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PROGRAMA NACIONAL DE FORMACIÓN EN SISTEMA DE CALIDAD Y
AMBIENTE.
Matemática Aplicada.
Ejercicios Unidad I. parte II
Integrante:
Yonaiker Morles C.I.: 25.814.152
Trayecto 3 Fase 2
Grupo-A
Febrero de 2021
Ejercicios propuestos 1.2:
Verifique si la ecuación diferencial es exacta, separable, homogénea o lineal.
𝟏)
𝒅𝒚
𝒅𝒙
=
𝒙𝒚 + 𝟑𝒙 − 𝒚 − 𝟑
𝒙𝒚 − 𝟐𝒙 + 𝟒𝒚 − 𝟖
Solución:
Verificamos si la ecuación diferencial es variable separable si se cumple:
𝑓(𝑦)𝑑𝑦 = 𝑔(𝑥)𝑑𝑥 ; 𝐻(𝑥, 𝑦)
𝑔(𝑥)
𝑓(𝑥)
Despejamos:
(𝑥𝑦 − 2𝑥 + 4𝑦 − 8)𝑑𝑦 = (𝑥𝑦 + 3𝑥 − 𝑦 − 3)𝑑𝑥
𝑑𝑦
𝑥𝑦 + 3𝑥 − 𝑦 − 3
=
𝑑𝑥
𝑥𝑦 − 2𝑥 + 4𝑦 − 8
Por lo que la ecuación diferencial
𝑑𝑦
𝑑𝑥
=
𝑥𝑦+3𝑥−𝑦−3
𝑥𝑦−2𝑥+4𝑦−8
es separable.
𝟑) 𝒚´ = 𝟐𝒚 + 𝒙𝟐
+ 𝟓
Solución:
Veamos si la ecuación diferencial 𝑦´ = 2𝑦 + 𝑥2
+ 5 se puede expresar en la forma estándar.
𝑑𝑦
𝑑𝑥
+ 𝑝(𝑥)𝑦 = 𝑓(𝑥)
Sabemos que 𝑦´ =
𝑑𝑦
𝑑𝑥
;
𝑑𝑦
𝑑𝑥
= 2𝑦 + 𝑥2
+ 5
Por lo que
𝑑𝑦
𝑑𝑥
+ 𝑝(𝑥)𝑦 = 𝑓(𝑥) 𝑑𝑜𝑛𝑑𝑒 𝑝(𝑥) = 0
𝑦𝑓(𝑥) = 2𝑦 + 𝑥2
+ 5
Asi la ecuación 𝑦´ = 2𝑦 + 𝑥2
+ 5 es lineal.
𝟖) (𝒚𝟐
+ 𝒚𝒙)𝒅𝒙 − 𝒙𝟐
𝒅𝒚 = 𝟎
Solución:
Sabemos que 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0
Utilizando la def. 2: 𝑀(𝑡𝑥, 𝑡𝑦) = 𝑡𝑎1 𝑀(𝑥, 𝑦) 𝑦 𝑁(𝑡𝑥, 𝑡𝑦) = 𝑡𝑎2 𝑁(𝑥, 𝑦) 𝑎1 = 𝑎2
Verificamos que 𝑀(𝑥, 𝑦) = 𝑦2
+ 𝑦𝑥 𝑦 𝑁(𝑥, 𝑦) = 𝑥2
Son homogéneas del mismo grado
𝑀(𝑡𝑥, 𝑡𝑦) = (𝑡𝑦)2
+ (𝑡𝑦)(𝑡𝑥) ; 𝑁(𝑡𝑥, 𝑡𝑦) = (𝑡𝑥)2
= 𝑡2
𝑦2
+ 𝑡2
𝑦𝑥 = 𝑡2
𝑥2
= 𝑡2
(𝑦2
+ 𝑦𝑥) = 𝑡2
𝑁(𝑥, 𝑦)
= 𝑡2
𝑀(𝑥, 𝑦) 𝑎1 = 2 𝑎2 = 2
Así tenemos que 𝑎1 = 𝑎2 , 𝑀(𝑥, 𝑦) 𝑦 𝑁(𝑥, 𝑦) son homogéneas del mismo grado.
Por lo tanto (𝑦2
+ 𝑦𝑥)𝑑𝑥 − 𝑥2
𝑑𝑦 = 0 es homogénea.
𝟏𝟏) (𝒚𝟑
− 𝒚𝟐
𝒔𝒆𝒏(𝒙) − 𝒙)𝒅𝒙 + (𝟑𝒙𝒚𝟐
+ 𝟐𝒚𝒄𝒐𝒔(𝒙)) 𝒅𝒚 = 𝟎
Solución:
En la ecuación diferencial (𝑦3
− 𝑦2
𝑠𝑒𝑛(𝑥) − 𝑥)𝑑𝑥 + (3𝑥𝑦2
+ 2𝑦𝑐𝑜𝑠(𝑥))𝑑𝑦 = 0
𝑀(𝑥, 𝑦) = 𝑦3
− 𝑦2
𝑠𝑒𝑛(𝑥) − 𝑥 𝑌 𝑁(𝑥, 𝑦) = 3𝑥𝑦2
+ 2𝑦𝑐𝑜𝑠(𝑥)
Derivando 𝑀 con respecto a 𝑦 y a 𝑁 con respecto a 𝑥, se tiene que:
𝜕
𝜕𝑦
𝑀(𝑥, 𝑦) = 𝑦3
− 𝑦2
𝑠𝑒𝑛(𝑥) − 𝑥 = 3𝑦2
− 2𝑦𝑠𝑒𝑛(𝑥)
𝜕
𝜕𝑥
𝑁(𝑥, 𝑦) = 3𝑥𝑦2
+ 2𝑦𝑐𝑜𝑠(𝑥) = 3𝑦2
− 2𝑦𝑠𝑒𝑛(𝑥)
𝑐𝑜𝑚𝑜
𝜕
𝜕𝑦
𝑀(𝑥, 𝑦) =
𝜕
𝜕𝑥
𝑁(𝑥, 𝑦)
Por lo tanto (𝑦3
− 𝑦2
𝑠𝑒𝑛(𝑥) − 𝑥)𝑑𝑥 + (3𝑥𝑦2
+ 2𝑦𝑐𝑜𝑠(𝑥))𝑑𝑦 = 0 es exacta.

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Yoaniker morles2

  • 1. PROGRAMA NACIONAL DE FORMACIÓN EN SISTEMA DE CALIDAD Y AMBIENTE. Matemática Aplicada. Ejercicios Unidad I. parte II Integrante: Yonaiker Morles C.I.: 25.814.152 Trayecto 3 Fase 2 Grupo-A Febrero de 2021
  • 2. Ejercicios propuestos 1.2: Verifique si la ecuación diferencial es exacta, separable, homogénea o lineal. 𝟏) 𝒅𝒚 𝒅𝒙 = 𝒙𝒚 + 𝟑𝒙 − 𝒚 − 𝟑 𝒙𝒚 − 𝟐𝒙 + 𝟒𝒚 − 𝟖 Solución: Verificamos si la ecuación diferencial es variable separable si se cumple: 𝑓(𝑦)𝑑𝑦 = 𝑔(𝑥)𝑑𝑥 ; 𝐻(𝑥, 𝑦) 𝑔(𝑥) 𝑓(𝑥) Despejamos: (𝑥𝑦 − 2𝑥 + 4𝑦 − 8)𝑑𝑦 = (𝑥𝑦 + 3𝑥 − 𝑦 − 3)𝑑𝑥 𝑑𝑦 𝑥𝑦 + 3𝑥 − 𝑦 − 3 = 𝑑𝑥 𝑥𝑦 − 2𝑥 + 4𝑦 − 8 Por lo que la ecuación diferencial 𝑑𝑦 𝑑𝑥 = 𝑥𝑦+3𝑥−𝑦−3 𝑥𝑦−2𝑥+4𝑦−8 es separable. 𝟑) 𝒚´ = 𝟐𝒚 + 𝒙𝟐 + 𝟓 Solución: Veamos si la ecuación diferencial 𝑦´ = 2𝑦 + 𝑥2 + 5 se puede expresar en la forma estándar. 𝑑𝑦 𝑑𝑥 + 𝑝(𝑥)𝑦 = 𝑓(𝑥) Sabemos que 𝑦´ = 𝑑𝑦 𝑑𝑥 ; 𝑑𝑦 𝑑𝑥 = 2𝑦 + 𝑥2 + 5 Por lo que 𝑑𝑦 𝑑𝑥 + 𝑝(𝑥)𝑦 = 𝑓(𝑥) 𝑑𝑜𝑛𝑑𝑒 𝑝(𝑥) = 0 𝑦𝑓(𝑥) = 2𝑦 + 𝑥2 + 5 Asi la ecuación 𝑦´ = 2𝑦 + 𝑥2 + 5 es lineal.
  • 3. 𝟖) (𝒚𝟐 + 𝒚𝒙)𝒅𝒙 − 𝒙𝟐 𝒅𝒚 = 𝟎 Solución: Sabemos que 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0 Utilizando la def. 2: 𝑀(𝑡𝑥, 𝑡𝑦) = 𝑡𝑎1 𝑀(𝑥, 𝑦) 𝑦 𝑁(𝑡𝑥, 𝑡𝑦) = 𝑡𝑎2 𝑁(𝑥, 𝑦) 𝑎1 = 𝑎2 Verificamos que 𝑀(𝑥, 𝑦) = 𝑦2 + 𝑦𝑥 𝑦 𝑁(𝑥, 𝑦) = 𝑥2 Son homogéneas del mismo grado 𝑀(𝑡𝑥, 𝑡𝑦) = (𝑡𝑦)2 + (𝑡𝑦)(𝑡𝑥) ; 𝑁(𝑡𝑥, 𝑡𝑦) = (𝑡𝑥)2 = 𝑡2 𝑦2 + 𝑡2 𝑦𝑥 = 𝑡2 𝑥2 = 𝑡2 (𝑦2 + 𝑦𝑥) = 𝑡2 𝑁(𝑥, 𝑦) = 𝑡2 𝑀(𝑥, 𝑦) 𝑎1 = 2 𝑎2 = 2 Así tenemos que 𝑎1 = 𝑎2 , 𝑀(𝑥, 𝑦) 𝑦 𝑁(𝑥, 𝑦) son homogéneas del mismo grado. Por lo tanto (𝑦2 + 𝑦𝑥)𝑑𝑥 − 𝑥2 𝑑𝑦 = 0 es homogénea. 𝟏𝟏) (𝒚𝟑 − 𝒚𝟐 𝒔𝒆𝒏(𝒙) − 𝒙)𝒅𝒙 + (𝟑𝒙𝒚𝟐 + 𝟐𝒚𝒄𝒐𝒔(𝒙)) 𝒅𝒚 = 𝟎 Solución: En la ecuación diferencial (𝑦3 − 𝑦2 𝑠𝑒𝑛(𝑥) − 𝑥)𝑑𝑥 + (3𝑥𝑦2 + 2𝑦𝑐𝑜𝑠(𝑥))𝑑𝑦 = 0 𝑀(𝑥, 𝑦) = 𝑦3 − 𝑦2 𝑠𝑒𝑛(𝑥) − 𝑥 𝑌 𝑁(𝑥, 𝑦) = 3𝑥𝑦2 + 2𝑦𝑐𝑜𝑠(𝑥) Derivando 𝑀 con respecto a 𝑦 y a 𝑁 con respecto a 𝑥, se tiene que: 𝜕 𝜕𝑦 𝑀(𝑥, 𝑦) = 𝑦3 − 𝑦2 𝑠𝑒𝑛(𝑥) − 𝑥 = 3𝑦2 − 2𝑦𝑠𝑒𝑛(𝑥) 𝜕 𝜕𝑥 𝑁(𝑥, 𝑦) = 3𝑥𝑦2 + 2𝑦𝑐𝑜𝑠(𝑥) = 3𝑦2 − 2𝑦𝑠𝑒𝑛(𝑥) 𝑐𝑜𝑚𝑜 𝜕 𝜕𝑦 𝑀(𝑥, 𝑦) = 𝜕 𝜕𝑥 𝑁(𝑥, 𝑦) Por lo tanto (𝑦3 − 𝑦2 𝑠𝑒𝑛(𝑥) − 𝑥)𝑑𝑥 + (3𝑥𝑦2 + 2𝑦𝑐𝑜𝑠(𝑥))𝑑𝑦 = 0 es exacta.