Caso 1 Mate Ejemplo Bs (1).docx

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EJEMPLO: CASO 1 (πŸπ’šπŸ + πŸ‘π’™)𝒅𝒙 + πŸπ’™π’šπ’…π’š = 𝟎 (2𝑦 2 + 3π‘₯)𝑑π‘₯ 2π‘₯𝑦𝑑𝑦 + =0 𝑀(π‘₯, 𝑦) 𝑁(π‘₯, 𝑦) 𝑀(π‘₯, 𝑦) = 2𝑦 2 + 3π‘₯ π‘΄π’š = πŸ’π’š

𝑁(π‘₯, 𝑦) = 2π‘₯𝑦 𝑡𝒙 = πŸπ’š π‘΄π’š β‰  𝑡𝒙

𝐌𝐲 βˆ’ 𝐍𝐱 4𝑦 βˆ’ 2𝑦 = = 𝑡 2π‘₯𝑦 𝑒(π‘₯) = 𝑒 ∫ 𝑔(π‘₯)𝑑π‘₯ 𝑒(π‘₯) = 𝑒

1 ∫ ( )𝑑π‘₯ π‘₯

2𝑦 𝟏 = 2π‘₯𝑦 𝒙

= 𝑒 𝐿𝑛 (π‘₯) = 𝒙

𝒙(πŸπ’šπŸ + πŸ‘π’™)𝒅𝒙 + 𝒙(πŸπ’™π’š)π’…π’š = 𝟎 (2π‘₯𝑦 2 + 3π‘₯ 2 )𝑑π‘₯ (2π‘₯ 2 𝑦)𝑑𝑦 + =0 𝑀(π‘₯, 𝑦) 𝑁(π‘₯, 𝑦) 𝑀(π‘₯, 𝑦) = 2π‘₯𝑦 2 + 3π‘₯ 2 π‘΄π’š = πŸ’π’™π’š

𝑁(π‘₯, 𝑦) = 2π‘₯ 2 𝑦 𝑡𝒙 = πŸ’π’™π’š

π‘΄π’š = 𝑴𝒙 𝒅𝒇 𝒅𝒙

= 𝑴(𝒙, π’š) = πŸπ’™π’šπŸ + πŸ‘π’™πŸ

𝑓(π‘₯, 𝑦) = ∫ (2π‘₯𝑦 2 + 3π‘₯ 2 )𝑑π‘₯ + 𝑐(𝑦) 2π‘₯ 2 2 3π‘₯ 3 = 𝑦 + + 𝐢(𝑦) 2 3 = π‘₯ 2 𝑦 2 + π‘₯ 3 + 𝐢(𝑦) 𝒅𝒇 = πŸπ’™πŸ π’š + π‘ͺ(π’š) = 𝑡(𝒙, π’š) = πŸπ’™πŸ π’š π’…π’š 2π‘₯ 2 𝑦 + 𝐢´(𝑦) = 2π‘₯ 2 𝑦 𝐢´(𝑦) = 0 𝐢(𝑦) = 0 𝐹(π‘₯, 𝑦) = π‘₯ 2 𝑦 2 + π‘₯ 3 + 𝐢1 = 𝐢 𝐱 𝟐 𝐲 𝟐 + 𝐱 πŸ‘ = π‚πŸ

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