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Ecuaciones exactas y factor integrante
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ECUACIONES EXACTAS Y FACTOR INTEGRANTE Instrucciones: Resuelve lo que te pide en cada apartado. I. Encuentra la solución general de las siguientes ecuaciones diferenciales exactas. 1. [𝐭𝐚𝐧 𝒙 − 𝐬𝐢𝐧 𝒙 𝐬𝐢𝐧 𝒚]𝒅𝒙 + [𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒚]𝒅𝒚 = 𝟎 𝑀 = [tan 𝑥 − sin 𝑥 sin 𝑦]𝑑𝑥 𝑁 = [cos 𝑥 cos 𝑦]𝑑𝑦 Comprobando que sean ecuaciones exactas: 𝜕𝑀 𝜕𝑦 = 𝜕 𝜕𝑦 tan 𝑥 − (sin 𝑥 𝜕 𝜕𝑦 sin 𝑦 + sin 𝑦 𝜕 𝜕𝑦 sin 𝑥) → 0 − (sin 𝑥 cos 𝑦 + 0) → − 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒚 𝜕𝑁 𝜕𝑥 = cos 𝑥 𝜕 𝜕𝑥 cos 𝑦 + cos 𝑦 𝜕 𝜕𝑥 cos 𝑥 → 0 − (sin 𝑥 cos 𝑦 + 0) → − 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒚 ∫[tan 𝑥 − sin 𝑥 sin 𝑦]𝑑𝑥 + ∫[𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 𝑦 ]𝑑𝑦 = ∫ 0 → ∫ tan 𝑥 𝑑𝑥 − sin 𝑥 ∫ sin 𝑦 𝑑𝑥 + ∫[cos 𝑥 cos 𝑦]𝑑𝑦 = ∫ 0 → 𝐥𝐧|𝐬𝐞𝐜 𝒙| − 𝐬𝐢𝐧 𝒚 𝐜𝐨𝐬 𝒙 − 𝐜𝐨𝐬 𝒙 𝐬𝐢𝐧 𝒚 = 𝑪 2. [ 𝒙 (𝒙𝟐+𝒚𝟐) 𝟑 𝟐 ] 𝒅𝒙 + [ 𝒚 (𝒙𝟐+𝒚𝟐) 𝟑 𝟐 ] 𝒅𝒚 = 𝟎 𝑀 = [ 𝑥 (𝑥2+𝑦2) 3 2 ] 𝑑𝑥 𝑁 = [ 𝑦 (𝑥2+𝑦2) 3 2 ] 𝑑𝑦 Comprobando que sean ecuaciones exactas: 𝜕𝑀 𝜕𝑦 = 𝑥 𝜕 𝜕𝑦 ((𝑥2 + 𝑦2) 3 2) → − 3 2(𝑥2+𝑦2) 6 2 𝜕 𝜕𝑦 (𝑥2 + 𝑦2) → − 3 2(𝑥2+𝑦2) ∗ 2𝑦 → − 𝟑𝒙𝒚 (𝒙𝟐+𝒚𝟐) 𝜕𝑁 𝜕𝑥 = 𝑦 𝜕 𝜕𝑦 ( 1 (𝑥2+𝑦2) 3 2 ) → 𝑦 𝜕 𝜕𝑦 [(𝑥2 + 𝑦2) 3 2] → − 3 2(𝑥2+𝑦2) 6 2 𝜕 𝜕𝑦 (𝑥2 + 𝑦2) → − 3 2(𝑥2+𝑦2) ∗ 2𝑥 → − 𝟑𝒙𝒚 (𝒙𝟐+𝒚𝟐) ∫ [ 𝑥 (𝑥2+𝑦2) 3 2 ] 𝑑𝑥 + ∫ [ 𝑦 (𝑥2+𝑦2) 3 2 ] 𝑑𝑦 = ∫ 0 𝒖 = 𝒙𝟐 + 𝒚𝟐 𝑑𝑢 = 2𝑦 𝒅𝒖 𝟐𝒖 = 𝒚 𝑣 = 𝑥2 + 𝑦2 𝑑𝑣 = 2𝑥 𝒅𝒗 𝟐𝒗 = 𝒙 → ∫ 𝑑𝑢 2(𝑢) 3 2 + ∫ 𝑑𝑣 2(𝑣) 3 2 = ∫ 0 → 1 2 ∫ 𝑑𝑢 (𝑢) 3 2 + 1 2 ∫ 𝑑𝑣 (𝑣) 3 2 = ∫ 0 → 1 2 ∫(𝑢) 3 2 + 1 2 ∫(𝑣) 3 2 = ∫ 0 → 1 2 ( 1 𝑢 −1 2 − 1 2 ) + 1 2 ( 1 𝑣 −1 2 − 1 2 ) = 𝐶 → 1 2 ( −2 𝑢 1 2 ) + 1 2 ( −2 𝑣 1 2 ) = 𝐶 → 1 2 ( −2 𝑢 1 2 ) + 1 2 ( −2 𝑣 1 2 ) = 𝐶 → 1 𝑢 1 2 + 1 𝑣 1 2 = 𝐶 → 1 (𝑥2+𝑦2) 1 2 + 1 (𝑥2+𝑦2) 1 2 = 𝐶 → 𝟏 (𝒙𝟐+𝒚𝟐) 𝟏 𝟐 = 𝑪 3. (𝒆𝒕𝒚 + 𝒕𝒆𝒕𝒚)𝒅𝒕 + (𝒕𝒆𝒕 + 𝟐)𝒅𝒚 = 𝟎 𝑀 = (𝑒𝑡𝑦 + 𝑡𝑒𝑡𝑦)𝑑𝑡 𝑁 = (𝑡𝑒𝑡 + 2)𝑑𝑦 Comprobando que sean ecuaciones exactas: 𝜕𝑀 𝜕𝑦 = (𝑒𝑡 𝜕 𝜕𝑦 𝑦 + 𝑦 𝜕 𝜕𝑦 𝑒𝑡) + 𝑡𝑒𝑡 𝜕 𝜕𝑦 𝑦 → 𝒆𝒕 + 𝒆𝒕𝒕 𝜕𝑁 𝜕𝑡 = (𝑡 𝜕 𝜕𝑡 𝑒𝑡 + 𝑒𝑡 𝜕 𝜕𝑡 𝑡) + 𝜕 𝜕𝑦 2 → 𝒆𝒕 + 𝒆𝒕𝒕 ∫(𝑒𝑡𝑦 + 𝑡𝑒𝑡𝑦)𝑑𝑡 + ∫(𝑡𝑒𝑡 + 2)𝑑𝑦 = ∫ 0 → 𝑦 ∫ 𝑒𝑡𝑑𝑡 + 𝑦 ∫ 𝑡𝑒𝑡𝑑𝑡 + ∫ 𝑡𝑒𝑡𝑑𝑦 + ∫ 2𝑑𝑦 = ∫ 0 → 𝑦 ∫ 𝑒𝑡𝑑𝑡 + 𝑦 ∫ 𝑡𝑒𝑡𝑑𝑡 + ∫ 𝑡𝑒𝑡𝑑𝑦 + ∫ 2𝑑𝑦 = ∫ 0 → 𝑒𝑡𝑦 + 𝑡𝑒𝑡𝑦 − 𝑡𝑒𝑡𝑦 + 2𝑦 = 𝐶 → 𝒆𝒕𝒚 + 𝒕𝒆𝒕𝒚 + 𝟐𝒚 = 𝑪 II. Determina el factor integrante para las siguientes ecuaciones diferencial. 1. 𝒚(𝟒𝒙 + 𝒚)𝒅𝒙 − 𝟐(𝒙𝟐 − 𝒚)𝒅𝒚 = 𝟎 𝑀 = 𝑦(4𝑥 + 𝑦)𝑑𝑥 𝑁 = 𝑦(4𝑥 + 𝑦)𝑑𝑦 𝑀𝑦 = [ 𝜕 𝜕𝑦 𝑦(4𝑥 + 𝑦)] → [𝑦 𝜕 𝜕𝑦 (4𝑥 + 𝑦)] → 4𝑥 + 𝑦 + 𝑦 = 𝟒𝒙 + 𝟐𝒚 𝑁𝑥 = 2 𝜕 𝜕𝑥 (𝑥2 − 𝑦) → 2(2𝑥) = 𝟒𝒙 𝑝(𝑦) = (−4𝑥) − (4𝑥 + 2𝑦) 𝑦(4𝑥 + 𝑦) → (−4𝑥) − (4𝑥 + 2𝑦) 𝑦(4𝑥 + 𝑦) → −8𝑥 − 2𝑦 𝑦(4𝑥 + 𝑦) → −2(4𝑥 + 𝑦) 𝑦(4𝑥 + 𝑦) → − 𝟐 𝒚 𝐹(𝑦) = 𝑒 ∫ 2 𝑦 𝑑𝑦 → 𝑒−2 ln|𝑦| → 1 𝑦2 → ( 1 𝑦2 ) [𝑦(4𝑥 + 𝑦)𝑑𝑥 − 2(𝑥2 − 𝑦)𝑑𝑦] = 0 → 𝟒𝒙+𝒚 𝒚 𝒅𝒙 − 𝟐(𝒙𝟐−𝒚) 𝒚𝟐 𝒅𝒚 = 𝟎 𝑀𝑦 = [ 𝜕 𝜕𝑦 4𝑥+𝑦 𝑦 ] → − 4𝑥 𝑦2 𝑁𝑥 = 2 𝜕 𝜕𝑥 (𝑥2−𝑦) 𝑦2 𝑑𝑦 → − 4𝑥 𝑦2 ∫ 4𝑥+𝑦 𝑦 𝑑𝑥 − ∫ 2(𝑥2−𝑦) 𝑦2 𝑑𝑦 = ∫ 0 → ∫ 4𝑥 𝑦 𝑑𝑥 + ∫ 𝑦 𝑦 𝑑𝑥 − 2 ∫ (𝑥2−𝑦) 𝑦2 𝑑𝑦 = ∫ 0 → 𝑥 + 2𝑥2 𝑦 + 2𝑥2 𝑦 + 2 ln 𝑦 = 𝐶 → 𝒙 + 𝟐𝒙𝟐 𝒚 + 𝟐 𝐥𝐧 𝒚 = 𝑪 2. 𝒚(𝒙 + 𝒚 + 𝟏)𝒅𝒙 + (𝒙 + 𝟐𝒚)𝒅𝒚 = 𝟎 𝑀 = 𝑦(𝑥 + 𝑦 + 1)𝑑𝑥 𝑁 = (𝑥 + 2𝑦)𝑑𝑦 𝑀𝑦 = 𝑦 𝜕 𝜕𝑦 (𝑥 + 𝑦 + 1) + (𝑥 + 𝑦 + 1) 𝜕 𝜕𝑦 𝑦 → 𝑦 + 𝑥 + 𝑦 + 1 = 𝟐𝒚 + 𝒙 + 𝟏 𝑁𝑥 = 𝜕 𝜕𝑥 𝑥 = 𝟏 𝑝(𝑥) = (2𝑦 + 𝑥 + 1) − (1) (𝑥 + 2𝑦) → 2𝑦 + 𝑥 + 1−1 (𝑥 + 2𝑦) → 2𝑦 + 𝑥 (𝑥 + 2𝑦) = 1 𝐹(𝑥) = 𝑒∫ 1 𝑑𝑥 → 𝑒ln 𝑥 = 𝒆𝒙 → (𝑒𝑥)[𝑦(𝑥 + 𝑦 + 1)𝑑𝑥 + (𝑥 + 2𝑦)𝑑𝑦 = 0] → (𝑒𝑥𝑦𝑥 + 𝑒𝑥𝑦 + 𝑒𝑥𝑦)𝑑𝑥 + (𝑒𝑥𝑥 + 2𝑒𝑥𝑦)𝑑𝑦 = 0 𝑀𝑦 = 𝜕 𝜕𝑦 (𝑒𝑥𝑦𝑥 + 𝑒𝑥𝑦 + 𝑒𝑥𝑦)𝑑𝑥 = 𝒙𝒆𝒙 + 𝟐𝒆𝒙𝒚 + 𝒆𝒙 𝑁𝑥 = 𝜕 𝜕𝑥 (𝑒𝑥𝑥 + 2𝑒𝑥𝑦)𝑑𝑦 = 𝒙𝒆𝒙 + 𝟐𝒆𝒙𝒚 + 𝒆𝒙 → ∫(𝑒𝑥𝑦𝑥 + 𝑒𝑥𝑦 + 𝑒𝑥𝑦)𝑑𝑥 + ∫(𝑒𝑥𝑥 + 2𝑒𝑥𝑦)𝑑𝑦 = ∫ 0 → 𝑦𝑒𝑥𝑥 + 𝑦2𝑒𝑥 + 𝑦𝑒𝑥𝑥 + 𝑦2𝑒𝑥 = 𝐶 → 𝒚𝒆𝒙𝒙 + 𝒚𝟐𝒆𝒙 = 𝑪 3. 𝟐𝒙𝒚𝒙´ = 𝟑𝒙𝟐 − 𝒚𝟐 → 2𝑥𝑦 𝑑𝑥 = 3𝑥2 − 𝑦2𝑑𝑦 → 2𝑥𝑦 𝑑𝑥 − (3𝑥2 − 𝑦2)𝑑𝑦 = 0 𝑀 = 2𝑥𝑦 𝑑𝑥 𝑁 = −(3𝑥2 − 𝑦2)𝑑𝑦 𝑀𝑦 = 2 𝜕 𝜕𝑦 𝑥𝑦 = 𝟐𝒙 𝑁𝑥 = − 𝜕 𝜕𝑥 3𝑥2 − 𝜕 𝜕𝑥 𝑦2 = −𝟔𝒙 𝑝(𝑦) = −6𝑥 − 2𝑥 2𝑥𝑦 → −8𝑥 2𝑥𝑦 → −8 2𝑦 → −2∗4 2𝑦 → − 4 𝑦 𝐹(𝑦) = 𝑒 ∫ − 4 𝑦 𝑑𝑦 → 𝑒 −4 ∫ 1 𝑦 𝑑𝑦 → 𝑒−4 ln 𝑦 = 1 𝑦4 → ( 1 𝑦4 ) [2𝑥𝑦 𝑑𝑥 − (3𝑥2 − 𝑦2)𝑑𝑦 = 0 ] → 1 𝑦4 (2𝑥𝑦)𝑑𝑥 − 1 𝑦4 (3𝑥2 − 𝑦2)𝑑𝑦 = 0 → 2𝑥 𝑦3 𝑑𝑥 − 3𝑥2−𝑦2 𝑦4 𝑑𝑦 = 0 𝑀𝑦 = 2𝑥 𝑦3 𝑑𝑥 = − 𝟔𝒙 𝒚𝟒 𝑁𝑥 = − 𝜕 𝜕𝑥 3𝑥2−𝑦2 𝑦4 = − 𝟔𝒙 𝒚𝟒 → ∫ 2𝑥 𝑦3 𝑑𝑥 − ∫ 3𝑥2−𝑦2 𝑦4 𝑑𝑦 = ∫ 0 → ∫ 2𝑥 𝑦3 𝑑𝑥 − ∫ 1 𝑦2 𝑑𝑦 + ∫ 3𝑥2 𝑦4 𝑑𝑦 = ∫ 0 → 𝒙𝟐 𝒚𝟑 − 𝟏 𝒚𝟒 (−𝒚𝟐𝒙 + 𝒙𝟑) = 𝑪
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