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UNIVERSIDAD AUTÓNOMA DENUEVO LEÓNFacultad de ingeniería Mecánica y EléctricaProducto Integrador de AprendizajeMatemáticas IIIEquipo 12Grupo: 038Hora: V6 Salón: 2304Docente: Ing. Christian Fabian Chapa ArceAlumno Matrícula CarreraAxel Alfredo Guerra Díaz 1911515 IMAJan Carlo Licon Tijerina 1913897 IM...

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UNIVERSIDAD AUTÓNOMA DE NUEVO LEÓN Facultad de ingeniería Mecánica y Eléctrica

Producto Integrador de Aprendizaje

Matemáticas III Equipo 12 Grupo: 038 Hora: V6 Salón: 2304 Docente: Ing. Christian Fabian Chapa Arce

Alumno Matrícula Axel Alfredo Guerra Díaz 1911515 Jan Carlo Licon Tijerina 1913897 Diego Alberto Vasconcelos Guajardo 1919447 Ivannia Abigail Chávez Martínez 1922922 Jorge Antonio Bailey Cantú 1924643

Carrera IMA IMA IMA IAS IMTC

CU, San Nicolas de los Garza, Nuevo León, Mexico, jueves 03 de noviembre del 2020

Actividad Fundamental #1

Obtención del Orden y Grado de una Ecuación Diferencial

1.

ⅆ2 y ⅆx2

ⅆy 3

ⅆy

+ 13 ( ) + x2 = ( ) ⅆx ⅆx

Tipo: Ordinario Orden: 2 Grado: 1 Linealidad: No lineal

2. √ ⅆx3 − 5x = 8 ( ⅆx) ⅆy

ⅆ3 y

=√ ⅆx3 = 8 ( ) + 5x = (√ ⅆy

ⅆ3 y

=

ⅆ3 y

ⅆx3

ⅆx

3

ⅆy

= (8 ( ⅆx) + 5x)

Tipo: Ordinario Orden: 3 Grado: 1 Linealidad: No lineal

ⅆ4 y

3 3 ⅆ y

3. ( ⅆx4 )

ⅆx

+

ⅆ2 y

ⅆx2

− y2 = 0

Tipo: Ordinario Orden: 4 Grado: 3 Linealidad: No lineal

ⅆ3 y

2

2

) + 5x) 3 ) = (8 ( ⅆx

ⅆx

ⅆy

4. √

ⅆ2 y

ⅆx2

=(√

5

+ 3x = √( ⅆx3 ) ⅆ2 y

ⅆx2

ⅆ3 y

10

+ 3x)

= ( √(

5

ⅆ2 y

5

ⅆ3 y

ⅆ3 y

ⅆx3

3

10

) )

= (

ⅆ2 y

ⅆx2

5

ⅆ3 y

+ 3x) = ((

ⅆx3

3 2

) )

6

=( ⅆx2 + 3x) = ( ⅆx3 ) Tipo: Ordinario Orden: 3 Grado: 6 Linealidad: No lineal

5.

ⅆ2 y

ⅆx2

+ k2y = O

Tipo: Ordinario Orden: 2 Grado: 1 Linealidad: No Lineal

6. (x 2 + y 2 )dx − 2 xy dy = 0 = (x 2 + y 2 )dx = 2 xy dy =

(x2 +y2 ) ⅆx 2xy

= dy =

(x2 +y2 ) 2xy

Tipo: Ordinario

=

ⅆy

ⅆx

Orden: 1 Grado: 1 Linealidad: Lineal

Comprobación de la solución general de una Ecuación Diferencial

1) 𝑦 = 𝑐 2 + 𝑐𝑥−1

𝑦 + 𝑥𝑦′ = 𝑥 4(𝑦 ′) 2

𝐷𝑥(𝑐) = 0

𝑦′ = 0 + (−𝑐𝑥−2)

𝑦′ = −𝑐𝑥−2

𝑦 + 𝑥𝑦′ = 𝑥4(𝑦′)2

(𝑐2 + 𝑐𝑥−1) + 𝑥(−𝑐𝑥−2) = 𝑥4(−𝑐𝑥−2)2 𝑐2 + 𝑐𝑥−1 − 𝑐𝑥−1 = 𝑥4(𝑐2𝑥−4)

𝑐2 = 𝑐2

Si es solución

2) 𝑒𝑐𝑜𝑠𝑥( 1 − 𝑐𝑜𝑠𝑦) = 𝑐

𝐷𝑥(𝑐) = 0

𝐷𝑥(𝑢 ∙ 𝑣) = 𝑢𝑑𝑣 + 𝑣𝑑𝑢 𝑑𝑦 𝑒𝑐𝑜𝑠𝑥𝑆𝑒𝑛𝑦 + (1 − 𝑐𝑜𝑠𝑦)𝑒𝑐𝑜𝑠𝑥(−𝑆𝑒𝑛𝑥) = 0 𝑑𝑥 𝑑𝑦 𝑒𝑐𝑜𝑠𝑥𝑆𝑒𝑛𝑦 − (1 − 𝑐𝑜𝑠𝑦)𝑒𝑐𝑜𝑠𝑥𝑆𝑒𝑛𝑥 = 0 𝑑𝑥 𝑑𝑦 𝑒𝑐𝑜𝑠𝑥𝑆𝑒𝑛𝑦 = (1 − 𝑐𝑜𝑠𝑦)𝑒𝑐𝑜𝑠𝑥𝑆𝑒𝑛𝑥 𝑑𝑥 𝑑𝑦 = (1 − 𝑐𝑜𝑠𝑦)𝑒𝑐𝑜𝑠𝑥𝑆𝑒𝑛𝑥 𝑑𝑥 𝑑𝑦 𝑑𝑥

= (1 − 𝑐𝑜𝑠𝑦)𝑆𝑒𝑛𝑥 𝑆𝑒𝑛𝑦

𝑠𝑒𝑛𝑦(

𝑑𝑦 𝑑𝑥

) + 𝑠𝑒𝑛𝑥𝑐𝑜𝑠𝑦 = 𝑠𝑒𝑛𝑥

𝑑𝑦 𝑆𝑒𝑛𝑦 (𝑑𝑥 ) + 𝑆𝑒𝑛𝑥𝐶𝑜𝑠𝑦 = 𝑆𝑒𝑛𝑥 (1 − 𝐶𝑜𝑠𝑦)𝑆𝑒𝑛𝑥

𝑆𝑒𝑛𝑦

) + 𝑆𝑒𝑛𝑥𝐶𝑜𝑠𝑦 = 𝑆𝑒𝑛𝑥

(1 − 𝐶𝑜𝑠𝑦)𝑆𝑒𝑛𝑥 + 𝑆𝑒𝑛𝑥𝐶𝑜𝑠𝑦 = 𝑆𝑒𝑛𝑥

𝑆𝑒𝑛𝑥 − 𝑆𝑒𝑛𝑥𝐶𝑜𝑠𝑦 + 𝑆𝑒𝑛𝑥𝐶𝑜𝑠𝑦 = 𝑆𝑒𝑛𝑥

𝑆𝑒𝑛𝑥 = 𝑆𝑒𝑛𝑥

Si es solución

3) 𝑦 = 8𝑥5 + 3𝑥2 + 𝑐 𝑑𝑦

𝑑𝑥

(

= 40𝑥4 + 6𝑥

𝑑2𝑦

𝑑𝑥2

𝑑2𝑦

𝑑𝑥2

𝑑 2𝑦 ) − 6 = 160𝑥3 𝑑𝑥 2

= 160𝑥3 + 6 − 6 = 160𝑥3

160𝑥3 + 6 − 6 = 160𝑥3

160𝑥3 = 160𝑥3

Si es solución

4)𝑦 = 𝑐1𝑠𝑒𝑛3𝑥 + 𝑐2𝑐𝑜𝑠3𝑥 𝑑𝑦 𝑑𝑥

= 3𝑐1𝑐𝑜𝑠3𝑥 − 3𝑐2𝑠𝑒𝑛3𝑥

𝑑 2𝑦

𝑑𝑥2

+ 9𝑦 = 0

𝑑2𝑦

𝑑𝑥2

𝑑2𝑦

𝑑𝑥2 𝑑2𝑦

𝑑𝑥2 𝑑2𝑦

𝑑𝑥2

= −9𝑐1𝑠𝑒𝑛3𝑥 − 9𝑐2𝑐𝑜𝑠3𝑥 = −9 (𝑐1𝑠𝑒𝑛3𝑥 + 𝑐2𝑐𝑜𝑠3𝑥) y

= −9𝑦 + 9𝑦 = 0

−9𝑦 + 9𝑦 = 0 Si es solución 𝑑𝑦

5)𝑦 = (𝑥 + 𝑐)𝑒−𝑥

𝑑𝑥

𝐷𝑥(𝑐) = 0

𝐷𝑥(𝑥) = 1

𝐷𝑥(𝑢 ∙ 𝑣) = 𝑢𝑑𝑣 + 𝑣𝑑𝑢 𝑑𝑦 = −(𝑥 + 𝑐)𝑒−𝑥 + 𝑒−𝑥 𝑑𝑥 𝑑𝑦

𝑑𝑥 𝑑𝑦

𝑑𝑥

y

= −𝑦 + 𝑒−𝑥 + 𝑦 = 𝑒−𝑥

−𝑦 + 𝑒−𝑥 + 𝑦 = 𝑒−𝑥

𝑒−𝑥 = 𝑒−𝑥

Si es solución

+ 𝑦 = 𝑒−𝑥

𝑑𝑦 𝑑𝑥 − 5𝑦 = 0

6) 𝑦 = 𝑐𝑒5𝑥 𝐷𝑥(𝑐) = 0

𝐷𝑥(𝑢 ∙ 𝑣) = 𝑢𝑑𝑣 + 𝑣𝑑𝑢 𝑑𝑦 = 5𝑐𝑒5𝑥 𝑑𝑥 𝑑𝑦

y

𝑑𝑥 = 5𝑦 𝑑𝑦

𝑑𝑥

− 5𝑦 = 0

5𝑦 − 5𝑦 = 0

Si es solución 7)𝐼𝑛𝑦 = 𝑐1 𝑠𝑒𝑛𝑥 + 𝑐 2𝑐𝑜𝑠𝑥 𝐷𝑥(𝐼𝑛𝑢) =

1

𝑢

𝐷𝑥𝑢

1 𝑑𝑦 ∙ 𝑦 𝑑𝑥 = 𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥 𝑑𝑦 𝑑𝑥

= 𝑦(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)

𝐷𝑥(𝑢 ∙ 𝑣) = 𝑢𝑑𝑣 + 𝑣𝑑𝑢

2 𝑑𝑦 2 𝑦 ( 𝑑 𝑦 ) − ( ) = 𝑦2𝐼𝑛𝑦

𝑑𝑥2

𝑑𝑥

𝑑𝑦 2

𝑑𝑥2 𝑑2𝑦

𝑑𝑥2 𝑑2𝑦

= 𝑦(−𝑐1𝑠𝑒𝑛𝑥 − 𝑐2𝑐𝑜𝑠𝑥) + (𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)

𝑑𝑦

𝑑𝑥

= 𝑦(−𝑐1𝑠𝑒𝑛𝑥 − 𝑐2𝑐𝑜𝑠𝑥) + (𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)(𝑦(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥))

𝑑𝑥2

= 𝑦(−𝑐1𝑠𝑒𝑛𝑥 − 𝑐2𝑐𝑜𝑠𝑥) + 𝑦(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)2

𝑑𝑦 2 𝑑2𝑦 𝑦 ( 2 ) − ( ) = 𝑦2𝐼𝑛𝑦 𝑑𝑥 𝑑𝑥

𝑦[𝑦(−𝑐1𝑠𝑒𝑛𝑥 − 𝑐2𝑐𝑜𝑠𝑥) + 𝑦(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)2] − [𝑦(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)]2 = 𝑦2𝐼𝑛𝑦

𝑦2[(−𝑐1𝑠𝑒𝑛𝑥 − 𝑐2𝑐𝑜𝑠𝑥) + (𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)2] − [𝑦2(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)2] = 𝑦2𝐼𝑛𝑦 𝑦2(−𝑐1𝑠𝑒𝑛𝑥 − 𝑐2𝑐𝑜𝑠𝑥) + 𝑦2(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)2 − 𝑦2(𝑐1𝑐𝑜𝑠𝑥 − 𝑐2𝑠𝑒𝑛𝑥)2 = 𝑦2𝐼𝑛𝑦 −𝑦2(𝑐1𝑠𝑒𝑛𝑥 + 𝑐2𝑐𝑜𝑠𝑥) = 𝑦2𝐼𝑛𝑦 Iny

−𝑦2𝐼𝑛𝑦 = 𝑦2𝐼𝑛𝑦

No es solución

Obtención de la ecuación diferencial a partir de Ecuaciones Diferenciales

1) 𝑦 = 7𝑥2 + 8𝑥 + 𝑐 𝑑𝑦

𝑑𝑥

= 14𝑥 + 8

2) 𝑦 = 𝑐1𝑥2 + 𝑐2 𝑦′ = 2𝑐1𝑥

𝑦′′ = 2 𝑐1

𝑦′ = 𝑥(𝑦 ′′) 𝑆𝑖 𝑦 = ′

𝑑𝑦 𝑑𝑥

, 𝑦 𝑦 ′′ =

𝑑2𝑦

𝑑𝑥 2

𝑑2𝑦

𝑑𝑦 𝑑𝑥

3)𝑦 = 𝑐1𝑠𝑒𝑛8𝑥 + 𝑐2𝑐𝑜𝑠8𝑥

𝑦′ = 8𝑐1𝑐𝑜𝑠8𝑥 − 8𝑐2𝑠𝑒𝑛8𝑥

𝑦′′ = −64𝑐1𝑠𝑒𝑛8𝑥 − 64𝑐2𝑠𝑒𝑛8𝑥

𝑦′′ = −64(𝑐1𝑠𝑒𝑛8𝑥 + 𝑐2𝑐𝑜𝑠8𝑥)

𝑦′′ = −64 𝑦

y

𝑦′′ + 64 𝑦 = 0 𝑑2𝑦 + 64𝑦 = 0 𝑑𝑥2

4)𝑦 = tan (3x + c)

𝑡𝑎𝑛−1𝑦 = 3𝑥 + 𝑐 𝐷𝑥𝑢 𝐷𝑥[𝐴𝑟𝑐𝑇𝑎𝑛𝑢] = 1 + 𝑢2 1

1+

𝑦2

∙

𝑑𝑦

𝑑𝑥

=3 𝑑𝑦

𝑑𝑥 5) 𝑦 = 𝑐1𝑒3𝑥 + 𝑐2𝑒−5𝑥

𝐷𝑥(𝑢 ∙ 𝑣) = 𝑢𝑑𝑣 + 𝑣𝑑𝑢

𝑦′ = 3𝑐1𝑒3𝑥 − 5𝑐2𝑒−5𝑥

𝑦′′ = 9𝑐1𝑒3𝑥 + 25𝑐2𝑒−5𝑥 5(𝑦′ = 3𝑐1𝑒3𝑥 − 5𝑐2𝑒−5𝑥) 𝑦′′ = 9𝑐1𝑒3𝑥 + 25𝑐2𝑒−5𝑥

5𝑦′ = 15𝑐1𝑒3𝑥 − 25𝑐2𝑒−5𝑥 𝑦′′ = 9𝑐1𝑒3𝑥 + 25𝑐2𝑒−5𝑥

5𝑦′ + 𝑦′′ = 24𝑐1𝑒 3 𝑥

5(𝑦 = 𝑐1𝑒3𝑥 + 𝑐2𝑒−5𝑥)

𝑦′ = 3𝑐1𝑒3𝑥 − 5𝑐2𝑒−5𝑥

5𝑦 = 5𝑐1𝑒3𝑥 + 5𝑐2𝑒−5𝑥 𝑦′ = 3𝑐1𝑒3𝑥 − 5𝑐2𝑒−5𝑥

5𝑦 + 𝑦′ = 8𝑐1𝑒3𝑥

−3(5𝑦 + 𝑦′ = 8𝑐1 𝑒3𝑥 )

5𝑦′ + 𝑦 ′′ = 24𝑐1𝑒 3 𝑥

−15𝑦−3𝑦′ = −24𝑐1𝑒3𝑥

= 3(1 + 𝑦2)

𝑦′′ + 2𝑦 ′ − 15𝑦 = 0

6) 𝑦 = 𝑥𝑡𝑎𝑛(𝑥 + 𝑐) 𝑦

𝑥

𝑑2𝑦 𝑑𝑦 + 2 ( ) − 15𝑦 = 0 2 𝑑𝑥 𝑑𝑥

= tan(𝑥 + 𝑐)

𝑡𝑎𝑛−1

𝑦 ()=𝑥+𝑐 𝑥

𝐷𝑥[𝐴𝑟𝑐𝑇𝑎𝑛𝑢] = 𝐷𝑥𝑢 1 + 𝑢2 𝑢 𝐷 𝑥 ( ) = 𝑣𝑑𝑢 − 𝑢𝑑𝑣 𝑣 𝑣2 𝑥𝑦′ − 𝑦 𝑥2 =1 2 𝑦 1+ 𝑥2

𝑥𝑦′ − 𝑦 𝑥2 =1 𝑥2 𝑦2 + 𝑥2 𝑥2

𝑥𝑦′ − 𝑦 𝑥2 =1 𝑥 2 + 𝑦2

𝑥2 𝑥𝑦 ′ − 𝑦 =1 𝑥2 + 𝑦2

𝑥𝑦′ − 𝑦 = 1(𝑥2 + 𝑦2) 𝑑𝑦 𝑥 − 𝑦 = 𝑥2 + 𝑦2 𝑑𝑥 𝑥

𝑑𝑥

= 𝑥2 + 𝑦 2 + 𝑦

7)𝑦 = 𝑐1𝑠𝑒𝑛ℎ(𝑥) + 𝑐2cosh (𝑥) 𝑦′ = 𝑐1𝑐𝑜𝑠ℎ(𝑥) + 𝑐2senh (𝑥)

*Derivamos una segunda vez 𝑐1 𝑠𝑒𝑛ℎ(𝑥) + 𝑐2cosh (𝑥) 𝑦 ′′ = 𝑦

Actividad Fundamental #2

Método para la Ecuación Diferencial de separación de Variables

1) 5x4 dx + 20y14 dy=0 =∫5 x4 dx +∫ 20y14 dy =5∫ x4 dx + 20 ∫ y4 dy =5(

𝑥5 5

) + 20 (

=x5 + y20 + C

𝑦20 20

)

2) 2 sen (2x) dx + 3eey dy = 2xdx = 2 sen (2x)dx – 2xdx = -3 eey dy = dx (2 sen (2x) dx – 2 x dx) = -3 e3y dy = ∫ (2 sen (2x) – 2x) dx) = ∫ -3 e3y dy = ∫ 2 sen (2x) dx - ∫ 2x dx = -∫ eey 3dy 𝑥2

= - cos (2x) – (-y) ( 𝑦 ) = -e3y

= e3y = cos (2x) + x2 + C

3) dy/ds = r ; (dr= rds= (1/r) = dr/r -ds = 0 = ∫ dr/r - ∫ ds = ∫ 0 = In |r| - s = c = ln |r| = c + s = r = ec+s = r = c/ es = res = c 4) dx + dy + xdx = ydx = dy + xdy = ydx -dx ; = dy(1 + x) = dy(y -1) = dy(1+x) – dx (y – 1) = 0; (dy (1 + x) – dx (y – 1) -0) (

1

(1+𝑋)(𝑦−1)

= dy / y-1 – dx / x+1 = 0 ; = ∫ dy / y – 1 - ∫ dx / x-1 = 0 ; = ∫ dy / y – 1 - ∫dx / x + 1 = ∫0 = ln |y – 1| - ln |x + 1| = C 𝑦−1 𝑦−1 = ln | 𝑥+1 | = C ; = ec 𝑥+1

= y – 1 = ec (x + 1)

)

5) x sen (y) dx + (x2 + 1) cos (y) dy= 0 = x sen (y) dx = - (x2 + 1) cos (y) dy =

𝑋𝑥𝑑

− cos(𝑦) dy 𝑠𝑒𝑛 (𝑦) 2𝑥𝑑 cos(𝑦)

= (𝑥2+1)

= 1/ 2 ∫

𝑥2+1

+∫

𝑠𝑒𝑛 (𝑦)

dx = ∫0

= C 1/2 ln |x2 + 1| + ln |sen (y)| = C (2) = ln |x2 + 1| + 2ln |sen (y) | = C = ln |(x + 1) (sen y)2 | = C = (x + 1) (sen y)2 = ec 6) (xy + 2x + y + 2) dx + ( x2 + 2x ) dy = 0 = ecuación diferencial no exacta decido a que se encuentra las 2 variables y es la misma derivada.

7) Dy / dx = x + 1 / y + 1 = dy (y2 + 1) = dx (x + 1) = ∫ ( y2 + 1 dy) = ∫ (x + 1) dx = =

𝑦5

𝑦

𝑥2

2 1 5 𝑥2+𝑥 𝑦5+𝑦 5

+

=

=

2

𝑥

+1

+C

= (y5 + y)2 = (x2 + x)5 + C 8) xy2 dx + (y2 + 2) e-3x dy = 0 =ecuación diferencial no exacta debido a que se encuentran las 2 variables y es misma derivada 9) x sen (x2) e-y2 dx – y dx = 0 =

𝑥 𝑠𝑒𝑛 (𝑥2) (𝑒−𝑦2)

= y dx

= ½ ∫ 2x sen (x2) dx = ½ ∫ ey2 2y dy = ( -1/2 cos (x2) ) = ½ ey2 + C) (2) = ey2 = cos (x2) + C

10) dy / dx = e3x+2y = dy / dx = e3x – e2y = dy / e2y = e3x dx = e-2y dy = e3x dx = ∫ e-2y dy - ∫ e3x dx = 0 = ½ ∫ e-2y – 2dy – 1/ 3 ∫ e3x 3dx = 0 = -1/2 e-2y - 1/3 e3x = C = (3e-2y -2e3y = C ) ln = 3 (-2y) – 2 (3x) = C = -6y – 6x = C 11) dy/dx = xy + 3x – y – 3 / xy – 2x + 4y – 8 = dy/dx = (x - 1) (y + 3) / (x + 4) (y – 2) = (y – 2) dy / y + 3 = (x -1) dx/ (x + 4) = (y – 2) (x – 1) dy/ (x + 3) = (x + 1 + 4 -4) dy/ (x + 4) = (y + 4) - ∫ dy/ (y + 3) = ( (x + 4) -∫) dx/ (x + 4) ) = ∫ y 3 / y + 3 dy - ∫ 5dy / ( 5+2) = ∫ (x+4) / (x+4) dx - ∫ 5dx/ (x+4) = y – 5 ln |y+3| = x-5 ln |x+4| = y – 5 ln | y+3| = x-5 ln |x+4| = C = y – x ln |(

(𝑥+4) 3 )| (𝑦+3)

=C

12) dy/dx = Sen x (cos 2y – cos2 y) =dy/dx= sen x (cos2y – cos2 y – sen2y) =dy/dx= sen x (-sen2y) = dy= sen x dx (-sen2y) =dy/-sen2y = senxdx = -csc2y dy = sen xdx = (– cot y) – (- cos x ) = C = coty + cosx = C 13) (x+1) dy + (y-1) dx = 0 ¨y¨=3 cuando ¨x¨= 0 = ( (x-1) dy + (y-1) dx = 0 ) (1/ cx + 1) (y-1) = ∫(y – 1) dy + ∫(x + 1) dx = 0 ∫ydy - ∫1dy + ∫xdx + ∫1dx = 0 = y2/2 – y + x2/2 + x = C = 32/2 -3 = C = 4.5 – 3 = C = 1.5 = C 14) X3 dy + xy = x2 dy + 2y dx ; ¨y¨ = e cuando ¨x¨ = 2 = x3dy – x2 dy = 2y dx – xydx = ( ( x3 – x2) dy = (2 – x) y dy) ( 1 / (X3 – X2 ) y ) = dy/y = 2-x / x3 – x2 dx = ∫dy/y = ∫ 2-x / x2 (x – 1) dx

= ∫dy/y = ∫2 – x/ x2 (x – 1) dx = 2 – x/ x2 ( x – 1) A/ x + B/ x2 + C/ x – 1 = ∫dy / y = ∫2dx / x2 + ∫ dx/ x – 1 = 2 – x = Ax2 (x + 1) / x + Bx2 (x + 1) / x2 + C2 (x – 1) / x – 1 = ln |y | = -2 (x-1/-1 ) + ln|x – 1|= ln|y| = 2/x + ln|x – 1| = ln| e |= 2/2 + ln| 2- 1 | = 1 = 1 + 0 + C = 1 – 1 = C C=0

Métodos para Ecuaciones Diferenciales Exactas 1) (6x2 – 7y2) dx – 14 xy dy = 0 = 6x2 – 7(vx)2 dx – 14 x (vx) (vdx + xdv) = 6x2 – 7(vx)2 dx – 14 x ( v2 xdx + vx2 dv) = 0 = 6x2 – 7(vx)2 dx – 14 v2x2 dx + 14 vx3 dv) = 0 = 6 x2 dx – 21 v2 x2 dx – 14 vx3 dv = 0 = (3x2 dx (3 – 7v2) – 14 vx3 dv = 0 ) ( 1/ )c – 7v2) v3) =

3𝑥2 𝑑𝑥

=3

𝑥3 𝑑𝑥 ∫ 𝑥

–

14 𝑣𝑑𝑥

3 𝑑𝑥

𝑥 2−7𝑣2 (−1)(14 𝑣𝑑𝑥) (-∫ ( 2−7𝑣2)𝑥3

=∫

- ∫

14 𝑣𝑑𝑥

2−7𝑣2

)=0

= 3 ln |x| + ln |2 – 7v2| = C = ln|x3| + ln |2 – 7 v2| = ln |x2 ( 2 – 7v2) = C = x3 (2 – 7v2) = C

2) 2(√ xy – y ) dx – xdy = 0 = 2 ( √ x(vx) – vx) dx – x (vdx + xdv) = 0 = 2 ( √ vx2 – vx) dx – xv dx – x2 dv = 0 = 2x2 v1/2 dx – 2vx dx – xv dx – x2 dv = 0 = 2 x2 √v1/2 dx – 3 vx dx – x2 dv = 0 = (c dx (2√ v – 3 v) – x2 dv = 0) =( 1 / (2√ v – 3 v) x2) =

−𝑥𝑑𝑥 𝑥2

𝑑𝑦

- 2𝑣⅓−3𝑣 = 0 = 1/2 ∫

2𝑥 𝑑𝑥 𝑥2

-∫

𝑑𝑣

2√v−3v

=0

=1/2 ln |x2| - ∫ 2 1/2 T – 3 T2 = -∫ 2T dt / T( 2-3T) = -∫ 2 dt / 2-3T = - 2/ 3 ∫ 3dt / 2-3 T = -2/3 ln |2 – 3T| = 1/2 ln |x2| - 1/ 3 ln |2 – 3 √ v| = 0

= 3ln |x2| - 4 ln |2 – 3 √ v | = C ln |x6 / (2-3√ v4| = C = x6 / (2 – 3 √ v )4 = x6 = C (2 – 3 √ v )4 3) (y + x cot (y / x) ) dx- xdy= 0 = ( vx + x cot (v x/x dx – x ( vdx + xdv) = v xdx + x cot (v) dx – xvdx – x2 dv= 0 = (x xot (v) dx – x2 dv = 0 ) ( 1 cot (v) (x4) ) 2𝑥𝑑𝑥 𝑑𝑣 𝑥𝑑𝑥 = ∫ 0 = 1/2 ∫ -∫ =∫ - ∫ tan (v) = ∫0 cot (𝑣) 𝑥2 1

2

2

2

𝑥2

= /2 ln|x / sec v| = C = x / sec2 v = C = x2 = C (sec2 v)

4) (y + √ x2 y2 dx – x dy = ( vx + √ x2 + (vx2 dx – x (v dx + x dv) ) = ( vx + (x2 + v2 x2)1/2 ) dx – x vx + x2 dv = ( vx + x2 (1 + v2)1/2 dx – x2 dv = 0) = (1 / (1 + v2) ½ x2 ) = ∫ dx = ∫ dv / (1 + v2)1/2 = 0 = v – sen h-1 (v/1) = C = x – 1 / sen h (v) = C 5) dy/dx = 2y4 + x4 / xy3 = (xy3) dy = ( 2y4 + x4 ) dx = x(vx)3 (vdx + xdv) = (2 (xv)4 + x4) dx = v3x4 (vdx + xdv) = 2v4x4 dx + x4 dx = v4x4 dx + v3x5 dv = 2v4x4 dx + x4 dx = v3x5 dv= v4x4 dx + x4 dy = (v3 x5 dv) = (v4 + 1) (x4 dx) (1/ x5 (v4 + 1) = ∫ v3 dv / v4 dx/ x5 (𝑥4+1)^5 (𝑥4+1)^5 = 1/4 ln| 𝑥^20 | = C = 𝑥^20 = C = (x4 + 1)5 = Cx20 6) dy/dx = 2xy(x/y)^2 / y2 + y2 e(x/y)^2 + 2x2 e(x/y)^2 = (y2 + y2 e(x/y)^2 + 2x2 e(x/y)^2 dy = 2xy e(x/y)^2 dx = (y2 + y2 e(x/y)^2 + 2(vy)2 e(x/y)^2 dy= 2(vy + ydv) =y2 dy + y2 + y2 ev2 dy + 2 v2y2ev2 dy = 2vy2 ev2(vdy + ydv) 2𝑣𝑒𝑣2 𝑑𝑣 𝑦2𝑑𝑦 =∫ =∫ 𝑦3 𝑦3

= ln| =C 𝑒3𝑣2 3 = y = Ce3v2

𝑒𝑣2

7) dy/dx = 2xy / x2 – y2 = (x2 – y2) dy = (2xy) dx = ( ( vy)2 – y2) dy = (2vy) (vdy + ydv) = v2y2 dy – y2 dy = 2 vy3 dv = (y2 dy (-1 – v2) = 2vy3 dy)

3𝑦2 𝑑𝑦

(−1)𝑒 𝑣 𝑑𝑣 =1/3 ∫ = ∫ (−1−𝑣2) 3 𝑦3 =ln |y | + ln |(-1 – v2) 3| = C = ln |(y3) (-1 – v2)3| = C = (y3) (-1 – v2)3 = C

8) (x2 + y2) dx + ( x2 – xy) dy = 0 = (x2 + (vx)2) dx + (x2 – x(vx) (vdx + xdv) = 0 = x2 dx + v2 x2 dx + x2 v dx + v3dv + x2 dx – x3 vdv = 0 1

= ( (1 + v) x2 dx +(1 – v) x3 dv = 0) ((1+𝑣)𝑥^3 ) =

𝑥^2𝑑𝑥 𝑥^3

+

1−𝑣 𝑑𝑣 1+𝑣

=0=C

Actividad Fundamental #3

Ecuaciones Reducibles a Lineales + 1. 𝒅𝒚 𝒅𝒙

𝒙𝒚 = 𝒙𝒚² 1

y= v𝑒 − ∫ 𝑃𝑑𝑥 = 𝑣𝑒 − ∫ 𝑥𝑑𝑥 = 𝑣𝑒 − 2𝑥² 1

y= v𝑒

− 2𝑥²

dy= 𝑒

−2𝑥²

1 dy= (-x𝑒2𝑥² ) (v) + (𝑒 −12𝑥² ) (dv)

(𝑒 𝑒

1

−2𝑥² 1

−2𝑥²

1

1

dv – v𝑒

𝑑𝑣 − 𝑣𝑒 1

−

1 2𝑥²

−

1 2𝑥²

) + x (v𝑒 1

−

1 2𝑥²

)dx = x (v𝑒

- v𝑒 − 2𝑥² + xv𝑒 − 2𝑥² dx= x (v²𝑒 −𝑥²) dx

[𝑒 −2𝑥² dv= v²x𝑒 −𝑥² 𝑑𝑥] [ 𝑑𝑣 𝑥𝑒 −𝑥²𝑑𝑥 = −1 𝑣² 𝑒 2𝑥²

=∫ 𝑥 𝑒 𝑢 (−

𝑑𝑢 ) 𝑥

𝑢𝑛+1

𝑛+1

1

1 − (𝑒 2𝑥2 )(𝑣 2 )

]=

−

−1

1 𝑦𝑒2𝑥²

1 FACTOR: [ -y𝑒2𝑥² ]−

=𝑒

1 1 𝑦𝑒2𝑥²

−

+C 1

− 2𝑥²

+C

∫ 𝑢−2 𝑑𝑢

dx 𝟏

𝟐𝒙²

dx= −

1 2𝑥²

=𝑒

1

− 2𝑥²

du=dv 𝑢1 = 𝑒 𝑢 + Cd

1

)² dv

1

u= −

+C

• Sustituir: −

1 2𝑥²

(𝑒 2𝑥2 )(𝑣2 )

u= v

∫ 𝑒 𝑢 𝑑𝑢 = 𝑒 𝑢 + C

−

)(𝑣2 )

𝑣²𝑥𝑒 −𝑥²𝑑𝑥

∫ 𝑣 −2 𝑑𝑣 = ∫ 𝑥𝑒

∫ 𝑢−2 𝑑𝑢= ∫ 𝑒 𝑢 𝑑𝑢 ∫ 𝑢𝑛 𝑑𝑢 =

1

(𝑒− 1 2 2𝑥

Factor=

𝒅𝒖 𝒙

1

1

− = 𝑒 −2𝑥² + C 𝑣

1=

1 (-y𝑒2𝑥²)

(𝑒

−

1

2𝑥²

1

) + C (−𝑦𝑒2𝑥² )

1= - y – C (𝑦𝑒12𝑥²)

𝟏 ➢ 1= y (1 + c𝒆𝟐𝒙² )

Solución General

2. [dx + (2𝑥𝑦)dy = 2x²y² dy] 𝑑𝑦1 𝑑𝑥 𝑑𝑦

+

2𝑥 𝑦

𝟐

P= 𝒚

∫ 𝒅𝒖/𝒖= ln|u| + C

= 2x² y²

2𝑥

dx + ( ) dy = 2x²y² dy

𝒆𝒍𝒏𝒖 = 𝒖

𝑦

x= v𝑒 − ∫ 𝑃𝑑𝑦 = v𝑒 − ∫ 𝑑𝑥/𝑦 dy = v𝑒 −2 ∫ 𝑑𝑦/𝑦 = v𝑒 −2 ∫ 𝑑𝑢/𝑢 x= v𝑒 −2𝑙𝑛𝑦 = v𝑒 𝑙𝑛𝑦 x= v𝑦 −2

−2

dx= 𝑦 −2 + (-2𝑦 −3 )(v)

Factor:

dx= 𝑦 −2dv – 2v𝑦 −3 dy

1

(𝑦 −2 )(𝑣 2 )

2𝑥 𝑦

• Sustituir: dx + ( )dy =2x²y² (dy) 2(𝑣𝑦−2 )𝑥

𝑦 −2 dv – 2v𝑦 −3 dy + (

𝑦

)dy = 2(v𝑦 −2 )² y²(dy)

𝑦 −2 dv – 2v𝑦 −3 dy + 2v𝑦 −3dy = 2v²𝑦 −4 y²dy

[𝑦 −2dv = 2v²𝑦 −2 dy]

𝑦 −2 𝑑𝑣

(𝑦 −2 )(𝑣 2 )

2𝑣²𝑦−2 𝑑𝑦

(𝑦 −2 )(𝑣 2 )

𝑑𝑣 𝑣²

= 2dy

𝑣 −1

∫ 𝑣 −2 𝑑𝑣 = 2∫ 𝑑𝑦

−1

x = v𝑦 −2 v=

1

− 𝑥𝑦²

= 2y + C

= 2y + C

𝑥

𝑦 −2

v= xy² −𝑥𝑦² −𝑥𝑦²

= 2y (-xy²) + C (-xy²)

1= -2xy³ - cxy²

➢ 1 + 2xy³ = cxy²

Solución General

3. [dx – 2xy dy = 6x³ y²𝑒 −2𝑦2 dy] dy= 𝑑𝑦 – 2xy = 6x³y²𝑒 −2𝑦 𝑑𝑥

2

x= v𝑒 − ∫ −2𝑦𝑑𝑦 = v𝑒 +2 ∫ 𝑦𝑑𝑦 = v𝑒22𝑦 = v𝑒 𝑥

x= v𝑒 𝑦

2

2

2

dx= (du) (𝑒 𝑦 ) + (v) 2

dx= v𝑒 −2𝑦 dv – 2yv𝑒 −2𝑦 dy – 2yv𝑒 −2𝑦 dy = 6(v𝑒 −2𝑦 y²𝑒 −2𝑦 )dx 2

2

v𝑒 −2𝑦 dv = 6(v𝑒 −2𝑦 y²𝑒 −2𝑦 ) dx 2

2

2

2

[v𝑒 −2𝑦 dv – 6(v𝑒 −2𝑦 )³ y²𝑒 −2𝑦 dx = 0] [ 2

2

6(v𝑒 −2𝑦 )² ∫ 𝑦 2 𝑑𝑣 = ∫ −𝑒 −2𝑦 2

2

2

2

1

𝑣𝑒 −2𝑦

2

]

Factor

6(v𝑒 −2𝑦 )² 2

2

3

y³ =

(v𝑒 −2𝑦 )² 4y³ = 2

2 −𝑒𝑐2𝑦

𝑐

2 −𝑒 2𝑦

(v𝑒 −2𝑦 )² 4y³ + C = 2

𝑒 2𝑦 = (c – 4y³) (v𝑒 −2𝑦 )² 2

2

1 2 𝑒 2𝑦

➢ 𝒆𝟐𝒚 = [c – 4y³] x² 𝟐

Solución General

4. 𝑑𝑥1 [ (12𝑒 2𝑥 y² - y) dx = dy]; y = 1 cuando x= 0 𝑑𝑦 𝑑𝑥

+ y = 12𝑒 2𝑥 y²

y= v𝑒 − ∫ 𝑃𝑑𝑥 = v𝑒 − ∫ 1𝑑𝑥 = v𝑒 − ∫ 𝑑𝑥 = v𝑒 −𝑥 v= y𝑒 𝑥

dy= v (−𝑒 −𝑥 𝑑𝑥) + 𝑒 −𝑥 dv

dy= −𝑣𝑒 −𝑥 𝑑𝑥 + 𝑒 −𝑥 𝑑𝑣

• Sustituir: dy + y dx = 12𝑒 2𝑥 𝑦 2 𝑑𝑥

𝑒 −𝑥 𝑑𝑣 − 𝑣𝑒 −𝑥 𝑑𝑥 + 𝑣𝑒 −𝑥 𝑑𝑥 = 12𝑒 2𝑥 (𝑣 2 𝑒 −2𝑥 )dx [𝑒 −𝑥 𝑑𝑣 = 12𝑣 2 𝑑𝑥] [ 𝑒 −𝑥𝑑𝑣 (𝑒 −𝑥)𝑣 2

=

12𝑣 2

𝑒 −𝑥(𝑣 2 )

1

(𝑒 −𝑥)𝑣 2

]

Factor ∫ 𝑣 −2 𝑑𝑣 = ∫ 12𝑒 𝑥 𝑑𝑥

∫ 𝑣 −2 𝑑𝑣 = 12 ∫ 𝑒 𝑥 𝑑𝑥 = 1

− = 12𝑒 𝑥 + 𝐶 𝑣

𝑣 −1

= 12 𝑒 𝑥 + 𝐶

−1

1

− 𝑦𝑒 𝑥 = 12𝑒 𝑥 − 𝐶

(−𝑦1 𝑒 −𝑥 = 12𝑒 𝑥 − 𝐶) − 1

𝑦 −1 𝑒 𝑥 = −12𝑒 𝑥 + 𝐶

𝑦 −1 𝑒 −𝑥 = 𝐶 − 12𝑒 𝑥

1−1 𝑒 −0 = 𝐶 − 12𝑒 0 1(1) = C – 12(1) 1= C – 12 C= 1 + 12 C= 13

𝑦 −1 𝑒 −𝑥 = 13 − 12𝑒 𝑥 ➢ 𝒚−𝟏 𝒆−𝒙 = 𝒄 − 𝟏𝟐𝒆𝒙 Solución General

5. ( 3𝑦1 2) (3𝑦2 (𝑑𝑦𝑑𝑥) + 𝑑𝑦 𝑑...