Caso 1 Mate Ejemplo Bs (1).docx
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EJEMPLO: CASO 1 (πππ + ππ)π
π + ππππ
π = π (2π¦ 2 + 3π₯)ππ₯ 2π₯π¦ππ¦ + =0 π(π₯, π¦) π(π₯, π¦) π(π₯, π¦) = 2π¦ 2 + 3π₯ π΄π = ππ
π(π₯, π¦) = 2π₯π¦ π΅π = ππ π΄π β π΅π
ππ² β ππ± 4π¦ β 2π¦ = = π΅ 2π₯π¦ π’(π₯) = π β« π(π₯)ππ₯ π’(π₯) = π
1 β« ( )ππ₯ π₯
2π¦ π = 2π₯π¦ π
= π πΏπ (π₯) = π
π(πππ + ππ)π π + π(πππ)π π = π (2π₯π¦ 2 + 3π₯ 2 )ππ₯ (2π₯ 2 π¦)ππ¦ + =0 π(π₯, π¦) π(π₯, π¦) π(π₯, π¦) = 2π₯π¦ 2 + 3π₯ 2 π΄π = πππ
π(π₯, π¦) = 2π₯ 2 π¦ π΅π = πππ
π΄π = π΄π π π π π
= π΄(π, π) = ππππ + πππ
π(π₯, π¦) = β« (2π₯π¦ 2 + 3π₯ 2 )ππ₯ + π(π¦) 2π₯ 2 2 3π₯ 3 = π¦ + + πΆ(π¦) 2 3 = π₯ 2 π¦ 2 + π₯ 3 + πΆ(π¦) π π = πππ π + πͺ(π) = π΅(π, π) = πππ π π π 2π₯ 2 π¦ + πΆΒ΄(π¦) = 2π₯ 2 π¦ πΆΒ΄(π¦) = 0 πΆ(π¦) = 0 πΉ(π₯, π¦) = π₯ 2 π¦ 2 + π₯ 3 + πΆ1 = πΆ π± π π² π + π± π = ππ
π(π₯, π¦) = 2π₯π¦ π΅π = ππ π΄π β π΅π
ππ² β ππ± 4π¦ β 2π¦ = = π΅ 2π₯π¦ π’(π₯) = π β« π(π₯)ππ₯ π’(π₯) = π
1 β« ( )ππ₯ π₯
2π¦ π = 2π₯π¦ π
= π πΏπ (π₯) = π
π(πππ + ππ)π π + π(πππ)π π = π (2π₯π¦ 2 + 3π₯ 2 )ππ₯ (2π₯ 2 π¦)ππ¦ + =0 π(π₯, π¦) π(π₯, π¦) π(π₯, π¦) = 2π₯π¦ 2 + 3π₯ 2 π΄π = πππ
π(π₯, π¦) = 2π₯ 2 π¦ π΅π = πππ
π΄π = π΄π π π π π
= π΄(π, π) = ππππ + πππ
π(π₯, π¦) = β« (2π₯π¦ 2 + 3π₯ 2 )ππ₯ + π(π¦) 2π₯ 2 2 3π₯ 3 = π¦ + + πΆ(π¦) 2 3 = π₯ 2 π¦ 2 + π₯ 3 + πΆ(π¦) π π = πππ π + πͺ(π) = π΅(π, π) = πππ π π π 2π₯ 2 π¦ + πΆΒ΄(π¦) = 2π₯ 2 π¦ πΆΒ΄(π¦) = 0 πΆ(π¦) = 0 πΉ(π₯, π¦) = π₯ 2 π¦ 2 + π₯ 3 + πΆ1 = πΆ π± π π² π + π± π = ππ
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