Tarea analsis - Ejercicios Espinoza

7/25/2019 Tarea analsis - Ejercicios Espinoza

1/23

Ejercicio 2

xy y=y3

Solucin:

dy

dx=y (y2+1)

xy y=y3x Separando las variables

dy

y (y2+1)=

dx

x

(y3+1y2 ) dy

y (y2+1) =

dx

x

dy

y

ydy

y3+1

= ln (x)

Hacemos:

u=y2+1 du=2ydy ln (c )+ ln (y )=1

2ln (y2+1)+ln (x )

y2+1

x

y2+1

ln (Cy )=ln

Ejercicio 6

ex+y

Sen(x ) dx+(2y+1 ) ey2

dy=0

Solucin:

ex+y

Sen(x ) dx+(2y+1 ) ey2

dy=0 ex ey Sen (x )dx+(2y+1)ey2

dy=0

Separando las variables

ex

Sen (x )dx+(2y+1 ) ey2y

dy=C

ex

Sen (x ) dx+ey2y (2y+1 ) dy=0

ex

2( Sen(x )cos (x ))ey

2y=C

ex (Sen (x )cos (x ))2ey

2y=C


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3/23

Ejercicio 14

exy

dx+eyx dy=0

Solucin

exy dx+eyx dy=0

Separando variables

ex

ey

dx+e

y

ex

dy=0

e2x

dx+e2y dy=0 Ine!rando " e2x

dx+e2y dy=0

e

2x

2 dx+

e2y

y dy=C e2x+e2y=C

Ejercicio 16

y#= 10x+y

Solucin

y#= 10x+y

Separando las variables

dy

dx=10x 10y 10x dx=10y dy

" 10x

d$=" 10y

dy 10

x

ln (10 )=

10y

ln (10 )+c 10x+10y=c


4/23

Ejercicio 2%

dy

dx=(y1)(x2)(y+3)(x1)(y2)(x+3)

Solucin:

Separando variables

(y2 )dy(y1)(y+3)

= (x2)dx

(x1)(x+3)

&racciones parciales en cada 'rmino:

(y2 )(y1)(y+3)

= A

y1+

B

y+3y2=A(y+3 )+B (y1)

y=() * ( %=+,(4- * += %.4 /=1 * (1=4 * =(1.4

Ine!rando

5

4

dy

y+3

1

4

dy

y+1=5

4

dx

x+3

1

4

dx

x1

5

4ln (y+3 )

1

4ln (y1)=

5

4ln (x+3 )

1

4ln (x1 )+ ln (c )

* 5 ln (y+3) ln (y1 )=5 ln (x+3 ) ln (x1 )+ ln ( c )

ln((y+3 )3

y1)=ln(C(x+3 )3

x1 ) (x1 ) (y+3 )2=C(y1 ) (x+3 )5


5/23

Ejercicio 2

(xy2

x) dx+(yx2

y ) dy=0

Solucin

Separando variables

x (1y2 ) dx+y (1x2 ) dy=0

Ine!rando

xdx1x2

+ ydy1y2

=0 12ln (1x2 )1

2ln (1y2 )=ln(c )

ln (1x2 )3

2 (1y2 )1

2=ln ( c ) (1x2 )3

2 (1y2 )3

2=C (1x2 ) (1y2 )=C

Ejercicio )%

(xy+2x+y+2 ) dx+(x2+2x ) dy=0

Solucin

[x (y+2)+y ] dx+ (x2+2x ) dy=0 (y+2 ) (x+1 ) dx+ (x2+2x ) dy=0

Separando variables

x+1

x2+2x

dx+ dy

y+2=0

Ine!rando

1

22 (x+1 ) dx

x2+2x

+ dy

y+2=0

En la primera ine!ral cambio de variables

u=x2+2x du=2 (x+1 ) dx du

u+2 ln (y+2 )=0

ln (u )+2 ln (y+2 )=ln (c ) ln [u (y+2 )2

]=ln (c )


6/23

d (x2+2x )(y+2 )2=C


7/23

Ejercicio )3

xdx1x4 dy=x21x4 dy

xdx=(x2+1)1x4 dy

Solucin

( Separando variables

xdx

(1+x2)1x4=dy

xdx

(1+x2)1x4=dy

( Cambio de variable:

u=x2 du=2xdx ;1

2

du

(1+u)1u2

( Susiucin ri!onom'rica:

$=Sen,- * d$=Cos,-d 5 1( u2

= cos2 ( )

y=1

2

cos ()d

(1Sen ( ))cos2()

=1

2

d

1Sen ( )=

1

2

(1Sen ( )) d1Sen2 ( )

y=1

2

(1Sen ( )) dcos

2 ( ) =

1

2tg ( )

1

2cos ( )+C


8/23

Ejercicio 42

,4$ xy2dx+(y+y x 2 ) dy=0

Solucin

* $,4 xy2dx+y (1+x2 ) dy=0

Separando variables:xdx

x2+1

+ ydy

y2+1

=0 ydy

y2+1

= xdx

x2+1

En la primera ine!ral

= y2+4 dt=2ydy

En la se!unda ine!ral

u= 4+x2

du=2xdx

2dt

t+2

du

u=0 ln ( t)+ ln (u )=ln (C)

y

( 2+4)(x2+1)=C ln (tu)=ln (C)


9/23

Ejercicio 44

dy

dx=

3x26x2y2

yx2y

Solucin

Separando variables:

12y 2

3x

2 dy

dx=

ydy

12y2=

3y2

dx

1y3

ydy

12y3=

3x2

dx

1x3

Hacemos

u=12y2 du=4ydy t=1x3 dt=3x2 dx

du/4

u =du/3

u 1

4ln (u

3

4

)=ln (Ct1

3

) u3

4=ct

1

3

Se deermina C si $=) 5 y=1

(12)3=C(127)4 C=1 /263

(1x3)4=264(12y2)3


10/23

7are 2

x (x6+1 ) dx+y2 (x4+1) dx=0 ; y (0 )=1

Solucin

Separando variables:

xdx

x4+1

+y

2dy

y4+1

=0 xdx

x4+1

+ y

2dy

y6+1

=0

En la primera ine!ral:

u=x2 du=2xdx

En la se!unda ine!ral:

t=y3 dt=3y3 dy

du /2

u2+1

+dt/3

t2+1

=01

2Arctg (u )+1

3Arctg ( t)=C

3Arctg (x2 )+2Arctg (y3 )=C

8a consane de ine!racin: y,0-=1

3Arctg (0 )+2Arctg (1 )=C C=

2

8ue!o:

3

Arctg (x2

)+2

Arctg (y3

)=

2


11/23

Ejercicio 9

x3

dy+xydx=x2 dy+2ydx=0 ; y (2 )=e

Solucin:

(x3x2 ) dy+ (xy2y )dx=0

(x3x2 ) dy+y (x2 ) dx=0

Separamos las variables e ine!ramos:

dy

y+

(x2 )dx

x3x2

=0 ln (y )+(x2 ) dx

x2 (x1 )

=0

7or racciones parciales

x2

x2 (x1 )

=A

x+

B

x2

C

x1x2=Ax (x1 )+B (x1 )+C x2

7unos cr;icos:

$=0 * (2=,0-+,(1-C,0-- *+=2

$=1 * (1=,0-+,0-C,1- *C=(1

$=(1 * ()=,2-+,(2-C,1- *()=,2-(4(1 * =1

8ue!o:

ln (y )+dx

x+2

dx

x2

dx

x1=0

ln (y )+ln (x )2

xln (x1 )=C

Hallamos la consane: y,2-=e

8n,e-8n,2-(1(8n,1-=C * 8n,2-


12/23


13/23

Ine!rando: dy

y+6yTg (2x ) dx=0L n (y )3ln [cos (2x )]=ln(C)

ln y

cos3(2x)

=ln (C) Y=cCos3(2x)

Hallamos C usando $=0 5 y=(2

2=cCos3 (0 ) C=2y=2cos3(2x )

Ejercicio 1

(1+ex )y y=ey ; y (0 )=0

Solucin:

(1+ex )ydydx

=ey Separando variables y ey dy= dx

1+ex

Ine!rando:

yey=

dx

1+exIntegramos por partes la primera integral

u=y * du=dy 5 v=" ey

dy=ey

yey

ey

=xln (1+ey

)+C

Se deermina la consane: $=0 5 y=0

0e0= ln (1+e0 )+C C= ln (2 )1

y eyey=x ln (1+ex )+ln (2 )1 ln( 1+ex

2 )x+1=(1+y )ey


14/23


15/23

7racica )

Ejercicio %

y

=(x+y )2

Solucin:

dy

dx=(x+y )2 acemos :u=x+y

dy

dx=

du

dx1

eempla>amos:

du

dx 1

=u2

du

dx =u2

+1

Separamos variables e ine!rando:

dx= du

u2+1

x+C=Arctg (u )x+y=Tg (x+C)

Ejercicio 11

y

=Sen (xy )

Solucin

Hacemos: u=$(y * u=1y

du

dx=1Sen (u )

Separamos variables e ine!rando:

dx= du1Sen(u) dx+[1+Sen (u )

]du

1Sen2 (u ) =0x+ ducos2 (u )+Sen

(u

)du

cos2 (u ) =0

x+Tg (u )+Sec (u )=Cx+Tg (xy )+Sec (xy )=C


16/23

Ejercicio 1)

(x2y3+y+x2) dx+(x3 y2+x ) dy=0

Solucin

Hacemos: u =$y * du=$dyyd$

Susiuyendo:

(u2 y+y+x2 ) dy+(x u2+x )(duydx )x

=0

(u2 y+y+x2) dx+ (u2+1 ) du (u2+1)ydx=0

(x2 ) dx+ (u2+1 ) du=0 (x2 )dx+(u2+1) du=0

x2

22x+

u3

3+ u=C3x212x+2x3y3+6xy=C

Ejercicio 20

[xy2x ln2 (y )+yLn (y )] dx+ [2x2 ln (y )+x ] dx=0 Sug;!=xLn (y )

Solucin:

Hacemos: >=$8n,y- * d> = 8n,y-d$$dy.y

$dy = yd> ( y8n,y-d$

(xy2y!2+y! /x)dx+(2!+1) [yd!yLn (y ) dx ]=0

(x2!2+! /x)dx+(2!+1) [ d!ln (y ) dx ]=0

(x2!2+! /x ) dx+ (2!+1 ) d!(2x!+1)(! /x )=0

(x2!2+! /x ) dx+ (2!+1 ) d!(2!2+! /x ) dx=0

x

2

2+ (2

!+1

)

2

4 =C2x2

+[2xLn (y )+1 ]2

+C


17/23


18/23

Ejercicio 26:

(2+4x2y )ydx+x3y dy=0

Solucin:

(2+4x2y )ydx+x3y dy=0


19/23


20/23


21/23

7racica 4

Ejercicio %:

x y=y+2x ey/x

Solucin:

8a ecuacin es ?omo!'nea de 1@ !radoA Hacemos: y = u$ * dy = ud$ $du

eempla>ando:

x (udx+xdu)=(ux+2x exx /x) d$

SimpliBcamos $: udx+xdu=udx+2ex

dx xdu=2ex dx

ex

dx=2dx

x ex=2ln (x )+ ln (#)pero u=

y

x e

yx =ln ( # x2 )


22/23

Ejercicio 6:

dy= y

xCsc2( yx) dx

Solucin:

8a ecuacin es ?omo!enea de !rado 0A

Hacemos:

y = u$ * dy= ud$ $du

udx+xdu=[uCsc2 (u )] dx xdu=Csc

2 (u ) dx

Separamos variables:

Sen2 (u ) du+ dx

x=0 Sen2 (u ) du+

dx

x=0

Sen2 (u )du+dx

x=0

1cos (2u )2

du+ln (x)=0

u Sen(u )2 +2 ln (x )=#2uSen(2u )+4 ln (x )=#

7ero u=y

x;

2y

xSen( 2yx)+4 ln (x )=#2yxSen( 2yx)+4xLn (x )=#

x


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Tarea analsis - Ejercicios Espinoza