Universidad de las Fuerzas ArmadasEcuaciones Diferenciales Ordinarias

Deber 6: Ecuaciones no exactas - primer y segundo caso

C. Benalcazar y T. Villacres

Para las siguientes ecuaciones diferenciales, identifique el caso de factor integrante y

encuentre una solucion general.

1. (3x + 2y2)dx + 2xydy = 0

My

= 4y ; Nx

= 2yPrimer Caso:My N

x

N

f(x) =4y 2y

2xy=

1

xu(x) = elnx = x

Aplicamos u(x)(3x2 + 2xy2)dx + 2x2ydy = 0

My

= 4yx ; Nx

= 4yx

F (x, y)/F1 =

(3x2 + 2xy2)dx = x3 + x2y2 y F2 =

2x2ydy = x2y2

F (x, y) = x3 + x2y2 = C

3. ydx + (3 + 3x y)dy = 0My

= 1 ; Nx

= 3Segundo Caso:

Nx M

y

M

g(y) =3 1y

=2

yu(y) = eln(y

2) = y2

Aplicamos u(y)y3dx + (3y2 + 3xy2 y3)dy = 0

My

= 3y2 ; Nx

= 3y2

F (x, y)/F1 =

y3dx = xy3 y F2 =

(3y2 + 3xy2 y3)dy = y3 + xy3 y

4

4

F (x, y) = y3 + xy3 y4

4= C

1

4. (3x2 + y + 3x3y)dx + xdy = 0

My

= 1 + 3x3 ; Nx

= 1Primer Caso:My N

x

N

f(x) =1 + 3x3 1

x= 3x2

u(x) = ex3

Aplicamos u(x)ex

3(3x2 + y + 3x3y)dx + xex

3dy = 0

My

= ex3

+ 3x3ex3

; Nx

= ex3

+ 3x3ex3

F (x, y)/F1 =

(3x2ex

3+ yex

3+ 3x3yex

3)dx

= ex3

+ yex

3dx + 3y

x3ex

3

si u = x3; x = u

1

3 ; dx =du

3u23

= ex3

+y

3

euduu

23

+ y ueudu

u23

= ex3

+y

3

euduu

23

+ yu

1

3 eudu

= yu

1

3 eu y euduu

13

+ yu

1

3 eu

F1 = ex3 + xyex

3y

F2 =xex

3dy = xyex

3

F (x, y) = ex3(xy + 1) = C

5. ydx (x3 3x)dy = 0My

= 1 ; Nx

= 3Segundo Caso:

Nx M

y

M

g(y) =3 1y

=2

yu(y) = eln(y

2) = y2

Aplicamos u(y)y3dx (y5 3y2x)dy = 0

My

= 3y2 ; Nx

= 3y2

F (x, y)/F1 =

y3dx = y3x y F2 =

(y5 + 3y2x) = y3x y

6

6

F (x, y) = y3x y6

6= C

6. (x2y + y3)dy xdx = 0

2

My

= 0 ; Nx

= 0Segundo Caso:

Nx M

y

M

g(y) =2xy

x = 2yu(y) = ey

2= x

Aplicamos u(y)(xey2)dx + ey2(x2y + y3)dy = 0

My

= 2yxey2

; Nx

= 2yxey2

F (x, y)/F1 =

(xey

2)dx =

xey22

y

u = y2 ; du = 2ydy ; dv =yey

2; v =

ey22

F2 =ey

2(x2y + y3)dy

= x2ey

2

2 y

2ey2

+

ydyey2

=ey2

2(x2 + y2 + 1) F (x, y) = e

y2

2(x2 + y2 + 1) = C

7. (x + x3sen(2y))dy 2ydx = 0My

= 2 ; Nx

= 1 + 3x2sen(2y)Primer Caso:My N

x

N

f(x) =2 1 3x2sen(2y)

x + x3sen(2y)=3(1 + x2sen(2y))x(1 + x2sen(2y)

=3x

u(x) = e3lnx =1

x3Aplicamos u(x)

2yx3

dx + (1

x2+ sen(2y))dy = 0

My

=2x3

; Nx

=2x3 F (x, y)/

F1 = 2yx3dx =

y

x2y F2 =

(

1

x2+ sen(2y))dy =

y

x2 cos(2y)

2

F (x, y) = yx2 cos(2y)

2= C

8. (y2cos(x) y)dx + (x + y2)dy = 0My

= 2ycos(x) 1 ; Nx

= 1Segundo Caso:

Nx M

y

M

g(y) =1 + 1 2ycos(x)y(ycos(x) 1) =

2(ycos(x) 1)y(ycos(x) 1) =

2y

u(y) = e2lny =1

y2

3

Aplicamos u(y)

(cos(x) 1y

)dx + (x

y2+ 1)dy = 0

My

=1

y2; Nx

=1

y2

F (x, y)/F1 =

(cos(x) 1

y)dx = sen(x) x

yy F2 =

2x2ydy = sen(x) x

y+ y

F (x, y) = sen(x) xy

+ y = C

12. Una ecuacion diferencial puede tener mas de un factor integrante. Pruebe que u1 =1

xy;u2 =

1

y2;u3 =

1

x2 + y2son factores integrantes de la ecuacion diferencial ydx + xdy = 0.

Demuestre que las soluciones son formalmente equivalentes.

Si aplicamos u1

dx

x dy

y= 0

My

= 0 ; Nx

= 0

F (x, y)/Fa1 =

dxx

= ln|x| y Fb1 = dy

= ln|y|

F1(x, y) =x

y= C

Si aplicamos u2

dx

ydx xdy

y2dy = 0

My

=1y2

; Nx

=1y2

F (x, y)/Fa2 =

dxydx =

x

yy Fb2 =

xdyy2

dy =x

y

F2(x, y) =x

y= C

Si aplicamos u3

ydx

x2 + y2 xdyx2 + y2

= 0

My

=x2 y2x2 + y2

; Nx

=x2 y2x2 + y2

F (x, y)/

Fa3 = ydxx2 + y2

= ln(x2 + y2)

1

2 y Fb3 = xdyx2 + y2

= ln(x2 + y2)

12

F3(x, y) = ln(1) = 0 = CF1(x, y) = F2(x, y) = F3(x, y)

4