Ejercicio 5- Primer Parcial PDF

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Ejercicio 5Ecuaciones diferenciales exactasArely del Refugio Sepúlveda GarcíaIngeniería biomédicaDeterminar si las siguientes ecuaciones diferenciales son exactas; resolverlas si lo son.1.(���� − ���� + ��)���� +(�� − ���� − ����)���� = ����(��,��)= 2�� − 5�� + 2 ��(��,��)= 1 − 6�� − 5����������= −�...

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Ejercicio 5

Ecuaciones diferenciales exactas Arely del Refugio Sepúlveda García Ingeniería biomédica Determinar si las siguientes ecuaciones diferenciales son exactas; resolverlas si lo son. 1. (𝟐𝐱 − 𝟓𝐲 + 𝟐)𝐝𝐱 + (𝟏 − 𝟔𝐲 − 𝟓𝐱)𝐝𝐲 = 𝟎

𝑀(𝑥, 𝑦) = 2𝑥 − 5𝑦 + 2 𝑁(𝑥, 𝑦) = 1 − 6𝑦 − 5𝑥 𝜕𝑁 𝜕𝑀 = −5 = −5 ∴ 𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠 𝜕𝑦 𝜕𝑥

∫ 𝑀 (𝑥, 𝑦) = ∫(2𝑥 − 5𝑦 + 2)𝑑𝑥 →

𝑓 = 𝑥 2 − 5𝑥𝑦 + 2𝑥 + 𝑓(𝑦)

𝑓𝑦 = −5𝑥 + 𝑓 ′ (𝑦) −5𝑥 + 𝑓 ′ (𝑦) = 1 − 6𝑦 − 5𝑥 𝑓 ′ (𝑦) = 1 − 6𝑦

∫ 𝑓 ′ (𝑦) = ∫(1 − 6𝑦)𝑑𝑦 → 𝑓(𝑦) = 𝑦 − 3𝑦 2 + 𝑐 𝑥 2 − 5𝑥𝑦 + 2𝑥 + 𝑦 − 3𝑦 2 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙

2. (𝟐𝐱𝐲 𝟑 − 𝟒𝐲 + 𝟒𝐱 − 𝟑)𝐝𝐱 + (𝟑𝐱 𝟐𝐲 𝟐 − 𝟒𝐱)𝐝𝐲 = 𝟎

𝑁(𝑥, 𝑦 ) = 3𝑥 2 𝑦 2 − 4𝑥 𝑀(𝑥, 𝑦) = 2𝑥𝑦 3 − 4𝑦 + 4𝑥 − 3 𝜕𝑁 𝜕𝑀 = 6𝑥𝑦 2 − 4 = 6𝑥𝑦 2 − 4 ∴ 𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠 𝜕𝑦 𝜕𝑥

∫ 𝑀(𝑥, 𝑦 ) = ∫(2𝑥𝑦 3 − 4𝑦 + 4𝑥 − 3)𝑑𝑥 →

𝑓 = 𝑥 2 𝑦 3 − 4𝑥𝑦 + 2𝑥2 − 3𝑥 + 𝑓(𝑦)

𝑓𝑦 = 3𝑥 2 𝑦 2 − 4𝑥 + 𝑓 ′ (𝑦) 3𝑥 2 𝑦 2 − 4𝑥 + 𝑓 ′ (𝑦) = 3𝑥 2 𝑦 2 − 4𝑥 𝑓 ′ (𝑦) = 0 ∫ 𝑓 ′ (𝑦) = ∫ 0 → 𝑓(𝑦) = 𝑐

𝑥 2 𝑦 3 − 4𝑥𝑦 + 2𝑥 2 − 3𝑥 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙

3. (𝟏𝟔𝐱𝐲 − 𝟑𝐱 𝟐)𝐝𝐱 + (𝟖𝐱 𝟐 + 𝟐𝐲)𝐝𝐲 = 𝟎

𝑀(𝑥, 𝑦 ) = (16xy − 3x 2 ) 𝜕𝑀 𝜕𝑁 = 16𝑥 = 16𝑥 𝜕𝑥 𝜕𝑦

∫ 𝑀(𝑥, 𝑦) = ∫(16xy − 3x2 )𝑑𝑥 →

𝑁(𝑥, 𝑦) = 8x 2 + 2y ∴

𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠

𝑓 = 8𝑥 2 𝑦 − 𝑥3 + 𝑓(𝑦)

𝑓𝑦 = 8𝑥 2 + 𝑓 ′ (𝑦) 8𝑥 + 𝑓 ′ (𝑦) = 8x2 + 2y 𝑓 ′ (𝑦) = 2𝑦 2

∫ 𝑓 ′ (𝑦) = ∫ 2𝑦 → 𝑓(𝑦) = 𝑦 2 + 𝑐

8𝑥2 𝑦 − 𝑥 3 + 𝑦 2 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙

4. (−𝟐𝟎𝐱𝐲 𝟐 + 𝟔𝐱)𝐝𝐱 + (𝟑𝐲 𝟐 − 𝟐𝟎𝐱 𝟐𝐲)𝐝𝐲 = 𝟎

𝑁(𝑥, 𝑦 ) = 3y2 − 20x 2 y 𝑀(𝑥, 𝑦) = −20xy2 + 6x 𝜕𝑁 𝜕𝑀 = −40𝑥𝑦 = −40𝑥𝑦 ∴ 𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠 𝜕𝑦 𝜕𝑥

∫ 𝑀(𝑥, 𝑦 ) = ∫ (−20xy2 + 6x ) 𝑑𝑥 →

𝑓 = −10𝑥 2 𝑦 2 + 3𝑥 2 + 𝑓(𝑦)

𝑓𝑦 = −20𝑥 2 𝑦 + 𝑓 ′ (𝑦) −20𝑥 2 𝑦 + 𝑓 ′ (𝑦) = 3y2 − 20x 2 y 𝑓 ′ (𝑦) = 3𝑦 2

∫ 𝑓 ′ (𝑦) = ∫ 3𝑦 2 → 𝑓(𝑦) = 𝑦 3 + 𝑐

−10𝑥 2 𝑦 2 + 3𝑥2 + 𝑦 3 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙

5. (𝐞𝐱 + 𝐲)𝐝𝐱 + (𝐞𝐲 + 𝐱)𝐝𝐲 = 𝟎

𝑁(𝑥, 𝑦 ) = ey + x 𝑀(𝑥, 𝑦) = ex + y 𝜕𝑀 𝜕𝑁 = 𝑒𝑦 + 1 ∴ 𝑛𝑜 𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠 = 𝑒𝑥 + 1 𝜕𝑥 𝜕𝑦

Determinar si las siguientes ecuaciones diferenciales son exactas; resolverlas si lo son. 1. (𝟒 + 𝟓𝐲)𝐝𝐱 + (𝟏 + 𝟓𝐱)𝐝𝐲 = 𝟎

𝐩𝐚𝐫𝐚 𝐲(−𝟏) = −𝟏

𝑀(𝑥, 𝑦) = 4 + 5𝑦 𝜕𝑁 𝜕𝑀 =5 = 5 𝜕𝑦 𝜕𝑥

𝑁(𝑥, 𝑦) = 1 + 5x

∫ 𝑀(𝑥, 𝑦 ) = ∫(4 + 5y )𝑑𝑥 →

∴

𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠

𝑓 = 4𝑥 + 5𝑥𝑦 + 𝑓(𝑦)

𝑓𝑦 = 5𝑥 + 𝑓 ′ (𝑦) 5𝑥 + 𝑓′ (𝑦) = 1 + 5x 𝑓 ′ (𝑦) = 1

∫ 𝑓 ′ (𝑦) = ∫ 1 → 𝑓(𝑦) = 𝑦 + 𝑐

4𝑥 + 5𝑥𝑦 + 𝑦 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 4(−1) + 5(−1)(−1) + (−1) = 0 4𝑥 + 5𝑥𝑦 + 𝑦 = 0 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟

2. 𝐞𝐱 𝐬𝐞𝐧 𝐲𝐝𝐱 + (𝐞𝐱 𝐜𝐨𝐬 𝐲 + 𝐞 𝐲)𝐝𝐲 = 𝟎 𝑀(𝑥, 𝑦) = 𝑒 𝑥 𝑠𝑒𝑛𝑦

𝜕𝑀 = 𝑒 𝑥 𝑠𝑒𝑛𝑦 + 𝑒 𝑥 𝑐𝑜𝑠𝑦 𝜕𝑦

3. (√𝐲 + 𝟏)𝐝𝐱 + (

𝐱

𝟐√𝐲

𝐩𝐚𝐫𝐚 𝐲(𝟎) = 𝟎

𝑁(𝑥, 𝑦) = ex 𝑐𝑜𝑠𝑦 + 𝑒 𝑦

𝜕𝑁 = 𝑒 𝑥 𝑐𝑜𝑠𝑦 + 𝑒 𝑥 − 𝑠𝑒𝑛𝑦 𝜕𝑥

𝑛𝑜 𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠

𝐩𝐚𝐫𝐚 𝐲(𝟏) = 𝟒

+ 𝟏) 𝐝𝐲 = 𝟎 𝑀(𝑥, 𝑦) = √y + 1

𝜕𝑀 1 = 𝜕𝑦 2√𝑦

∴

𝑁(𝑥, 𝑦) =

𝜕𝑁 1 = 𝜕𝑥 2√𝑦

∴

∫ 𝑀(𝑥, 𝑦) = ∫(√y + 1 )𝑑𝑥 →

x

2√y

+1

𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠 𝑓 = 𝑥 + 𝑓(𝑦)

𝑓𝑦 = 1 + 𝑓 ′ (𝑦) x 1 + 𝑓′ (𝑦) = +1 2√y x 𝑓 ′ (𝑦) = 2√y x ∫ 𝑓 ′ (𝑦) = ∫ → 𝑓(𝑦) = 𝑥 √𝑦 + 𝑐 2 √y 𝑥 + 𝑥√𝑦 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 (1) + (1)√(4) = 3 𝑥 + 𝑥√𝑦 = 3 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟

4. 𝐲𝐜𝐨𝐬 𝐱𝐲𝐝𝐱 + (𝐱𝐜𝐨𝐬 𝐱𝐲 + 𝐬𝐞𝐧 𝐲)𝐝𝐲 = 𝟎

𝐩𝐚𝐫𝐚 𝐲(𝟏) = 𝟎

𝑀(𝑥, 𝑦) = ycos xy 𝑁(𝑥, 𝑦 ) = xcos xy + sen y 𝜕𝑀 𝜕𝑁 = −𝑠𝑒𝑛𝑥𝑦 ∴ 𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠 = −𝑠𝑒𝑛𝑥𝑦 𝜕𝑥 𝜕𝑦 ∫ 𝑀(𝑥, 𝑦 ) = ∫(ycos xy )𝑑𝑥 → 𝑓 = 𝑥𝑠𝑒𝑛𝑥𝑦 + 𝑓(𝑦) 𝑓𝑦 = 𝑐𝑜𝑠𝑥𝑦 + 𝑓 ′ (𝑦) 𝑥𝑐𝑜𝑠𝑥𝑦 + 𝑓 ′ (𝑦) = xcos xy + sen y 𝑓 ′ (𝑦) = sen y

∫ 𝑓 ′ (𝑦) = ∫ 𝑠𝑒𝑛𝑦 → 𝑓(𝑦) = −𝑐𝑜𝑠𝑦 + 𝑐

𝟏

𝟏

𝑥𝑠𝑒𝑛𝑥𝑦 − 𝑐𝑜𝑠𝑦 = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 (1)𝑠𝑒𝑛(1)(0) − cos(0) = −1 𝑥𝑠𝑒𝑛𝑥𝑦 − 𝑐𝑜𝑠𝑦 = −1 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟

5. ( 𝐱 + 𝐲𝐞𝐱𝐲) 𝐝𝐱 + (𝐲 + 𝐱𝐞𝐱𝐲 ) 𝐝𝐲 = 𝟎

𝟏

𝐩𝐚𝐫𝐚 𝐲 ( ) = 𝟐 𝟐

1 1 𝑀(𝑥, 𝑦) = + yexy 𝑁(𝑥, 𝑦) = + xexy x y 𝜕𝑀 𝜕𝑁 = 𝑒 𝑥𝑦 + 𝑥𝑦𝑒 𝑥𝑦 ∴ 𝑠𝑜𝑛 𝑒𝑥𝑎𝑐𝑡𝑎𝑠 = 𝑒 𝑥𝑦 + 𝑥𝑦𝑒 𝑥𝑦 𝜕𝑥 𝜕𝑦 1 ∫ 𝑀(𝑥, 𝑦 ) = ∫( + yexy)𝑑𝑥 → 𝑓 = ln|𝑥| + 𝑒 𝑥𝑦 + 𝑓(𝑦) x 𝑓𝑦 = 𝑥𝑒 𝑥𝑦 + 𝑓 ′ (𝑦) 1 𝑥𝑒 𝑥𝑦 + 𝑓 ′ (𝑦) = + xexy y 1 𝑓 ′ (𝑦) = 𝑦 1 ∫ 𝑓 ′ (𝑦) = ∫ 𝑑𝑦 → 𝑓(𝑦) = ln|𝑦| + 𝑐 𝑦 ln|𝑥| + 𝑒 𝑥𝑦 + ln |𝑦| = 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙

1 1 1 ( )(2) + ln|2| = 𝑐 → ln |( ) (2)| + 𝑒1 = 𝑐 → ln|1| + 𝑒 = 𝑐 → 𝑒 = 𝑐 ln | | + 𝑒 2 2 2

ln|𝑥𝑦| + 𝑒 𝑥𝑦 = 𝑒...