Simulacro Examen Final
of 20
-
Upload
fran-poma-castellanos -
Category
Documents
-
view
8 -
download
0
Transcript of Simulacro Examen Final
-
FRANKLIN POMA CASTELLANOS
RESPONSABLE DEL CURSO: ORTEGA VARGAS, JORGE LUIS
DESARROLLO DEL
SIMULACRODEL EXAMEN FINAL
DE EDO
-
P g i n a | 1
ASIGNATURA: ANALISIS MATEMTICO IV
1. Resolver las siguientes ecuaciones diferenciales lineales o reducibles a ella:
a) . (2). = ( 2). .
. (2). = ( 2). .
. = ( 2). . + (2).
. = (( 2). + (2))
.
= ( 2). + (2)
=
( 2).
+
(2)
(2)
=
( 2).
+ (
2
) =
( 2).
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= (2). [ (
2).. (
( 2).
) . + ]
= 2 [ 2
. (
( 2).
) . + ]
= 2ln () [ 2ln (). ( 2
) . + ]
= 2 [1
2. ( 2
) . + ]
= 2 [ (
2 2
3) . + ]
-
P g i n a | 2
ASIGNATURA: ANALISIS MATEMTICO IV
= 2 [ (
2) . 2 (
3) . + ]
:
= =
3
= . =1
2.2
= 2 [ (
2) . 2 ( . (
1
2. 2) (
1
2
2. )) + ]
= 2 [
2. . (
1
2)
2. + ]
= 2 [(
2) +]
= + 2.
b)
6 = 10. (2)
6 = 10. (2)
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= 6 . [ 6 . (10. (2)). + ]
= 6 [ 6. (10. (2)). + ]
= 6 [10 6. ((2)). + ]
:
-
P g i n a | 3
ASIGNATURA: ANALISIS MATEMTICO IV
6. ((2)).
= (2) = 6
= 2 cos(2) . = 6
6
6. (2). = 6
6(2) (
6
6) 2 cos(2) .
6. (2). = 6
6(2) +
1
3 6. cos(2) .
= (2) = 6
= 2 sen() . = 6
6
6(2). =6
6(2) +
1
3(
6
6(2) (
6
6) (2 sen(2) ))
6(2). =6
6(2) +
1
3(
6
6(2)
1
3 6 sen(2) )
6(2). =6
6(2)
6
18(2)
1
9 6 sen(2)
10
9 6 sen(2) =
6
6(2)
6
18(2)
6 sen(2) =36
20(2)
6
20(2)
:
= 6 [10 (36
20(2)
6
20(2)) + ]
= 6 [(36
2(2)
6
2(2)) + ]
=3
2(2)
1
2(2) + . 6
c) . + ( + 3) = 0
. + ( + 3) = 0
-
P g i n a | 4
ASIGNATURA: ANALISIS MATEMTICO IV
. = ( + 3)
.
= + 3
=
+ 3
= (1 +
1
) + 3
+ (1 +
1
) = 3
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= (1+
1). [
(1+1).. 3 + ]
= ln || [3 +ln ||. + ]
= . ln || [3 . ln ||. + ]
=3
. [ . . + ]
:
= = .
= =
=3
. [. . + ]
=3
. [. + ]
-
P g i n a | 5
ASIGNATURA: ANALISIS MATEMTICO IV
=3
. [. + ]
= 3 (1 1
+
. )
d) + (2. + (2)). = 0
+ (2. + (2)). = 0
= (2. + (2)).
= (2. + (2))
+ . 2. = (2)
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= 2.. [ 2... ((2)). + ]
= 2.. [ 2... ((2)). + ]
= 2 . [ 2 .. ((2)). + ]
= 2.|()| [ 2.|()|. ((2)). + ]
=1
2()[ 2(). ((2)). + ]
=1
2()[ 2(). 2(). cos (). + ]
-
P g i n a | 6
ASIGNATURA: ANALISIS MATEMTICO IV
=2
2()[4()
4+ ]
= 2()
2+
2
2()
e) (1 + 2) = 2(1 22)
(1 + 2) = 2(1 22)
( + 3)
= (2 42)
=
(2 42)
( + 3)
=
2
( + 3)
(42)
( + 3)
+
(42).
( + 3)=
2
( + 3)
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= (
(42)(+3)
).[
((42)
(+3)).
. (2
( + 3)) . + ]
= 2 (
2(1+2)
).[
2 (2
(1+2)).
. (2
( + 3)) . + ]
= 2.|1+2| [ 2.|1+
2|. (2
( + 3)) . + ]
=1
(1 + 2)2[(1 + 2)2. (
2
(1 + 2)) . + ]
-
P g i n a | 7
ASIGNATURA: ANALISIS MATEMTICO IV
=1
(1 + 2)2[(1 + 2). (
2
) . + ]
=2
(1 + 2)2[ (
1
+ ) . + ]
=2
(1 + 2)2[ (
1
) . + . + ]
=2
(1 + 2)2[|| +
2
2+ ]
f) 1()
+ 2() = 1
1()
+ 2() = 1
+
2()
1()=
1
1()
:
+ (). = ()
:
= (). [ ().. (). + ]
:
=
2()1()
.[
2()1()
.. (
1
1()) . + ]
=
2()1()
.[
2()1()
..
1()+ ]
g) .
= (1 . ()). 2. cos ()
= (1 . ()). . cos ()
-
P g i n a | 8
ASIGNATURA: ANALISIS MATEMTICO IV
(1 . ()). . cos() = 0
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= (1.())..cos(). [ (1.())..cos().. 0. + ]
= (.cos()2()) [ 0. + ]
:
= .++2+[0 + ]
= . .++2+
h) (2 + 2) ( + 2 + 3) = 0
(2 + 2) = ( + 2 + 3)
(2 + 2) = ( + 2 + 3)
(2 + 2)
= ( + 2 + 3)
=
( + 2 + 3)
(2 + 2)
=
2 + 2+
2 + 3
2 + 2
2 + 2=
2 + 3
2 + 2
(
2 + 2) =
2 + 3
2 + 2
-
P g i n a | 9
ASIGNATURA: ANALISIS MATEMTICO IV
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= (
2+2
).[
(
2+2).
. (2 + 3
2 + 2) . + ]
= 12 (
22+2
).[
12 (
22+2
).. (
2 + 3
2 + 2) . + ]
= 12|2+
2| [ 12|2+
2|. (2 + 3
2 + 2) . + ]
= 12|2+
2| [ 12|2+
2|. (2 + 3
2 + 2) . + ]
= |2+2| [
|2+21
|. (
2 + 3
2 + 2) . + ]
= 2 + 2 [1
2 + 2. (
2 + 3
2 + 2) . + ]
= 2 + 2 [ (2 + 3
(2 + 2)32
) . + ]
= 2 + 2 [ (2
(2 + 2)32
) . + (3
(2 + 2)32
) . + ]
:
= 2 = (2 + 2)3
2. .
= 2. =(2+2)
32
5
= 2 + 2 [2
2 + 2+ (
2. (2 + 2)32
5
2
5(2 + 2)
32 . 2. ) + ]
-
P g i n a | 10
ASIGNATURA: ANALISIS MATEMTICO IV
= 2 + 2 [2
2 + 2+ (
2. (2 + 2)32
5
2
5(
2
5. (2 + 2)
52)) + ]
= 2 + 2 [2
2 + 2+
2. (2 + 2)32
5
4
25. (2 + 2)
52 + ]
= 2 +2. (2 + 2)2
5
4
25. (2 + 2)3 + 2 + 2
i) (1 + 2) = ( )
(1 + 2) = ( )
(1 + 2)
= ( )
=
(1 + 2)
(1 + 2)
+
(1 + 2)=
(1 + 2)
:
+ (). = ()
:
= (). [ ().. (). + ]
:
=
1(1+2)
.[
1
(1+2).
.
(1 + 2). + ]
= [ .
(1 + 2). + ]
:
= =
(1+2).
-
P g i n a | 11
ASIGNATURA: ANALISIS MATEMTICO IV
=
1+2 =
= [(. .
1 + 2) + ]
= [(. ) + ]
= 1 + .
j) (1 + ) = [2 ( + )].
(1 + ) = [2 ( + )].
(1 + )
= [2 ( + )]
= [
2
(1 + )
( + )
(1 + )]
+
( + )
(1 + )=
2
(1 + )
:
+ (). = ()
:
= (). [ ().. (). + ]
:
=
(+)(1+)
.[
(+)
(1+).
. (2
(1 + )) . + ]
=
(1+
)
(1+).
[
(1+
)
(1+).
. (2
(1 + )) . + ]
= sec (). [ sec ().. (2
(1 + )) . + ]
= ln|+| [ ln|+|. (2
1 + ) . + ]
-
P g i n a | 12
ASIGNATURA: ANALISIS MATEMTICO IV
=1
+ [( + ). (
2
1 + ) . + ]
=1
+ [ ( (
2
1 + ) + (
2
1 + )) . + ]
=1
+ [ ((
2
1 + ) + (
2.
1 + )) . + ]
=1
+ [ ((
2
1 + )) . + ((
2.
1 + )) . + ]
=1
+ [ ((
2 + 2.
1 + )) . + ]
=1
+ [ (
2(1 + )
1 + ) . + ]
=1
+ [ 2. + ]
=1
+ [2 + ]
=2 +
+
k) 2. cos() .
= 2 1 =
. =
2.cos() .
= 2 1
:
2.
= 2 1
=
(2 1)
2
=
2
1
2
-
P g i n a | 13
ASIGNATURA: ANALISIS MATEMTICO IV
2
=
1
2
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= (2) [
(2). (
1
2) . + ]
= 2.|| [ 2.||. (1
2) . + ]
= 2 [ (1
2) . (
1
2) . + ]
= 2 [ (1
4) . + ]
= 2 [ 4. + ]
= 2 [3
3+ ]
= [1
3+ 2]
= 2 1
3
= 2 1
3
= (2 1
3)
l) (3 3 2. ) + 32 = 0 : 3 =
-
P g i n a | 14
ASIGNATURA: ANALISIS MATEMTICO IV
32 =
(2 2. ) + = 0
(2 2. ) =
(2 2. ) =
(2 ) 2. =
+ (2 ) = 2.
+ ( 1) = .
:
+ (). = ()
:
= (). [ ().. (). + ]
:
= (1). [ (1).. (. ). + ]
= (1)2
2 [ (1)2
2 . (. ). + ]
= (1)2
2 [ 22+1
2 . (. ). + ]
= (1)2
2 [ 22+1
2 +. + ]
= (1)2
2 [ 2+1
2 . + ]
= (1)2
2 [1
2
2+1. + ]
= (1)2
2 [1
2. 2
2+1. 2 + ]
-
P g i n a | 15
ASIGNATURA: ANALISIS MATEMTICO IV
= (1)2
2 [1
2. 2
2+1 + ]
=
2+21
2[
1
2. 2
2+1 + ]
=24
2+ .
2+23
: 3 =
3
=
24
2+ .
2+23
3 =. 24
2+ . .
2+23
= . 24
2+ . .
2+233
2. Resolver las Edo coeficientes constantes:
a. 2 + 2 = 0
2 + 2 = 0
3 22 + 2 = 0
Factorizamos por Ruffini:
1 -2 -1
-1
-1 3
-2
2
1 -3
2
2
-2
0
1
1
-1
1
0
1 0
-
P g i n a | 16
ASIGNATURA: ANALISIS MATEMTICO IV
( + 1)( 2)( 1) = 0
1 = 1
2 = 2
3 = 1
= 1. +2.
2 + 3.
b. + 3 + 3 + = 0
+ 3 + 3 + = 0
+ 3 + 3 + = 0
3 + 32 + 3 + 1 = 0
Factorizamos por Ruffini:
1 3 3
-1
-1 -2
1
-1
1 2
-1
1
-1
0
-1
1
-1
1
0
1 0
( + 1)( + 1)( + 1) = 0
( + 1)3 = 0
1 = 1
2 = 1
3 = 1
-
P g i n a | 17
ASIGNATURA: ANALISIS MATEMTICO IV
= 1. +2.
+ 23
c. = 0
= 0
4 1 = 0
(2 1)(2 + 1) = 0
( 1)( + 1)(2 + 1) = 0
1 = 1
2 = 1
3; 4 =(0) (0)2 4(1)(1)
2(1)
3; 4 =4
2
3; 4 =2
2
3; 4 = 0
= 1. +2.
+ 3 cos() + 4()
d. 4 + 6 4 + = 0
4 + 6 4 + = 0
4 43 + 62 4 + 1 = 0
Factorizamos por Ruffini:
1 -4 6 -4
1
1 -3 3
-1
1
1 -3 3
1 -2
-1
1
0
-
P g i n a | 18
ASIGNATURA: ANALISIS MATEMTICO IV
1
1 -2
1
1
-1
0
1
-1
1
0
1 0
( 1)( 1)( 1)( 1) = 0
( 1)4 = 0
1 = 1
2 = 1
3 = 1
4 = 1
= 1. +2.
+ 23 + 33
e. 3 + 2 4 7 + + 5 = 0
3 + 2 4 7 + + 5 = 0
35 + 24 43 72 + + 5 = 0
Factorizamos por Ruffini:
3 2 -4 -7 1 5
-1
-3 1 3 4
-5
1
3 -1 -3 -4
3 2 -1
5
-5
0
3 2 -1 -5 0
( + 1)( 1)(33 + 22 5) = 0
-
P g i n a | 19
ASIGNATURA: ANALISIS MATEMTICO IV
DATOS DEL ALUMNO:
APELLIDOS: POMA CASTELLANOS, FRANKLIN
SEMESTRE: IV
ASIGNATURA: ANALISIS MATEMTICO IV
CATEDRTICO: JORGE LUIS ORTEGA VARGAS
