Universidad Autónoma de Baja California

Facultad de Ingeniería

Ecuaciones Diferenciales

TAREA 2

Alumna:

Jauregui Padilla Monica Giselle

Matricula:

01135400

Profesora:

Rivera Castellón Ruth Elba

Grupo:

324

Mexicali Baja California Lunes 22 de septiembre del 2015

Ejercicios sección 2.2

a) En los ejercicios siguientes determine si la función dada es homogénea. Si lo es indique

su grado de homogeneidad.

1.x

yxyx

32 2

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡2𝑥2 + 2𝑡2𝑥𝑦 −(𝑡3)𝑦3

(𝑡)𝑥

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡2(𝑥2 + 2𝑥𝑦 −𝑦3

𝑥)

Homogénea grado 2

2. )64( yxyx

𝑓(𝑡𝑥, 𝑡𝑦) = √𝑡𝑥 + 𝑡𝑦 (4𝑡𝑥 + 6𝑡𝑦)

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡32(√𝑥 + 𝑦)(4𝑥 + 6𝑦))

Homogénea grado 𝟑

𝟐

3. 2)1( yx

𝑥2 + 𝑦2 + 2𝑥𝑦 + 2𝑥 + 2𝑦 + 1

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡2𝑥2 + 𝑡2𝑦2 + 𝑡22𝑥𝑦 + 2𝑡𝑥+ 2𝑡𝑦 + 1

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡(𝑡𝑥2 + 𝑡𝑦2 + 2𝑡𝑥𝑦 + 2𝑥+ 2𝑦) + 1

No es Homogénea

4.yx

x

2

cos

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑐𝑜𝑠𝑡2𝑥2

𝑡𝑥 + 𝑡𝑦

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑐𝑜𝑠𝑡2(𝑥2)

𝑡(𝑥 + 𝑦)

No es Homogénea

5.yx

xsen

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑠𝑒𝑛𝑡𝑥

𝑡𝑥 + 𝑡𝑦

𝑓(𝑡𝑥, 𝑡𝑦) = 𝑠𝑒𝑛𝑡𝑥

𝑡(𝑥 + 𝑦)

Homogénea grado 0

6.yx

yxyx

8

223

𝑓(𝑡𝑥, 𝑡𝑦)

=((𝑡3𝑥3)(𝑡𝑦)) − ((𝑡2𝑥2)(𝑡2𝑦2)

((𝑡)(𝑥)) + (8(𝑡)(𝑦))

𝑓(𝑡𝑥, 𝑡𝑦) =𝑡4((𝑥3𝑦) − (𝑥2𝑦2))

𝑡((𝑥)) + (8𝑦)

Homogénea grado 3

7. 442 yxy

x

𝑓(𝑡𝑥, 𝑡𝑦) =𝑡𝑥

𝑡2𝑦2 + √𝑡4(𝑥4 + 𝑦4)

𝑓(𝑡𝑥, 𝑡𝑦) =𝑡𝑥

𝑡2((𝑦2 + √(𝑥4 + 𝑦4))

Homogénea grado 1

8. yLnxLn 22

𝐼𝑛𝑥2

𝑦2

𝑓(𝑡𝑥, 𝑡𝑦) = 𝐼𝑛𝑡2𝑥2

𝑡2𝑦2

𝑓(𝑡𝑥, 𝑡𝑦) = 𝐼𝑛(𝑡2)𝑥2

(𝑡2)𝑦2

Homogénea grado 0

b) Resuelva las ecuaciones diferenciales siguientes usando la sustitución más

conveniente:

9. 0)( xdydxyx

((𝑥 − 𝑢𝑥)𝑑𝑥 + 𝑥(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 0

𝑥𝑑𝑥 − 𝑢𝑥𝑑𝑥 + 𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = 0

∫ 𝑥

𝑥2𝑑𝑥 = ∫ 𝑑𝑢

∫ 𝑑𝑥

𝑥= ∫ 𝑑𝑢

Inǀxǀ = 𝑢+c

Sust. u=𝑦

𝑥

Inǀxǀ =𝑦

𝑥 +c

10. 03)( 233 dyxydxyx

(𝑥3 + 𝑢3𝑥3)𝑑𝑥 + 3𝑢2𝑥3(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 0

𝑥3𝑑𝑥 + 𝑢3𝑥3𝑑𝑥 + 3𝑢3𝑥3𝑑𝑥 + 3𝑢2𝑥4𝑑𝑢=0

𝑥3𝑑𝑥 + 4𝑢3𝑥3𝑑𝑥 + 3𝑢2𝑥4𝑑𝑢=0

𝑥3(1 + 4𝑢3)𝑑𝑥 = −3𝑢2𝑥4𝑑𝑢

𝑥3

𝑥4𝑑𝑥 =

−3𝑢2

(4𝑢3 + 1)𝑑𝑢

∫𝑑𝑥

𝑥= ∫

−3𝑢2

(4𝑢3 + 1)𝑑𝑢

∫𝑑𝑥

𝑥= ∫

1

12𝑢2

−3𝑢2

(4𝑢3 + 1)𝑑𝑢

∫𝑑𝑥

𝑥= −

1

4∫

𝑑𝑢

(4𝑢3 + 1)

𝐼𝑛ǀ𝑥ǀ = −1

4𝐼𝑛ǀ4𝑢3 + 1ǀ + 𝑐

Sust. u=𝑦

𝑥

𝐼𝑛ǀ𝑥ǀ = −1

4𝐼𝑛ǀ4

𝑦

𝑥3

3

+ 1ǀ + 𝑐

11. 0)( 22 dyxdxyxy

(𝑢2𝑥2 + 𝑢𝑥2)𝑑𝑥 − 𝑥2(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 0

𝑢2𝑥2𝑑𝑥 + 𝑢𝑥2𝑑𝑥 − 𝑢𝑥2𝑑𝑥 + 𝑥3𝑑𝑢 = 0

𝑢2𝑥2𝑑𝑥 = −𝑥3𝑑𝑢

−∫𝑥2

𝑥3𝑑𝑥 = ∫

𝑑𝑢

𝑢2

−∫𝑑𝑥

𝑥= ∫

𝑑𝑢

𝑢2

−𝐼𝑛ǀ𝑥ǀ = −1

𝑢+ 𝑐

𝐼𝑛ǀ𝑥ǀ =1

𝑢+ 𝑐

Sust. u=𝑦

𝑥

𝐼𝑛ǀ𝑥ǀ =𝑥

𝑦+ 𝑐

12. 02

2

2

dyy

xdx

y

x

2𝑥

𝑦𝑑𝑥 =

𝑥2

𝑦2𝑑𝑦

2𝑢𝑦

𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢) =

𝑢2𝑦2

𝑦2𝑑𝑦

2𝑢2𝑑𝑦 + 2𝑢𝑦𝑑𝑢 =𝑢𝑦2

𝑦2𝑑𝑦

2𝑢𝑦𝑑𝑢 = ( 𝑢2𝑦2

𝑦2− 2𝑢2) 𝑑𝑦

2𝑢𝑦𝑑𝑢 = ( 𝑢2 − 2𝑢2)𝑑𝑦

2𝑢𝑦𝑑𝑢 = −𝑢2𝑑𝑦

−2∫𝑑𝑢

𝑢𝑑𝑢 = ∫

𝑑𝑦

𝑦

−2𝐼𝑛ǀ𝑢ǀ + 𝑐 = 𝐼𝑛ǀ𝑦ǀ

Sust. u=𝑥

𝑦

−2𝐼𝑛ǀ𝑥

𝑦ǀ + 𝑐 = 𝐼𝑛ǀ𝑦ǀ

13. 22 xyydx

dyx

𝑥𝑑𝑦 = (√𝑦2 + 𝑥2 + 𝑦)𝑑𝑥

𝑥(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = (√𝑢2𝑥2 + 𝑥2 + 𝑢𝑥)𝑑𝑥

𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = (√𝑥2(𝑢2 + 1) + 𝑢𝑥)𝑑𝑥

𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = (𝑥√𝑢2 + 1)𝑑𝑥 + 𝑢𝑥𝑑𝑥

𝑢𝑥𝑑𝑥 + 𝑥2𝑑𝑢 = 𝑥√𝑢2 + 1𝑑𝑥 + 𝑢𝑥𝑑𝑥

∫𝑑𝑢

√𝑢2 + 1= ∫

𝑑𝑥

𝑥2

𝐼𝑛 ǀ𝑢 + √𝑢2 + 1ǀ + 𝑐 = −1

𝑥

Sust. u=𝑦

𝑥

𝐼𝑛 ǀ𝑦

𝑥+ √

𝑦

𝑥2

2

+ 1ǀ + 𝑐 = −1

𝑥

14. xydydxyex xy )( 2/2

(𝑥2𝑒−𝑢𝑥

𝑥 + 𝑢2𝑥2)𝑑𝑥 = 𝑢𝑥2(𝑢𝑑𝑥 + 𝑥𝑑𝑢)

(𝑥2𝑒−𝑢)𝑑𝑥 + (𝑢2𝑥2)𝑑𝑥 = (𝑢2𝑥2)𝑑𝑥 + (𝑢𝑥3)𝑑𝑢

(𝑥2𝑒−𝑢)𝑑𝑥 = (𝑢𝑥3)𝑑𝑢

𝑥2

𝑥3𝑑𝑥 =

𝑢

𝑒−𝑢𝑑𝑢

∫𝑑𝑥

𝑥= ∫ 𝑢𝑒𝑢𝑑𝑢

𝐼𝑛ǀ𝑥ǀ = 𝑢𝑒𝑢 − ∫ 𝑒𝑢𝑑𝑢

𝐼𝑛ǀ𝑥ǀ = 𝑢𝑒𝑢 − 𝑒𝑢 + 𝑐

𝐼𝑛ǀ𝑥ǀ = 𝑒𝑢(𝑢 − 1) + 𝑐

Sust. u=𝑦

𝑥

𝐼𝑛ǀ𝑥ǀ = 𝑒𝑦𝑥(

𝑦

𝑥− 1) + 𝑐

16. 02)25( ' yyxy

(5𝑦 − 2𝑥)𝑑𝑦

𝑑𝑥− 2𝑦 = 0

(5𝑦 − 2𝑥)𝑑𝑦 = 2𝑦𝑑𝑥 (5𝑦 − 2𝑢𝑦)𝑑𝑦 = 2𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢)

(5𝑦)𝑑𝑦 − (2𝑢𝑦)𝑑𝑦 = (2𝑢𝑦)𝑑𝑦 + (2𝑦2)𝑑𝑢

(5𝑦)𝑑𝑦 − (4𝑢𝑦)𝑑𝑦 = (2𝑦2)𝑑𝑢

𝑦(5 − 4𝑢)𝑑𝑦 = (2𝑦2)𝑑𝑢

∫𝑦

2𝑦2𝑑𝑦 = ∫

𝑑𝑢

(5 − 4𝑢)

∫ 𝑑𝑦

2𝑦= ∫

𝑑𝑢

(5 − 4𝑢)

1

2𝐼𝑛 ǀ𝑦 ǀ = 𝐼𝑛 ǀ 5 − 4𝑢 ǀ + 𝑐

Sust. u=𝑥

𝑦

1

2𝐼𝑛 ǀ𝑦 ǀ = 𝐼𝑛 ǀ 5 − 4

𝑥

𝑦 ǀ + 𝑐

18.dyyxydx )(2

𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢) = 2(𝑢𝑦 + 𝑢)𝑑𝑦

𝑢𝑦𝑑𝑦 + 𝑦2𝑑𝑢 = 2𝑢𝑦𝑑𝑦 + 2𝑢𝑑𝑦

𝑦2𝑑𝑢 = 𝑢𝑦𝑑𝑦 + 2𝑢𝑑𝑦

𝑦2𝑑𝑢 = 𝑢(𝑦 + 2)𝑑𝑦

∫𝑑𝑢

𝑢= ∫

𝑑𝑦

𝑦+ 2∫

𝑑𝑦

𝑦2

𝐼𝑛 𝑢 + 𝑐 = 𝐼𝑛 𝑦 −2

𝑦

Sust. u=𝑥

𝑦

𝐼𝑛 𝑥

𝑦+ 𝑐 = 𝐼𝑛 ǀ𝑦ǀ −

2

𝑦

19. xy

xy

dx

dy

(𝑦 + 𝑥)𝑑𝑦 = (𝑦 − 𝑥)𝑑𝑥

(𝑢𝑥 + 𝑥)(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = (𝑢𝑥 − 𝑥)𝑑𝑥

𝑥2𝑑𝑢 = −(𝑢2𝑥 − 𝑢𝑥 + 𝑥)𝑑𝑥

𝑥2𝑑𝑢 = −𝑥(𝑢2 − 𝑢 + 1)𝑑𝑥 𝑑𝑢

(𝑢2 − 𝑢 + 1)𝑑𝑢 = −

𝑥

𝑥2𝑑𝑥

∫𝑑𝑢

(𝑢2 − 𝑢 + 1)𝑑𝑢 = −∫

𝑑𝑥

𝑥𝑑𝑥

𝐼𝑛ǀ𝑢2 − 𝑢 + 1ǀ + 𝑐 = −𝐼𝑛ǀ𝑥ǀ

Sust. u=𝑦

𝑥

𝐼𝑛ǀ𝑦2

𝑥2−

𝑦

𝑥+ 1ǀ + 𝑐 = −𝐼𝑛ǀ𝑥ǀ

20. yx

yx

dx

dy

3

3

(3𝑥 + 𝑦)𝑑𝑦 = (𝑥 + 3𝑦)𝑑𝑥

(3𝑥 + 𝑢𝑥)(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 𝑥𝑑𝑥 + 3𝑢𝑥𝑑𝑥

(3𝑢𝑥𝑑𝑥 + 3𝑥2𝑑𝑢 + 𝑢2𝑥𝑑𝑥 + 𝑢𝑥2𝑑𝑢)= 𝑥𝑑𝑥 + 3𝑢𝑥𝑑𝑥

(3𝑥2𝑑𝑢 + 𝑢𝑥2𝑑𝑢) = 𝑥𝑑𝑥 − 𝑢2𝑥𝑑𝑥

𝑥2(𝑢 + 3)𝑑𝑢 = −𝑥(𝑢2 − 1)𝑑𝑥 (𝑢 + 3)

(𝑢2 − 1)= −

𝑥

𝑥2𝑑𝑥

∫𝑢

(𝑢2 − 1)𝑑𝑢 + 3∫

𝑑𝑢

(𝑢2 − 1)= −∫

𝑑𝑥

𝑥

1

2𝐼𝑛ǀ𝑢2 − 1ǀ +

3

2𝐼𝑛ǀ

1 + 𝑢

1 − 𝑢ǀ + 𝑐 = − 𝐼𝑛ǀ𝑥ǀ

Sust. u=𝑦

𝑥

1

2𝐼𝑛ǀ

𝑦2

𝑥2− 1ǀ +

3

2𝐼𝑛ǀ

1 +𝑦𝑥

1 −𝑦𝑥

ǀ + 𝑐 = − 𝐼𝑛ǀ𝑥ǀ

c) Resuelva cada ecuación diferencial sujeta a la condición inicial indicada.

21. 332 xydx

dyxy ; 2)1( y

𝑥𝑦2𝑑𝑦 = (𝑦3 − 𝑥3)𝑑𝑥

𝑢2𝑥3(𝑢𝑑𝑥 + 𝑥𝑑𝑢) = (𝑢3𝑥3 − 𝑥3)𝑑𝑥

𝑢3𝑥3𝑑𝑥 + 𝑢3𝑥4𝑑𝑢 = 𝑢3𝑥3𝑑𝑥 − 𝑥3𝑑𝑥

𝑢3𝑥4𝑑𝑢 = −𝑥3𝑑𝑥

𝑢3𝑑𝑢 = −𝑥3

𝑥4𝑑𝑥

∫ 𝑢3𝑑𝑢 = −∫𝑑𝑥

𝑥

𝑢4

4+ 𝑐 = −𝐼𝑛ǀ𝑥ǀ

𝑐 = −𝐼𝑛ǀ𝑥ǀ −𝑢4

4

Sust. u=𝑦

𝑥

𝑐 = −𝐼𝑛ǀ𝑥ǀ −

𝑦4

𝑥4

4

𝑐 = −𝐼𝑛ǀ𝑥ǀ −𝑦4

4𝑥4

𝒚(𝟏) = 𝟐

𝑐 = −(𝐼𝑛ǀ1ǀ +(2)4

4(1)4) 𝒄 = −4

22. dyyxydyxxydx 222 ; 1)0( y

𝑥𝑦𝑑𝑥 = (𝑦√𝑥2 + 𝑦2 + 𝑥2)𝑑𝑦

𝑢𝑦2(𝑢𝑑𝑦 + 𝑦𝑑𝑢) = (𝑦√𝑢2𝑦2 + 𝑦2 + 𝑢2𝑦2)𝑑𝑦

𝑢2𝑦2𝑑𝑦 + 𝑢𝑦3𝑑𝑢 = 𝑦√𝑢2𝑦2 + 𝑦2𝑑𝑦 + 𝑢2𝑦2𝑑𝑦

𝑢𝑦3𝑑𝑢 = 𝑦√𝑢2𝑦2 + 𝑦2𝑑𝑦

𝑢𝑦3𝑑𝑢 = 𝑦√𝑦2(𝑢2 + 1)𝑑𝑦

𝑢𝑦3𝑑𝑢 = 𝑦2√(𝑢2 + 1)𝑑𝑦

𝑢

√(𝑢2 + 1)𝑑𝑢 =

𝑦2

𝑦3𝑑𝑦

∫𝑢

√(𝑢2 + 1)𝑑𝑢 = ∫

𝑑𝑦

𝑦𝑑𝑦

1

2∫ (𝑢2 + 1)−

12𝑑𝑢 = ∫

𝑑𝑦

𝑦𝑑𝑦

√(𝑢2 + 1) + 𝑐 = 𝐼𝑛ǀ𝑦ǀ

𝑐 = 𝐼𝑛ǀ𝑦ǀ − √(𝑢2 + 1)

Sust. u=𝑥

𝑦

𝑐 = 𝐼𝑛ǀ𝑦ǀ − √𝑥2

𝑦2+ 1

𝒚(𝟎) = 𝟏

𝑐 = 𝐼𝑛ǀ1ǀ − √(0)2

(1)2 + 1 𝒄 = −𝟏

23. 0cos

dyx

y

xyydx ; 2)0( y

𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢) + ( 𝑦𝑐𝑜𝑠𝑢𝑦

𝑦− 𝑢𝑦 ) 𝑑𝑦

𝑢𝑦𝑑𝑦 + 𝑦2𝑑𝑢 + (𝑦𝑐𝑜𝑠𝑢)𝑑𝑦 − 𝑢𝑦𝑑𝑦

𝑦2𝑑𝑢 = −𝑦𝑐𝑜𝑠𝑢𝑑𝑦 𝑑𝑢

𝑐𝑜𝑠𝑢= −

𝑦

𝑦2𝑑𝑦

∫ 𝑠𝑒𝑐𝑢𝑑𝑢 = −∫𝑑𝑦

𝑦

𝑠𝑒𝑐𝑢𝑡𝑎𝑛𝑢 + 𝑐 = −𝐼𝑛ǀ𝑦ǀ 𝑐 = −(𝐼𝑛ǀ𝑦ǀ + 𝑠𝑒𝑐𝑢𝑡𝑎𝑛𝑢)

Sust. u=𝑥

𝑦

𝑐 = −(𝐼𝑛ǀ𝑦ǀ + 𝑠𝑒𝑐𝑥

𝑦𝑡𝑎𝑛

𝑥

𝑦)

𝒚(𝟎) = 𝟐

𝑐 = − (𝐼𝑛ǀ2ǀ + 𝑠𝑒𝑐0

2𝑡𝑎𝑛

0

2)

𝒄 = −𝟎. 𝟔𝟗𝟑𝟏

24. ydx

duxyyx )( 2 ; 1)2/1( y

(𝑥 + √𝑦2 − 𝑥𝑦) 𝑑𝑦 = 𝑦𝑑𝑥

(𝑢𝑦 + √𝑦2 − 𝑢𝑦2) 𝑑𝑦 = 𝑦(𝑢𝑑𝑦 + 𝑦𝑑𝑢)

𝑢𝑦𝑑𝑦 + √𝑦2 − 𝑢𝑦2𝑑𝑦 = 𝑢𝑦𝑑𝑦 + 𝑦2𝑑𝑢

√𝑦2(1 − 𝑢)𝑑𝑦 = 𝑦2𝑑𝑢

𝑦√(1 − 𝑢)𝑑𝑦 = 𝑦2𝑑𝑢

𝑦

𝑦2𝑑𝑦 =

𝑑𝑢

√(1 − 𝑢)

∫𝑑𝑦

𝑦= ∫

𝑑𝑢

√(1 − 𝑢)

𝐼𝑛ǀ𝑦ǀ =1

2√(1 − 𝑢) + 𝑐

𝐼𝑛ǀ𝑦ǀ −1

2√(1 − 𝑢) = 𝑐

Sust. u=𝑥

𝑦

𝐼𝑛ǀ𝑦ǀ −1

2√(1 −

𝑥

𝑦) = 𝑐

𝒚(𝟏𝟐⁄ ) = 𝟏

𝐼𝑛ǀ1ǀ −1

2√(1 −

12⁄

1) = 𝑐

𝒄 = −𝟎. 𝟑𝟓𝟑𝟓

25. 0)( dyxedxyex xy

xy

; 0)1( y

(𝑥 + 𝑦𝑒𝑦

𝑥⁄ ) 𝑑𝑥 = 𝑥𝑒𝑦

𝑥⁄ 𝑑𝑦

(𝑥 + 𝑢𝑥𝑒𝑢𝑥

𝑥⁄ )𝑑𝑥 = 𝑥𝑒𝑢𝑥

𝑥⁄ (𝑢𝑑𝑥 + 𝑥𝑑𝑢)

𝑥𝑑𝑥 + 𝑢𝑥𝑒𝑢𝑑𝑥 = 𝑥𝑢𝑒𝑢𝑑𝑥 + 𝑥2𝑒𝑢𝑑𝑢

𝑥𝑑𝑥 = 𝑥2𝑒𝑢𝑑𝑢 𝑥

𝑥2𝑑𝑥 = 𝑒𝑢𝑑𝑢

∫ 𝑑𝑥

𝑥= ∫ 𝑒𝑢𝑑𝑢

𝐼𝑛ǀ𝑥ǀ = 𝑒𝑢 + 𝑐

𝐼𝑛ǀ𝑥ǀ − 𝑒𝑢 = 𝑐

Sust. u=𝑦

𝑥

𝐼𝑛ǀ𝑥ǀ − 𝑒𝑦

𝑥⁄ = 𝑐

𝒚(𝟏) = 𝟎

𝐼𝑛ǀ1ǀ − 𝑒0

1⁄ = 𝑐 𝒄 = −𝟏