ECUACIONES DIFERENCIALES HOMOGENEAS

ECUACIONES DIFERENCIALES HOMOGENEAS ECUACIONES HOMOGÉNEAS. Si una función f  tiene la propiedad  f ( tx , ty ) = t f ( x , y ) α 

para algún número real α  , entonces se dice que es una función homogénea de grado α  Por ejemplo  f ( x, y) = x3 + y3 es homogénea de grado 3, por que 3

3

 f ( tx, ty ) = ( tx ) + ( ty ) = t

3

(x

3

+ y 3 ) = t 3 f ( x, y ) Mientras que  f ( x, y ) = x3 + y3 + 1 no es

homogénea. Una

ecuación

diferencial

de

primer

orden

en

forma

diferencial

 M ( x, y ) dx + N ( x , y ) dy = 0 se dice que es homogénea si los coeficientes M y N a la vez,

son funciones homogéneas del mismo grado. En otras palabras la ecuación  M ( x, y ) dx + N ( x , y ) dy = 0 es homogénea si:

 M ( tx, ty ) = t M ( x, y ) y N ( tx, ty ) = t N ( x, y ) α

MÉTODO

SOLUCIÓN:

α 

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Aplicamos las propiedades de los logaritmos para escribir la solución anterior en la forma ln

( x + y ) cx

2

=

y x

0⇒

2

( x + y ) = cxe y / x

Nota: Aunque se puede usar cualquiera de las sustituciones en toda ecuación diferencial homogénea, en la práctica probaremos con  x = vy Cuando la funcion M ( x, y ) sea más simple que  N ( x, y ) . También podría suceder que después de aplicar una sustitución, nos

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EJEMPLO 4. Resuelva la ecuación diferencial ( y + x 2 + y 2 ) dx − xdy = 0 2 2 2  y = ux ⇒ dy = udx + xdu ⇒ (ux + x + u y ) dx − x(udx + xdu) = 0

 x 1 + u 2 dx − x 2du = 0 ⇒

dx  x

−

du

1+ u

2

=0

ln  x − ln u + 1 + u 2 = c ⇒ u + 1 + u 2 = cx ⇒ y + y 2 + x 2 = cx 2

EJEMPLO 5. Resuelva la ecuación diferencial ( x3 + y 3 )dx − xy 2 dy = 0 con la condición

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EJERCICIOS RESUELTOS 1)  x  xdy =

dy dx

− y = x2 + y 2

)

(

x 2 + y 2 + y dx ⇒ y = vx ⇒ dy = vdx + xdv ⇒

dv

1 + v2

=

dx  x

Integrar lado izquierdo por su sustitución trigonométrica y lado derecho por tablas para obtener, después de revertir el cambio de variable.  y + x + y = cx 2

2

2

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5)

( y

2

− x 2 ) dx + xydy = 0

 y = vx ⇒ dy = udx + xdu ⇒

dx  x

=−

u

2u − 1 2

du ⇒ ln x = −

⎛ 2 y 2 ⎞  x ⎜ 2 − 1 ⎟ = c ⇒ 2 x 2 y 2 − x 4 = c ⎝  x ⎠ 4

t  ⎛ ⎞ y 6) e ( y − t ) dy + y ⎜ 1 + e ⎟ dt =   0 ⎜ ⎟ ⎝ ⎠ dy eu + 1 t uy ⇒ dt udy + ydu ⇒ ⇒ eu +u t

 y

z

1 4

ln 2u 2 − 1 + ln c

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u2

⇒  xdu =

u2 +1

Pero u =

 y  x

⇒−

u2 +1

⇒

u

2

 x 2 +  y 2  y

du =

+ 1n

dx  x

⇒⇒ −

 x 2 +  y 2  x

+

 y  x

1+ u2 u

= 1nx + c

11) ( x − y ) dx + xdy = 0

( x − y ) dx + xdy = 0 (

)

dy

0

dy

y−x

dy

y

+ 1n 1 + u 2 + u = 1nx + c

1

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⇒  x + ( y − 2 x )

dy dx

=0⇔

dy dx

=−

x y − 2x

⇔

dy dx

=

x

2x − y

⇔

dy dx

=

1 2−

 y

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⇒  x

dv dx

=

v

2 + 2v

−v ⇔ x

dv dx

=

v − 2v − 2v

2 + 2v

2

2 + 2v 1 −v − 2v 2 ⇔x = ⇔ = dv dx 2 + 2v dx x −v − 2v 2 dv

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⇒

dv v 2 + 2v

1

= − dx ⇔ ∫ x

⎡1 ⎤ 1 1 1 = −∫ dx ⇔ ∫ ⎢ − ⎥ dv = −∫ dx v ( v + 2) x x ⎣ 2v 2 ( v + 2) ⎦ dv

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⇒ 2udx = ( 3u + u ) dx + x ( 3 + u 3 ) du ⇔ 2udx − ( 3u + u 4 ) dx = x ( 3 + u 3 ) du ⇒ ( 2u

( 3v + u ) ) dx 4

(

x 3+u

3

) du ⇔ ( 2u

3u u 4 ) dx

(

x 3 +u

3

) du

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2

⎛ y⎞ ⇒ v = 2 ln x + c ⇒ v = ⇒ ⎜ ⎟ = 2 ln x + c  x ⎝x⎠ 2

 y

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