Post on 06-Jul-2018
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17.- Resuelva la siguiente ecuación integro diferencial:
y ´ (t )+ y (t )+4∫0
t
z (u )+10=0
y ´ (t )+ z ´ ( t )+ z (t )=0
z ´ (0 )=−6 ; y ´ (t )=12
Resolución:
L[ y ´ ( t )+ y ( t )+4∫0
t
z (u )+10]s= L [0 ]s
sy ( s)− y (0 )+ y ( s)+4 z (s )
s +
10
s =0
…1
L [ y ´ (t )+ z ´ (t )+ z (t ) ]= L [0 ]
sy ( s)− y (0 )+sz (s )− z (0 )+ z ( s)=0 … ..2
Para t=0
¿ y ´ (0 )+ y (0 )+4∫0
t
z (u )+10=0
12+ y (0 )+10=0
y (0 )=−22
¿ y ´ (0 )+ z ´ (0 )+ z (0 )=0
12−6+ z (0)=0
z (0 )=−6
Luego se tiene el sistema
(s+1 ) y ( s)+(4
s
) z (s )=−22−
10
s … . α
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sy ( s )+( s+1 ) z ( s )=30…β
Hallamos la determinante
Δ=
|s+1
4
s
s s+1|=( s+2)2−4
Δ=s2+2 s−3 ; Δ ≠0
Regla de cramer
¿ y ( s )=|−22−
10
s
4
s
−30 s+1
| Δ =(−22−
10
s ) (s+1 )+
120
s
Δ
y (s )=−32s +
12
s−1
L−1 [ y (s ) ]=−32+12e t
¿ z ( s)=|s+1 −22−10s
s −30 | Δ
=−30 (s+1 )+22 s+10
Δ
z ( s)= −8s−1
z ( t )=−8et
18.- Resuelva la siguiente ecuación diferencial:
y (t)+2y ´(t)+y(t)=sen(4t)+t {e} ^ {-t} + {t} ^ {2} {e} ^ {t} +8 {δ} rsub {2π} (t)
y (0 )=2
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y ´ (0)=1
Resolución:
s2
y ( s)−sy (0 )+2 sy ( s)−2 (2 )+ y (s )= 4
s2+16+
1
s+1 + 1
( s−1)2+8e2πs
s2 y ( s)−2 s+2 sy ( s)−4+ y ( s)=
4
s2+16
+ 1
s+1+
1
( s−1 )2+8 e−2πs
y (s ) ( s2+2 s+1 )=2s+5+ 4s2+16
+ 1
s+1+
1
(s−1 )2+8e2πs
y (s )= 2 s(s+1 )2
+ s(s+1 )2
+ 4
( s2+42 ) ( s+1 )2+
1
( s2+4 ) ( s+1 )2+
1
( s2+4 ) ( s−1 )2+8e−2 πs
( s+1 )2
y (s )=2 ( s+1 )
(s+1 )2 −
2
(s+1 )2+
5
(s+1 )2+
4
(s2+4 ) (s+1 )2+
1
(s2+4 ) ( s+1 )2+
1
(s−1 )2 (s+1 )2+8e
−2πs
(s+1)2
y (s )= 2(s+1 )
−3
( s+1 )2+
4
( s2+4) ( s+1 )2+
1
(s+1)3+
1
( s−1 )2 ( s+1 )2+
8 e−2 πs
(s+1 )2
y (s )= 2(s+1 )
+ 3
(s+1 )2−
50
51 (s2+16 )+
145
867 (s+1 )+
41
17 (s+1 )2+
1
(s+1 )3+
1
(s−1)2 ( s+1 )2+8e
−2πs
(s+1 )2
Reduciendo:
y (s )=0.8616 1
(s+1)+304852
1
( s+1 )2+8e
−2πs
(s+1 )2+0.1944
1
( s−1 )+0.25(
1
( s−1 ))2
50
51 ( s2+16 )+1
2( 2
s+1)3
ntonces:
y (s )=0.8616 e−t +304852 t e−t +0.1944 e t +0.25 e t +0.2450 sen4 t +0.5et t 2+8 (t −2 π ) et −2π ; t >2π
!
y (s )=0.8616 e−t +304852 t e−t +0.1944 e t +0.25 e t +0.2450 sen4 t +0.5et t 2+; t
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1".- y ´ (t )=!s4 t −2∫0
t
y (θ ) !s2 (θ−t )dθ; y (0=2)
Resolución
sy ( s)− y (0 )= s
s2+16
−2 y (s) s
s2+4
sy ( s)−2= s
s2+16
− y (s) 2s
s2+4
sy ( s)+ y (s ) 2 s
s2+4
= s
s2+16
+2
y (s )(s+ 2ss2+4 )= s
s2+16
+2
y (s )=
s
s2+16
s3+6 s
s2+4
+ 2
s3+6 s
s2+4
y (s )= s
2+4(s2+16 )(s2+6)
+2(s2+4)
s(s2+6)
y (s )= 3
10.
1
( s2+16 )+ 7
15. 1
(s2+6)+4
3. 1
s
ntonces:
y (t )= 3
10sen4 t +
7
15√ 6sen√ 6 t +
4
3
( ) ( ) ( ) ( )
( )
( )
#0$ Resuelva el P%&:
' 8
0 0
0 1
X t X t X t F t
X
X
′′ ′+ + =
=
′ =
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4224
2π 4π 6 π 8π 10π t
(onde ) es la función *eriódica est+ dada en la siguiente figura:
),t$
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( ) ( ) ( ) ( )
[ ]
( )
( )
#
#
# #
:
*licamos L al P%&:
' 8
, $ ,0$ ,0$ ' , $ ,0$ 8 , $ , $
, $ ' 8 , $ 1
, $ 1, $
' 8 ' 8
Pero:
, $
# , # $ , $
st
Solucion
L X t L X t L X t L F t
s x s sx x sx s x x s f s
x s s s f s
f s x s
s s s s
f s L F t
e t dt F t F t L F t π
−
′′ ′+ + = ′− − + − + =
+ + = +
= ++ + + +
=
+ = ⇒ = #
#
0
# #
# # # # #
0
#
#0
## # # , 1$
#0 0
##
#
,1 $
' # #, $
,1 $ ,1 $ ,1 $
/e sae ue:
1
1
11
si 2acemos n n-11
1
s
s s
s s s
n
n
sn
sn
s sn s s n
sn n
s s
s
e
e e f s
e s e s e s
ee
ee e e
e
ee
e
π
π
π π
π π π
π
π
π
π π π
π
π
π
π
π
π π
−
− −
− − −
+∞
=
+∞−
−=
− +∞ +∞− − − +
−= =
−−
−
−
−= − +
− − −
=−
=−
= = ⇒−
=−
∫
∑
∑
∑ ∑
1
# # #
# #1 1 0
# # #
# # #1 1 1
#
#1
#
#1
Reem*la3ando:
# #, $ '
# # 1, $ '
#, $ '
2ora:
#, $ '
, ' 8$
n
n
sn sn sn
n n n
sn sn sn
n n n
sn
n
sn
n
e e e f s
s s s
e e e f s
s s s s
e f s s s
e x s
s s s
π π π
π π π
π
π
π π
π π
π
+∞
=
− − −+∞ +∞ +∞
= = =
− − −+∞ +∞ +∞
= = =
−+∞
=
−+∞
=
= − − +
= − − + + ÷
= − +
= − ++ +
∑
∑ ∑ ∑
∑ ∑ ∑
∑
∑
[ ]
# # #
# # #
#1
1 # # #
0
1 # #
# 0 0
1.........., $
, ' 8$ ' 8
Hacemos:
1 1, $
' 8 , #$ #
#, $
#
, $ # # # 1
# 8 8 8
, $ # # 1
# 14 14 8
*
t
t u t t
u t t t
s s s s s
r s s s s
e Sen t L r s
r s e Sen u du e Sen t e Cos t L
s
r s e Sen u du e Cos t t L
s
α
π
−−
− − − −
− − −
++ + + +
= =+ + + +
⇒ =
⇒ = = − − +
⇒ = = − +
∫
∫ ∫
[ ] [ ]
1
#11 1 1
#1
# #1
#
1 , # $
#
licamos L , $ 5 reem*la3ando :
, $ # , $, $ ' , $
, $ # # 1 #, $ ' , $ , $
14 14 8 #
)inalmente:
, $
sn
n
t t
n
n t n
n
a
e r s r s L x s L L L r s
s s
r s e Cos t t e Sen t X t U t L
s
X t U
π
π
π
π
α
π
π
−
−+∞−− − −
=
− −+∞−
= −
= − + +
= − + − + +
=
∑
∑
#, # $ #, # $ # #
1
#, # $ #, # $ 1 1 # 1 #, $, $ , $
# # # 8 8 ' #
t n t n t t
n
e Sen t n e Cos t n e Cos t t e Sen t t
π π
π π
π
− − − − − −+∞
=
− −+ − + − + +∑
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