Universidad Tecnológica de Panamá
Lic. En Ingeniería Logística y Cadena de Suministro
Facultad de Ingeniería Industrial
Curso de Ecuaciones Diferenciales
Grupo: 1CL121
Código: 1423
Integrantes: Lynn Salazar (3-748-1444)
Fernando Gómez (8-987-899)
Stanley Berrio (8-991-696)
Taller #6 EDO
Profesor: Martín Peralta
Fecha de entrega: 3 de mayo del 2022
Resuelva las siguientes E.D exactas, demuestre que son exactas y presente todos los
procedimientos y escriba cada problema en el editor de ecuaciones
1. (𝑥 2 − 1)𝑦´ + 2𝑥𝑦 = 0 para y(2)=-1
𝑑𝑦
(𝑥 2 − 1)
+ 2𝑥𝑦 = 0
𝑑𝑥
(𝑥 2 − 1) 𝑑𝑦 = (−2𝑥𝑦)(𝑑𝑥)
(𝑥 2 − 1) 𝑑𝑦 + (2𝑥𝑦)(𝑑𝑥) = 0
2𝑀 2𝑁
=
2𝑥
2𝑦
𝑑 2
(𝑥 − 1) = 2𝑥
𝑑𝑥
𝑑
2𝑥𝑦 = 2𝑥
𝑑𝑦
exactas
∫(𝑥 2 − 1) 𝑑𝑦
∫ 𝑥 2 𝑑𝑦 − ∫ 1 𝑑𝑦
∫ 2𝑥𝑦(𝑑𝑥)
2𝑦 ∫ 𝑥𝑑𝑥
𝑥2𝑦 − 𝑦
2𝑦( 2 )
𝑥2
𝑥2𝑦
𝑥 2 𝑦 − 𝑦 = 𝑐 solución general
Reemplazamos para y(2)=-1
22 (−1) − (−1) = 𝑐
−4 + 1 = 𝑐
𝑐 = −3
𝑥 2 𝑦 − 𝑦 − 3 = 0 solución particular
𝟓
𝟐. (𝟓𝒚 − 𝟐𝒙)𝒚′ − 𝟐𝒚 = 𝟎 … … . 𝒚𝟐 − 𝟐𝒙𝒚 = 𝒄
𝟐
(5𝑥 − 2𝑥)
𝑑𝑦
− 2𝑦 = 0
𝑑𝑥
(5𝑥 − 2𝑥)
𝑑𝑦
= 2𝑦
𝑑𝑥
(5𝑥 − 2𝑥)𝑑𝑦 = 2𝑦𝑑𝑥
−2𝑦𝑑𝑥 + (5𝑥 − 2𝑥)𝑑𝑦 = 0 →
𝜕𝑀
= (−2𝑦)′ = −2(1) = −2
𝜕𝑦
𝜕𝑁
= (5𝑦 − 2𝑥)′ = −2(1) = −2
𝜕𝑥
𝑀 = −2𝑦 ; 𝑁 = (5𝑦 − 2𝑥)
𝜕𝑀 𝜕𝑁
=
𝑠𝑜𝑛 𝐸. 𝐷 𝑒𝑥𝑎𝑐𝑡𝑎𝑠
𝜕𝑦
𝜕𝑥
Resolvemos integrales:
∫ 𝑁(𝑥, 𝑦)𝑑𝑦 = ∫(5𝑥 − 2𝑥)𝑑𝑦
∫ 𝑀(𝑥, 𝑦)𝑑𝑥 = ∫ −2𝑦𝑑𝑥
= −2𝑦 ∫ 𝑑𝑥
∫ 5𝑦𝑑𝑦 − ∫ 2𝑥𝑑𝑦
= −2𝑦[𝑥]
5 ∫ 𝑦𝑑𝑦 − 2𝑥 ∫ 𝑑𝑦
= −2𝑥𝑦
𝑦2
5 [ ] − 2𝑥[𝑦]
2
5 2
𝑦 − 2𝑥𝑦
2
𝑓(𝑥, 𝑦) = ∫ 𝑀(𝑥, 𝑦)𝑑𝑥 + ∫ 𝑁(𝑥, 𝑦)𝑑𝑦 = 𝐶
5
𝑓(𝑥, 𝑦) = −2𝑥𝑦 + 𝑦 2 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖𝑜𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 "𝑓𝑎𝑚𝑖𝑙𝑖𝑎"
2
3. (2𝑥 + 𝑦 + 2𝑥𝑦 2 )𝑑𝑥 + (𝑥 + 2𝑥 2 𝑦)𝑑𝑦 = 0
𝑑
(𝑥 + 2𝑥 2 𝑦) = 1 + 4𝑥𝑦
𝑑𝑥
𝑑
(2𝑥 + 𝑦 + 2𝑥𝑦 2 ) = 1 + 4𝑥𝑦
𝑑𝑦
∫(2𝑥 + 𝑦 + 2𝑥𝑦 2 )𝑑𝑥
∫ 2𝑥𝑑𝑥 + ∫ 𝑦𝑑𝑥 + ∫ 2𝑥𝑦 2 𝑑𝑥
𝑥2
𝑥2
2 ( 2 ) + 𝑦𝑥 + 𝑦 2 (2) ( 2 )
(𝑥 2 ) + 𝑦𝑥 + 𝑥 2 𝑦 2
𝑥 2 + 𝑦𝑥 + 𝑥 2 𝑦 2 =c solución general
4.𝑦 ′ =
−(4𝑥 3 −4𝑥𝑦 2 +𝑦)
4𝑦 3 −4𝑥 2 𝑦+𝑥
∶ 𝑦(−1) = −1
∫ 𝑥 + 2𝑥 2 𝑦(𝑑𝑦)
∫ 𝑥𝑑𝑦 + ∫ 2𝑥 2 𝑦(𝑑𝑦)
𝑦2
𝑦𝑥 + 2𝑥 2 ( 2 )
𝑦𝑥 + 𝑥 2 𝑦 2
𝑑𝑦 −(4𝑥 3 − 4𝑥𝑦 2 + 𝑦)
=
; 𝑦(−1) = −1
𝑑𝑥
4𝑦 3 − 4𝑥 2 𝑦 + 𝑥
(4𝑦 3 − 4𝑥 2 𝑦 + 𝑥)𝑑𝑦 = −(4𝑥 3 − 4𝑥𝑦 2 + 𝑦)𝑑𝑥
(4𝑥 3 − 4𝑥𝑦 2 + 𝑦)𝑑𝑥 = (4𝑦 3 − 4𝑥 2 𝑦 + 𝑥)𝑑𝑦
𝜕𝑀
= (4𝑥 3 − 4𝑥𝑦 2 + 𝑦) = −4(𝑥)(2𝑦) + 1 = −8𝑥𝑦 + 1
𝜕𝑦
𝜕𝑁
= (4𝑦 3 − 4𝑥 2 𝑦 + 𝑥) = −4(2𝑥)(𝑦) + 1 = −8𝑥𝑦 + 1
𝜕𝑥
∫ 𝑀 𝑑𝑥 = ( 4𝑥 3 − 4𝑥𝑦 2 + 𝑦)
= ∫ 4𝑥 3 𝑑𝑥 − ∫ 4𝑥𝑦 2 𝑑𝑥 + ∫ 𝑦 𝑑𝑥
= 4 ∫ 𝑥 3 𝑑𝑥 − 4𝑦 2 ∫ 𝑥 𝑑𝑥 + 𝑦 ∫ 𝑑𝑥
= 4(
𝑥4
𝑥2
) − 4𝑦 2 ( ) + 𝑦(𝑥)
4
2
= 𝑥 4 − 2𝑦 2 𝑥 2 + 𝑥𝑦
∫ 𝑁 𝑑𝑦 = (4𝑦 3 − 4𝑥 2 𝑦 + 𝑥)
= 4 ∫ 𝑦 3 𝑑𝑦 − 4𝑥 2 ∫ 𝑦 𝑑𝑦 + 𝑥 ∫ 𝑑𝑦
= 4(
𝑦4
𝑦2
2
−
4𝑥
)
( ) + 𝑥(𝑦)
4
2
= 𝑦 4 − 2𝑥 2 𝑦 2 + 𝑥𝑦
𝑥 4 + 𝑦 4 − 2𝑥 2 𝑦 2 + 𝑥𝑦 = 𝐶 𝑠𝑜𝑙𝑢𝑐𝑖𝑜𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 "𝑓𝑎𝑚𝑖𝑙𝑖𝑎"
Para y(-1) = -1
𝑥 4 + 𝑦 4 − 2𝑥 2 𝑦 2 + 𝑥𝑦 = 0
(−1)4 + (−1)4 − 2(−1)2 (−1)2 + (−1)(−1) = 𝐶
1=𝐶
𝑓(𝑥, 𝑦) = 𝑥 4 + 𝑦 4 − 2𝑥 2 𝑦 2 + 𝑥𝑦 − 1 = 0 solución particular
2
2
5.(−2𝑥 3 + 2𝑥𝑒 𝑥 𝑦)𝑑𝑥 + (𝑒 𝑥 + 2𝑦)𝑑𝑦 = 0
𝑑
2
2
(−2𝑥 3 + 2𝑥𝑒 𝑥 𝑦) = 2𝑥𝑒 𝑥
𝑑𝑦
𝑑 𝑥2
2
(𝑒 + 2𝑦) = 2𝑥𝑒 𝑥
𝑑𝑥
2
∫(−2𝑥 3 + 2𝑥𝑒 𝑥 𝑦)𝑑𝑥
2
− ∫ 2𝑥 3 𝑑𝑥 + ∫ 2𝑥𝑒 𝑥 𝑦 (𝑑𝑥)
𝑥4
2
−2 ( 4 ) + 2𝑦 ∫ 𝑥𝑒 𝑥 (𝑑𝑥)
𝑡 = 𝑒𝑥
2
2
𝑑𝑡 = 2𝑥𝑒 𝑥 (𝑑𝑥)
4
𝑥
𝑑𝑡
) + 2𝑦 ∫ 𝑡 ( )
2
2𝑡
4
𝑥
𝑑𝑡
− ( ) + 2𝑦 ∫
2
2
4
𝑥
1
− ( ) + 2𝑦( 𝑡)
2
2
𝑥4
1 2
− ( ) + 2𝑦( 𝑒 𝑥 )
2
2
4
𝑥
2
− ( ) + 𝑦𝑒 𝑥
2
−(
2
∫(𝑒 𝑥 + 2𝑦)𝑑𝑦
2
∫ 𝑒 𝑥 𝑑𝑦 + ∫ 2𝑦𝑑𝑦
2
𝑒 𝑥 ∫ 𝑑𝑦 + 2 ∫ 𝑦𝑑𝑦
2
𝑒𝑥 𝑦 + 2 (
𝑦2
)
2
2
𝑒 𝑥 𝑦 + 𝑦2
𝑥4
2
(− ) + 𝑦𝑒 𝑥 + 𝑦 2 = 𝑐
2
6. 𝑒 2𝑦 𝑑𝑥 + 2𝑥 𝑒 2𝑦 𝑑𝑦 − 𝑦 cos(𝑥𝑦) 𝑑𝑥 − 𝑥𝑐𝑜𝑠(𝑥𝑦)𝑑𝑦 = −2𝑦 𝑑𝑦
𝑒 2𝑦 𝑑𝑥 + 2𝑥 𝑒 2𝑦 𝑑𝑦 − 𝑦 cos(𝑥𝑦) 𝑑𝑥 − 𝑥𝑐𝑜𝑠(𝑥𝑦)𝑑𝑦 + 2𝑦 𝑑𝑦 = 0
𝑒 2𝑦 𝑑𝑥 − 𝑦 cos(𝑥𝑦) 𝑑𝑥 + 2𝑥 𝑒 2𝑦 𝑑𝑦 − 𝑥𝑐𝑜𝑠(𝑥𝑦)𝑑𝑦 + 2𝑦 𝑑𝑦 = 0
[𝑒 2𝑦 − 𝑦 cos(𝑥𝑦)] 𝑑𝑥 + [2𝑥 𝑒 2𝑦 − 𝑥𝑐𝑜𝑠(𝑥𝑦) + 2𝑦] 𝑑𝑦 = 0
𝑑𝑥 =
𝑑𝑡
2
2𝑥𝑒 𝑥
Donde M= [𝑒 2𝑦 − 𝑦 cos(𝑥𝑦)] y N= 2𝑥 𝑒 2𝑦 − 𝑥𝑐𝑜𝑠(𝑥𝑦) + 2𝑦
𝜕𝑀 𝜕𝑁
=
𝑦
𝑥
𝜕𝑀
= [𝑒 2𝑦 − 𝑦 cos(𝑥𝑦)]′ = 2𝑒 2𝑦 − cos(𝑥𝑦)
𝑦
𝜕𝑁
= [2𝑥 𝑒 2𝑦 − 𝑥𝑐𝑜𝑠(𝑥𝑦) + 2𝑦]′ = 2𝑒 2𝑦 − cos(𝑥𝑦)
𝑥
2𝑒 2𝑦 − cos(𝑥𝑦) = 2𝑒 2𝑦 − cos(𝑥𝑦)
Es una E.D exacta
∫ 𝑀 𝑑𝑥
∫[𝑒 2𝑦 − 𝑦 cos(𝑥𝑦)] 𝑑𝑥
∫ 𝑒 2𝑦 𝑑𝑥 − ∫ 𝑦 cos(𝑥𝑦)] 𝑑𝑥
𝑒 2𝑦 ∫ 𝑑𝑥 − 𝑦 ∫ cos(𝑥𝑦) 𝑑𝑥
𝑒 2𝑦 (𝑥) − 𝑦 ∫ cos 𝑢
𝑢 = 𝑥𝑦
𝑑𝑢
𝑦
𝑑𝑢 = 𝑦 𝑑𝑥
𝑑𝑢
= 𝑑𝑥
𝑦
1
𝑒 2𝑦 (𝑥) − 𝑦. ∫ cos 𝑢 𝑑𝑢
𝑦
𝑥𝑒 2𝑦 − 𝑠𝑒𝑛 𝑢
𝑥𝑒 2𝑦 − 𝑠𝑒𝑛(𝑥𝑦)
∫ 𝑁 𝑑𝑦
∫[2𝑥 𝑒 2𝑦 − 𝑥𝑐𝑜𝑠(𝑥𝑦) + 2𝑦] 𝑑𝑦
∫ 2𝑥 𝑒 2𝑦 𝑑𝑦 − ∫ 𝑥𝑐𝑜𝑠(𝑥𝑦) 𝑑𝑦 + ∫ 2𝑦 𝑑𝑦
2𝑥 ∫ 𝑒 2𝑦 𝑑𝑦 − 𝑥 ∫ 𝑐𝑜𝑠(𝑥𝑦) 𝑑𝑦 + 2 ∫ 𝑦 𝑑𝑦
𝑢 = 2𝑦
1
2𝑥 ∫ 𝑒 𝑢 . 2 𝑑𝑢 − 𝑥 ∫ cos 𝑣
𝑑𝑣
𝑥
+ 2 ∫ 𝑦 𝑑𝑦
1
1
1
. 2 𝑥 [𝑒 𝑢 𝑑𝑢] − 𝑥. ∫ cos 𝑣 𝑑𝑣 + 2. (𝑦 2 )
2
𝑥
2
𝑥𝑒 𝑢 − 𝑠𝑒𝑛 𝑣 + 𝑦 2
𝑑𝑢 = 2 𝑑𝑦
1
𝑑𝑢 = 𝑑𝑦
2
𝑣 = 𝑥𝑦
𝑑𝑣 = 1(𝑥)𝑑𝑦
𝑑𝑣
= 𝑑𝑦
𝑥
𝑥𝑒 2𝑦 − 𝑠𝑒𝑛(𝑥𝑦) + 𝑦 2
𝑓(𝑥, 𝑦) = 𝑥𝑒 2𝑦 − 𝑠𝑒𝑛(𝑥𝑦) + 𝑦 2 = 𝐶 Solución general
𝑑𝑦
7. 𝑑𝑥 =
𝑥𝑦 2 −𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥
𝑦−𝑥 2 𝑦
(𝑦 − 𝑥 2 𝑦) 𝑑𝑦 = (𝑥𝑦 2 − 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥)𝑑𝑥
(𝑦 − 𝑥 2 𝑦) 𝑑𝑦 − (𝑥𝑦 2 − 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥)𝑑𝑥 = 0
(−𝑥𝑦 2 + 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥)𝑑𝑥 + (𝑦 − 𝑥 2 𝑦) 𝑑𝑦 = 0
Donde M=(−𝑥𝑦 2 + 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥) y N= (𝑦 − 𝑥 2 𝑦)
𝜕𝑀
= (−𝑥𝑦 2 + 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥) ′
𝑦
𝜕𝑀
𝑦
𝜕𝑀 𝜕𝑁
=
𝑦
𝑥
= −2𝑥𝑦
𝜕𝑁
= (𝑦 − 𝑥 2 𝑦)′
𝑥
−2𝑥𝑦 = −2𝑥𝑦
La E.D es exacta
𝜕𝑁
= −2𝑥𝑦
𝑥
∫ 𝑀 𝑑𝑥
∫(−𝑥𝑦 2 + 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥) 𝑑𝑥
− ∫ 𝑥 𝑦 2 𝑑𝑥 + ∫ 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥 𝑑𝑥
−𝑦 2 ∫ 𝑥 𝑑𝑥 + ∫ 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥
𝑥2
−𝑦 2 ( 2 ) − ∫ 𝑢 𝑑𝑢
2
−
1 2 2 𝑢
𝑥 𝑦 −
2
2
−
1 2 2 1
𝑥 𝑦 − 𝑐𝑜𝑠 2 𝑥
2
2
𝑢 = 𝑐𝑜𝑠𝑥
𝑑𝑢 = −𝑠𝑒𝑛𝑥
∫ 𝑁 𝑑𝑦
∫(𝑦 − 𝑥 2 𝑦) 𝑑𝑦
∫ 𝑦 𝑑𝑦 − ∫ 𝑥 2 𝑦 𝑑𝑦
∫ 𝑦 𝑑𝑦 − 𝑥 2 ∫ 𝑦 𝑑𝑦
𝑦2
𝑦2
− 𝑥2 ( )
2
2
1 2 1 2 2
𝑦 − 𝑥 𝑦
2
2
𝑓(𝑥, 𝑦) = −
1 2 2 1 2 1
𝑥 𝑦 + 𝑦 − 𝑐𝑜𝑠 2 𝑥 = 𝐶
2
2
2
− 𝑥 2 𝑦 2 + 𝑦 2 − 𝑐𝑜𝑠 2 𝑥 = 2𝐶
𝑦 2 (− 𝑥 2 + 1) − 𝑐𝑜𝑠 2 𝑥 = 2𝐶
𝑦 2 (1 − 𝑥 2 ) − 𝑐𝑜𝑠 2 𝑥 = 2𝐶
𝑦 2 (1 − 𝑥 2 ) − 𝑐𝑜𝑠 2 𝑥 = 𝑘 Solución general
Para el punto 𝑦(0) = 2
𝑥 = 0 ;𝑦 = 2
(2)2 (1 − 02 ) − 𝑐𝑜𝑠 2 0 = 𝑘
4(1) − (1)2 = 𝑘
4−1=𝑘
3=𝑘
𝑦 2 (1 − 𝑥 2 ) − 𝑐𝑜𝑠 2 𝑥 = 3 Solución Particular
8. (𝑦 2 cos 𝑥 − 3𝑥 2 𝑦 − 2𝑥)𝑑𝑥 + (2𝑦 sin 𝑥 − 𝑥 3 + 𝑙𝑛|𝑥|)𝑑𝑦 = 0 … … 𝑦(0) = 𝑒
𝑦 2 sin 𝑥 − 𝑥 3 𝑦 − 𝑥 2 + 𝑦𝑙𝑛|𝑦| − 𝑦 = 𝑘
{ 2
𝑦 sin 𝑥 − 𝑥 3 𝑦 − 𝑥 2 + 𝑦𝑙𝑛|𝑦| − 𝑦 = 0
−2𝑥𝑑𝑥 + 2𝑦 sin 𝑥 𝑑𝑦 − 3𝑥 2 𝑦𝑑𝑥 − 𝑥 3 𝑑𝑦 + 𝑦 2 cos 𝑥 𝑑𝑥 + 𝑙𝑛|𝑦|𝑑𝑦 = 0
(−2𝑥 − 3𝑥 2 𝑦 + 𝑦 2 cos 𝑥)𝑑𝑥 + (2𝑦 sin 𝑥 − 𝑥 3 + 𝑙𝑛|𝑦|)𝑑𝑦 = 0 →
𝑀 = −2𝑥 − 3𝑥 2 𝑦 + 𝑦 2 cos 𝑥
𝑁 = 2𝑦 sin 𝑥 − 𝑥 3 + 𝑙𝑛|𝑦|
𝜕𝑀
= (−2𝑥 − 3𝑥 2 𝑦 + 𝑦 2 cos 𝑥)′
𝜕𝑦
= −3𝑥 2 (1) + (2𝑦) cos 𝑥
𝜕𝑁
= (2𝑦 sin 𝑥 − 𝑥 3 + 𝑙𝑛|𝑦|)′
𝜕𝑥
= −3𝑥 2 + 2𝑦 cos 𝑥
2𝑦(cos 𝑥)(1) − 3𝑥 2
𝜕𝑀 𝜕𝑁
=
𝑒𝑠 𝑢𝑛𝑎 𝐸. 𝐷 𝑒𝑥𝑎𝑐𝑡𝑎
𝜕𝑦
𝜕𝑥
2𝑦 cos 𝑥 − 3𝑥 2
∫ 𝑀(𝑥, 𝑦)𝑑𝑥 = ∫(−2𝑥 − 3𝑥 2 𝑦 + 𝑦 2 cos 𝑥)𝑑𝑥
∫ −2𝑥𝑑𝑥 − ∫ 3𝑥 2 𝑦𝑑𝑥 + ∫ 𝑦 2 cos 𝑥 𝑑𝑥
= −2 ∫ 𝑥𝑑𝑥 − 3𝑦 ∫ 𝑥 2 𝑑𝑥 + 𝑦 2 ∫ cos 𝑥 𝑑𝑥
𝑥2
𝑥3
−2 [ ] − 3𝑦 [ ] + 𝑦 2 [sin 𝑥]
2
3
= −𝑥 2 − 𝑥 3 𝑦 + 𝑦 2 sin 𝑥
∫ 𝑁(𝑥, 𝑦)𝑑𝑦 = ∫(2𝑦 sin 𝑥 − 𝑥 3 + 𝑙𝑛|𝑦|)𝑑𝑦
∫ 2𝑦 sin 𝑥𝑑𝑦 − ∫ 𝑥 3 𝑑𝑦 + ∫ 𝑙𝑛[𝑦] 𝑑𝑦
2 sin 𝑥 ∫ 𝑦 𝑑𝑦 − 𝑥 3 ∫ 𝑑𝑦 + ∫ 𝑙𝑛|𝑦|𝑑𝑦 →
∫ 𝑙𝑛|𝑢|𝑑𝑢 = 𝑢𝑙𝑛|𝑢|-u
𝑦2
2 sin 𝑥 [ ] − 𝑥 3 [𝑦] + 𝑦𝑙𝑛|𝑦| − 𝑦
2
𝑦 2 sin 𝑥 − 𝑥 3 𝑦 + 𝑦𝑙𝑛|𝑦| − 𝑦
𝑓(𝑥, 𝑦) = −𝑥 2 − 𝑥 3 𝑦 + 𝑦 2 sin 𝑥 + 𝑦𝑙𝑛|𝑦| − 𝑦 = 𝑐 solución general “familia”