Parcial Ecuaciones Diferenciales 2015B 1

7/24/2019 Parcial Ecuaciones Diferenciales 2015B 1

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Profesor: CASTRO VIDAL RAUL PEDRO

Integrantes Cdigo

Armas Espinoza Christopher Darwin 1323220524

Baldeon Trinidad Victor Antonio 1323210036 Chipao Ac!a "is Al#erto 13232100$$ Cho%%e Csihaman &enzo 'saac 132322024( )into *dar +aldan Ale,is 132321025$ -ica!a .alazar /lio Ale,ander 1323210223 &amrez &aza /aime &al 100097G

TRABA O: Resolucin del examen arcial

4

Universidad

Nacional de Callao

Ciencia y Tecnologa rumbo al tercer milenio

ECACIO!E"#I$ERE!CIA%E"

&RPO:


2/15

PROBLEMA (1)

Considere el circito &"C %e se mestra en la ira con &110 ohmios "1+

C0001 7 ha#iendona #ateria %e proporciona E120cos83t9: *riinalmente no ha7

corriente en el circito ni cara en el condensador: En el instante t0 se cierra el interrptor

7 se de;a asi hallela intensidad de corriente resltante interprete eometricamente 7 hallelos


3/15

m2+110m+1000=0

m=bb24ac

2a

m=110110

24 (1000)2(1)

m1=55+90=35

m2=5590=145

xs=A em1 t+B em2 t

xs=A e35 t+B e145 t

.olci=n particlar

q' '+100q'+1000q=120cos(2t)

q=Acos (2t)+Bsen (2t)

q'=2Asen (2 t)+2Bcos(2 t)

q' '=4Acos (2 t)+4 Bsen (2t)

&eemplazando en la ecaci=n

[4Acos (2t)+4 Bsen(2 t)]+100 [2Asen (2 t)+2Bcos(2 t)]+1000 [Acos (2 t)+Bsen(2t) ]=120cos (

[4 B220A+1000 B ] sen (2t)+[4A+220B+1000A ] cos (2t)=120cos (2 t)

996 B220A=0


4/15

996A+220 B=120

A=0,115

B=0,0254

xp=0,115cos (2t)+0,0254 sen(2 t)

.olci=n eneral

xg=xs+xp

xg=A e35 t+B e145 t+0,115 cos (2t)+0,0254 sen(2 t)

q (t)=Ae35t+B e145t+0,115cos (2 t)+0,0254 sen(2 t)

q (0 )=A+B+0.115=0

A+B=0,115 .()

'q (t)t

=I(t)=35A e35t145B e145 t0,23cos (2 t)+0.050 sen (2t)

I(0 )=35A145 B+0,0508=0

35A145B=0,0508.()

&elacionando 8>9 7 8>>9

A+B=0,115

35A145B=0,0508

A=0,093

B=0,022


5/15

q (t)=0,093 e35 t0,022 e145 t+0,115 cos (2t)+0,0254sen(2 t)

PROBLEMA (2)

&esel


6/15

.olci=n particlar?

d' (x)

d x2 +2

d (x )dx

+ (x )=ex !nx+x3ex+e2xxsen(2x)

c (x ) :

b1

(x )=ex @b2 (x )=

x

ex

Calclando el wrosiano deb1 (x ) 7

b2 (x )

$(x )=

| e

x x

ex

d

d x(ex ) d

d x(xex )|$(x )=|e

x x

ex

ex exx

ex|=e2x

%(x )=ex !nx+x3ex+e2xxsen(2x)

&1 (x )=%(x )b2 (x )

$(x ) dx

&2 (x )=%(x)b1 (x )

$(x ) dx

"a solci=n particlar est dada por?

p(x )=&1 (x )b1 (x )+&2 (x )b2 (x )

Calclando&1 (x ) ?


7/15

&1 (x )=exx(x3

ex+ln (x)

ex +e2xxsen(2x))dx

169x2156x+46

2e3xcos (2x )

&1 (x )=

120

(x2(4x3+10ln (x )5))+

Calclando&2 (x ) ?

&2 (x )= ex(x3

ex+ln (x)

ex +e2xxsen(2x))dx

&2 (x )=x

4

4x+x!n (x )

1

169e3xcos2x (26x12 )+

1

338e

3x (78x10 ) sen2x

"a solci=n particlar %edara as?

p(x )=&1 (x )b1 (x )+&2 (x )b2 (x )

p(x )=1

exx(120(x2 (4x3+10ln (x )5))+2e

3xcos (2x )(169x2156x+46)

2197

e3x (507x2130x18) sen2x

2197 )+1ex

.impliicando?

p(x )=( 462197 92197 )e(2+2 )x+( 462197+ 9 2197 )e(22 )x+ x5

20ex

3x2

4ex( 6169 + 5 338 )e(2+2 )xx( 6169 5338 )e(2

(x )=c (x )+p (x )

(x )=c1

ex+

c2x

ex +( 462197 9 2197 )e(2+2 )x+( 462197 + 92197 )e(22 )x+ x

5

20ex

3x2

4 ex( 6169+ 5 338 )e(2+2 )xx( 6169 533


8/15

PROBLEMA (3)

.pona la estrctra representada por el siiente modelo

De la segunda ley de newton:

F=m.a

Solucin

!n!"al#


9/15

1000co$%0& ' 10 ' (%00) ' (0)* = 10)**

)** + ()* + (%0) = 100co$%0& , 9.-

Hallando Xc:

2 260 0

r1222

24(1)(260)2

r1 F1 32:1(;

r2 F1 G 32:1(;

Hallando Xp:

)ara la ecaci=n?

)** + ()* + (%0) = 100co$%0&/ 1/

)1= Aco$%0&/ + 2$!n%0&/

)1*= '%0A$!n%0&/ + %02co$%0&/

)1**= '3%00Aco$%0&/ ' 3%002$!n%0&/

)1** + ( )* + (%0 ) = 100co$%0&/

1(02'3340A/co$%0&/ + '33402'1(0A/$!n%0&/= 100co$%0&/

1(02'3340A=100

'33402'1(0A=0

A= '0.0(9 2= 0.001

)1= '0.0(9co$%0&/ +0.001$!n%0&/

)ara la ecaci=n?

)** + ()* + (%0) = '9.- (/

)c=


10/15

)(= '0.037

Solucin general:

t C1 etcos (32.18 t)+C2etsin(32.18 t) F0:02$cos860t9 F0:001sen860t9 F0:03H

Condiciones iniciales:

809 0

8o9 0

t 0.066e

tcos (32.18 t)0.065etsin (32.18 t) F0:02$cos860t9 F0:001sen860t9 F0:03H

PROBLEMA (4)

.i se tiene dos tan%es interconectados con los rios A:BC cada tan%e contiene 150" de

aa con 150r de sal diselta 7 el sendo contiene 60r? De sal diselta 7 por A inresa

solo aa a razon de 4" por minto: "a mezcla siempre se mantiene homoenea 7 los tres

ca!os se a#ren simltaneamente t0 determinar en %e tiempo ha#ra la ma,ima cantidad

de sal en el sendo tan%e si por B 7 C circla la solcion a razon de 4" por minto:

Solucin:

"la


11/15

)ara el tan%e JK 1?

&0=150 !

(0=150 g

Condiciones 'niciales Lt0M

C0=

(0

&0=

150

150=1

g

!

"eo sea?

a? raz=n de


12/15

)araq1=a .c1 dando? c1 7 c2 son concentraciones %e

entranq2=b . c2 7 salen respecti


13/15

"eo sa#emos?

#? raz=n de


14/15

q3=0,026(2(t)

d(2

dt =q

3q

2=0,026(

2 (t)0,026 [e0,026 t+5,01 ]

(*2+ (0,026(2)=0,026 e0,026t0.1302

&esol


15/15

Di:iimo$# /

2x

ln (a )(a)=

2x tg1 (x )+ ln (a)ln (a )(a)

2x

ln

(a

)(a)

dx= 1

ln (a)-=e

/

!na

2x

ln2a(a)

=2x tg1 (x )+!na

ln2a(a)

Su$&i&u!no2x

x2+ln2(a)

=d

dx( 1

!na)

/

ln (a)+

d

dx ( 1!na )=2x tg1x+!na

ln2a(a)

Alicano "!la ! "ouc&o# gd%

dx+%

dg

dx=d

( %g)dx

d

dx ( ln (a ))=2x tg1 (x )+tg(a)

ln2 (a )(a)

ddx

x

ln (a)dx=2x tg

1(x)ln ( a )(a)

!ndx

ln (x2+1)=

ln1(x )

ln(a)+C1

ECUACIO;#

=tg1

(x )+C1 ln (a)

ln (x2+1)=tg1 (x )+C1

Parcial Ecuaciones Diferenciales 2015B 1

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