITUTO POLIT@CNICO NACIONAL
UNIDAD PROFESIONALINTERDISCIPLINARIA DE IN;ENIERBA
CIENCIAS SOCIALES ADMINISTRATI
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Métodos Matemáticos /2-ene#o-/
M@TODOS MATEMTICOS DE LA IN;ENIERBA
S%!%n!i(> 2IV3A
In"%$(n"%>
CeNaos C#z And#és
TERCER CORTE
M@TODOS MATEMTICOS DE LA IN;ENIERBA
SERIE COMPLEJA DE FOURIER
f ( x )=a0
2 +∑
n=1
∞
[ancos nπ p x+bnsen nπ p x]
cos nπ
p x=
einπ
p x
+e−inπ
p x
2
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Métodos Matemáticos /2-ene#o-/
sen nπ
p x=
einπ
p x
+e−inπ
p x
2
ei@=cos@+isen@
e−i@=cos@−isen@
D%ni!in d% '%$i% !#*.%( d% F#$i%$
L( '%$i% !#*.%( d% F#$i%$ d% .( ,n!in , d%nid( %n n in"%$/(.#d% GHH %'" d(d( #$>
f ( x )=∑−∞
∞
C n einπ
p x
C n= 1
2 p∫− p
p
f ( x) e−inπ
p x
dx;n=0,±1,±2,±3,± …
En!#n"$($ .( '%$i% !#*.%( d% F#$i%$ d% .( ,n!in
1) f ( x )={−1 ;−2
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Métodos Matemáticos /2-ene#o-/
C n=−14 ∫
0
2
e
−inπ 2
x
dx+ 1
4∫0
2
e
−inπ 2
x
dx
u=−inπ
2 x du=
−inπ 2
dx
dx=−inπ 2
dx
C n=−14 ∫−2
0
eu[−2duinπ ]+ 14∫
0
2
eu[−2duinπ ]
C n= +24 inπ
[ eu ] 20−
2
4 inπ [ eu ] 2
0=
1
2inπ [e
−inπ 2
x ] 0−2
− 1
2inπ [e
−inπ 2
x ]20
C n= 1
2 inπ [e0−e
−inπ 2 (−2)]− 1
2inπ [e
−inπ 2 (−2)
−e0 ]
C n= 1
2 inπ [1−einπ ]− 1
2inπ [e−inπ −1 ]
2) f ( x )=e− x
;−π
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Métodos Matemáticos /2-ene#o-/
C n= 1
2(π )∫−π
π
e− x
∙ e
−inπ π
x
dx
e
(¿¿− x ∙ e−inx)dx
C n= 1
2 π ∫−π
π
¿
C n= 1
2 π ∫−π
π
e(− x−inx )
dx
C n= 1
2 π ∫−π
π
e(−1−¿) x
dx
u=−(1−¿) x du=−(1−¿)dx
dx=−du1+¿
C n= 1
2 π ∫−π
π
eu[ −du(1+¿) ]
C n= −12 π (1+¿)∫−π
π
eu du
C n= −12 π (1+¿ )
eu| π −π = −12π (1+¿ ) e−(1+¿) x| π −π
C n= −12 π (1+¿ )
C n= −12 π (1+¿ )
C n= −12 π (1+¿ )
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Métodos Matemáticos /2-ene#o-/
C n= −12 π (1+¿ )
L( '%$i% d% F#$i%$ !#*.%( d% .( ,n!in , d%nid( %n 0 L)%'" d(d( #$
f ( x )=∑n=−∞
∞
C n e2inπ
L x
C n=1
L
∫0
L
f ( x )e−2 inπ
L x
dx
En!#n"$($ .( '%$i% !#*.%( d% .( ,n!in f ( x )= x ;0
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Métodos Matemáticos /2-ene#o-/
T$(n',#$*(d( d% L(.(!%
S%( , ") n( ,n!in d%nid( %n %. in"%$/(.# d% K0 ∞ ) t 0
L( "$(n',#$*(d( d% L(.(!% d% , ") %' .( ,n!in f (%)
∫0
∞
f (t )e−st dt
F ( % )=∫0
∞
f (t )e−st dt
¿ limb #∞∫0
b
f (t )e−st
dt
L {f (t )}= F (%)
1) f ( t )=1
L {f (t )}= L {1 }=limb # ∞
∫0
b
1∙ e−st
dt
¿ limb #∞
∫0
b
e−st
dt
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Métodos Matemáticos /2-ene#o-/
¿ limb #∞
[−1s e−st
]b0
¿ limb #∞ [−1s e
−sb
−( 1s e0
)]
¿ limb #∞
[−1s e−sb
+limb #∞
1
s ]¿−1
s lim
b # ∞
e−sb+
1
s
¿−1
s
limb #∞
1
esb +
1
s
L {1}=1
s
D%$i/(d( d% n( "$(n',#$*(d(
Si L {f (t ) = F (%)
L {t n f ( t ) }=(−1 )n d( n)
d% (n )
F (%)
U'% .( d%$i/(d( d% n( "$(n',#$*(d( ($( %n!#n"$($
1) L {t e3t }= 1
(%−3)2
L {t eat }= 1(%−a)2
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Métodos Matemáticos /2-ene#o-/
L {t e3t }=(−1 )n d(n)
d%(n) F (%)
L {t e3t }=(−1 )1 dd%
F (%)
f ( t )=e3 t
L {f (t )}= L {e3t }= 1%−3
F ( % )= 1
%−3
L {te3t }=(−1) dd% [ 1%−3 ]
L {te3t }=−(%−3 ) (0 )−1(1)(%−3)2
¿− −1
(%−3)2
¿ 1
(%−3)2
P$i*%$ "%#$%*( d% "$('.(!in
Si L {f (t )}= F ( % ) + a∈ 0 , %n"#n!%'>
L {eat f (t ) }= F ( %−a )
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Métodos Matemáticos /2-ene#o-/
L {eat f (t ) }= L {f ( t )}
¿ F (%)
¿ F (%−a)
1) L {e5 t
t 3 }= 3 '
(%−5)4=
6
(%−5)4
L {e5 t t 3 }= L { f (t )}
¿ L {t 3 }
¿ 3 '(%−5)4
= 6(%−5)4
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