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ESCUELA DE INGENIERÍA CIVIL
MATEMÁTICA IV
Dr. FRANCISCO BAUTISTA LOYOLA
fbautista@unjfsc.edu.pe
SEMESTRE ACADEMICO 2022- I
𝑪𝑶𝑵𝑪𝑬𝑷𝑻𝑶𝑺 𝑩𝑨𝑺𝑰𝑪𝑶𝑺
𝐸𝑠 𝑎𝑞𝑢𝑒𝑙𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑞𝑢𝑒 𝑒𝑛 𝑒𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑎 𝑐𝑜𝑛𝑡𝑖𝑒𝑛𝑒 𝑙𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑜 𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑢𝑛𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛
𝑑𝑒𝑠𝑐𝑜𝑛𝑜𝑐𝑖𝑑𝑎 𝑙𝑙𝑎𝑚𝑎𝑑𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛
𝒂) 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑶𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂𝒔:
𝐸𝑗𝑒𝑚𝑝𝑙𝑜𝑠 𝑑𝑒 𝐸𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙𝑒𝑠:
1)
dy
dx
= 4X + 7 , 2) y2
dx − x2
dy = 0, 3) m
d2y
dt2 = mg − K
dy
dx
, 4)
d2y
dx2
3
− cos x
dy
dx
= sen x ,
5)
𝛿2
𝜔
𝛿𝑥2 +
𝛿2
𝜔
𝛿𝑦2 +
𝛿2
𝜔
𝛿𝑧2 = 0, 𝑑𝑜𝑛𝑑𝑒 𝜔 = 𝑓 𝑥, 𝑦, 𝑧 , 6)𝑥2
𝛿2
𝜔
𝛿𝑥2 + 𝑦2
𝛿2
𝜔
𝑑𝑦2 + 𝑧2
𝛿2
𝜔
𝑑𝑧2 = 0, 𝑑𝑜𝑛𝑑𝑒 𝜔 = 𝑓(𝑥, 𝑦, 𝑧)
𝑳𝒂𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝒔𝒆 𝒄𝒍𝒂𝒔𝒊𝒇𝒊𝒄𝒂𝒏 𝒆𝒏 𝒅𝒐𝒔 𝒕𝒊𝒑𝒐𝒔:
1) Ecuación Diferencial.
2) Clases de ecuaciones diferenciales
Es aquella ecuación cuya función incógnita depende de una sola variable independiente donde sólo
aparecen derivadas ordinarias. Son ecuaciones ordinarias 1, 2, 3 y 4 de los ejemplos anteriores.
𝒃) 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑷𝒂𝒓𝒄𝒊𝒂𝒍𝒆𝒔:
𝑬𝒍 𝒐𝒓𝒅𝒆𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒆𝒔𝒕á 𝒅𝒂𝒅𝒐 𝒑𝒐𝒓 𝒍𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒎𝒂𝒚𝒐𝒓 𝒒𝒖𝒆 𝒂𝒑𝒂𝒓𝒆𝒄𝒆 𝒆𝒏 𝒍𝒂 𝒆𝒄. 𝒅𝒊𝒇
𝑬𝒍 𝒈𝒓𝒂𝒅𝒐 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒆𝒔𝒕á 𝒅𝒂𝒅𝒐 𝒑𝒐𝒓 𝒆𝒍 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕𝒆 𝒅𝒆 𝒍𝒂 𝒎𝒂𝒚𝒐𝒓 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒒𝒖𝒆
𝒂𝒑𝒂𝒓𝒆𝒄𝒆 𝒆𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍.
𝑬𝒋𝒆𝒎𝒑𝒍𝒐𝒔:
𝑪𝒖𝒂𝒏𝒅𝒐 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏 𝒊𝒏𝒄ó𝒈𝒏𝒊𝒕𝒂 𝒅𝒆𝒑𝒆𝒏𝒅𝒆 𝒅𝒆 𝒗𝒂𝒓𝒊𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒊𝒆𝒏𝒕𝒆𝒔 𝒚 𝒍𝒂𝒔 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂𝒔 𝒔𝒐𝒏
𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂𝒔 𝒑𝒂𝒓𝒄𝒊𝒂𝒍𝒆𝒔. 𝑳𝒂𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝟓 𝒚 𝟔 𝒂𝒏𝒕𝒆𝒓𝒊𝒐𝒓𝒆𝒔, 𝒔𝒐𝒏 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑷𝒂𝒓𝒄𝒊𝒂𝒍𝒆𝒔
𝟑) 𝑶𝒓𝒅𝒆𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
𝟒) 𝑮𝒓𝒂𝒅𝒐 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
1) 𝑒𝑥
𝑑2𝑦
𝑑𝑥2 + 𝑠𝑒𝑛 𝑥
𝑑𝑦
𝑑𝑥
= 𝑥, 2𝑑𝑜 𝑜𝑟𝑑𝑒𝑛 𝑦 1𝑒𝑟 𝑔𝑟𝑎𝑑𝑜. 2)
𝑑3
𝑦
𝑑𝑥3
+ 2
𝑑2
𝑦
𝑑𝑥2
3
+
𝑑𝑦
𝑑𝑥
= 𝑡𝑔 𝑥, 3𝑒𝑟 𝑜𝑟𝑑𝑒𝑛 𝑦 1𝑒𝑟 𝑔𝑟𝑎𝑑𝑜
3)
𝑑𝑦
𝑑𝑥
+ 𝑝 𝑥 𝑦 = 𝑄 𝑥 , 1𝑒𝑟 𝑜𝑟𝑑𝑒𝑛 𝑦 1𝑒𝑟 𝑔𝑟𝑎𝑑𝑜.
4)
𝑑3𝑦
𝑑𝑥3
2
− 2
𝑑𝑥
𝑑𝑥
4
+ 𝑥𝑦 = 0, 𝑂𝑟𝑑𝑒𝑛 3 𝑦 2𝑑𝑜 𝑔𝑟𝑎𝑑𝑜.
𝑺𝒖𝒑𝒐𝒏𝒈𝒂𝒎𝒐𝒔, 𝒒𝒖𝒆 𝑮 𝒙 = 𝑭 𝒙 + 𝑪, 𝒅𝒐𝒏𝒅𝒆 𝑪 𝒆𝒔 𝒖𝒏𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆 𝒂𝒓𝒃𝒊𝒕𝒓𝒂𝒓𝒊𝒂 𝒚 𝒒𝒖𝒆 𝒍𝒂
𝒅 𝑮 𝒙 = 𝒅 𝑭 𝒙 + 𝒄 = 𝑭′
(𝒙)𝒅𝒙 = 𝒇 𝒙 𝒅𝒙
𝑺𝒊 𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒐𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂 𝒅𝒆 𝒍𝒂 𝒇𝒐𝒓𝒎𝒂
𝒅𝒚
𝒅𝒙
= 𝒇 𝒙
𝑳𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒅𝒆 𝒆𝒔𝒕𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒆𝒔 𝒖𝒏𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏 𝒅𝒆 𝒍𝒂 𝒇𝒐𝒓𝒎𝒂 𝒚 = 𝑮 𝒙
𝒅𝒆 𝒕𝒂𝒍 𝒎𝒂𝒏𝒆𝒓𝒂 𝒒𝒖𝒆 𝒗𝒆𝒓𝒊𝒇𝒊𝒒𝒖𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒂𝒅𝒂.
𝑬𝒏𝒕𝒐𝒏𝒄𝒆𝒔: 𝒚 = 𝑮 𝒙 = 𝑭 𝒙 + 𝒄, 𝒔𝒆 𝒍𝒍𝒂𝒎𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
𝑳𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍, 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒂 𝒖𝒏𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒖𝒓𝒗𝒂𝒔 𝒒𝒖𝒆 𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏 𝒅𝒆 𝒍𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆
𝒂𝒓𝒃𝒊𝒕𝒓𝒂𝒓𝒊𝒂.
𝟓) 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
𝑬𝒋𝒆𝒎𝒑𝒍𝒐𝒔:
𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏:
𝑺𝒊 𝒚 = 𝒆𝒙𝟐
න
𝟎
𝒙
𝒆−𝒕𝟐
𝒅𝒕 + 𝒆𝒙𝟐
𝑳𝒖𝒆𝒈𝒐, 𝒓𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒏𝒅𝒐, 𝒕𝒆𝒏𝒆𝒎𝒐𝒔:
𝒚′ − 𝟐𝒙𝒚 = 𝟐𝒙𝒆𝒙𝟐
න
𝟎
𝒙
𝒆−𝒕𝟐
𝒅𝒕 + 𝟏 + 𝟐𝒙𝒆𝒙𝟐
− 𝟐𝒙 𝒆𝒙𝟐
න
𝟎
𝒙
𝒆−𝒕𝟐
𝒅𝒕 + 𝒆𝒙𝟐
𝑬𝒏𝒕𝒐𝒏𝒄𝒆𝒔: 𝒚′ − 𝟐𝒙𝒚 = 𝟐𝒙𝒆𝒙𝟐
න
𝟎
𝒙
𝒆−𝒕𝟐
𝒅𝒕 + 𝟏 + 𝟐𝒙𝒆𝒙𝟐
− 𝟐𝒙𝒆𝒙𝟐
න
𝟎
𝒙
𝒆−𝒕𝟐
𝒅𝒕 − 𝟐𝒙𝒆𝒙𝟐
= 𝟏
𝒚′ = 𝟐𝒙𝒆𝒙𝟐
න
𝟎
𝒙
𝒆−𝒕𝟐
+ 𝟏 + 𝟐𝒙𝒆𝒙𝟐
𝟏. 𝑽𝒆𝒓𝒊𝒇𝒊𝒄𝒂𝒓 𝒒𝒖𝒆 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏 𝒚 = 𝒆𝒙𝟐
න
𝟎
𝒙
𝒆−𝒕𝟐
𝒅𝒕 + 𝒆𝒙𝟐
, 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒚′
− 𝟐𝒙𝒚 =
𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐: 𝒚′ − 𝟐𝒙𝒚 = 𝟏
𝑹𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒏𝒅𝒐 𝒆𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒂𝒅𝒂 𝒕𝒆𝒏𝒆𝒎𝒐𝒔
𝟐. 𝑽𝒆𝒓𝒊𝒇𝒊𝒄𝒂𝒓 𝒔𝒊 𝒇 𝒙 =
𝟐
𝝅
න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒅𝜽 , 𝒔𝒂𝒕𝒊𝒔𝒇𝒂𝒄𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
𝒇′′ 𝒙 +
𝒇′(𝒙)
𝒙
+ 𝒇 𝒙 = 𝟎
𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏:
𝑺𝒊 𝒇 𝒙 =
𝟐
𝝅
න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒅𝜽 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑡𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒:
𝒇′
𝒙 = −
𝟐
𝝅
න
𝟎
𝝅
𝟐
𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒅𝜽 𝒅𝒆 𝒅𝒐𝒏𝒅𝒆 𝒇′′ 𝒙 = −
𝟐
𝝅
න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏𝟐
𝜽 𝒅𝜽
𝑓′′
𝑥 +
𝑓′
𝑥
𝑥
+ 𝑓 𝑥 = −
2
𝜋
න
0
𝜋
2
𝑐𝑜𝑠 𝑥𝑠𝑒𝑛 𝜃 𝑠𝑒𝑛2
𝜃 𝑑𝜃 −
2
𝜋
න
0
𝜋
2 𝑠𝑒𝑛 𝑥 𝑠𝑒𝑛 𝜃 𝑠𝑒𝑛 𝜃
𝑥
𝑑𝜃
+
2
𝜋
න
0
𝜋
2
𝑐𝑜𝑠(𝑥 𝑠𝑒𝑛 𝜃) 𝑑𝜃
𝒇′′
𝒙 +
𝒇′
𝒙
𝒙
+ 𝒇 𝒙 =
𝟐
𝝅
න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝟏 − 𝒔𝒆𝒏𝟐
𝜽 𝒅𝜽 −
𝟐
𝝅
න
𝟎
𝝅
𝟐 𝒔𝒆𝒏 𝒙𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒅𝜽
𝒙
=
𝟐
𝝅
න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐𝜽 𝒅𝜽 −
𝟐
𝝅
න
𝟎
𝝅
𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒅𝜽
𝒙
𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒏𝒅𝒐 𝒑𝒐𝒓 𝒑𝒂𝒓𝒕𝒆𝒔 𝒍𝒂 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒍 න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐𝜽 𝒅𝜽
𝒉𝒂𝒄𝒆𝒎𝒐𝒔: 𝒖 = 𝒄𝒐𝒔 𝜽 → d𝒖 = − 𝒔𝒆𝒏 𝜽𝒅𝜽
𝒅𝒗 = 𝒄𝒐𝒔 𝒙𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔 𝜽 𝒅𝜽 → 𝒗 =
𝒔𝒆𝒏 (𝒙 𝒔𝒆𝒏𝜽)
𝒙
𝑟𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜: න
0
𝜋
2
𝑐𝑜𝑠 𝑥 𝑠𝑒𝑛 𝜃 𝑐𝑜𝑠2
𝜃 𝑑𝜃 =
𝑐𝑜𝑠𝜃. 𝑠𝑒𝑛(𝑥 𝑠𝑒𝑛𝜃)
𝑥
/0
𝜋
2
+ න
0
𝜋
2 𝑠𝑒𝑛 𝑥 𝑠𝑒𝑛 𝜃 𝑠𝑒𝑛 𝜃
𝑥
𝑑𝜃
𝒑𝒓𝒊𝒎𝒆𝒓𝒐 𝒂 𝒍𝒂 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒍𝒐 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒎𝒐𝒔 𝒄𝒐𝒎𝒐 න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝜽. 𝒄𝒐𝒔𝜽 𝒅𝜽
𝑹𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒏𝒅𝒐 𝒆𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝒒𝒖𝒆:
𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐
𝜽 𝒅𝜽 = 𝟎 − 𝟎 + න
𝟎
𝝅
𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽
𝒙
𝒅𝜽
න
𝟎
𝝅
𝟐
𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐
𝜽 𝒅𝜽 = න
𝟎
𝝅
𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽
𝒙
𝒅𝜽
𝒇′′ 𝒙 +
𝒇′(𝒙)
𝒙
+ 𝒇 𝒙 =
𝟐
𝝅
න
𝟎
𝝅
𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽
𝒙
𝒅𝜽 −
𝟐
𝝅
න
𝟎
𝝅
𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽𝒅𝜽
𝒙
= 𝟎
𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐 𝒇′′ 𝒙 +
𝒇′(𝒙)
𝒙
+ 𝒇 𝒙 = 𝟎
𝑬𝒋𝒆𝒎𝒑𝒍𝒐𝒔 𝟏. 𝑬𝒏𝒄𝒐𝒏𝒕𝒓𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒄𝒖𝒚𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒆𝒔: 𝒚 = 𝑨𝒔𝒆𝒏𝒙 + 𝑩𝒄𝒐𝒔𝒙
𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏:
𝑺𝒊 𝒔𝒆 𝒕𝒊𝒆𝒏𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒖𝒓𝒗𝒂𝒔, 𝒔𝒆 𝒑𝒖𝒆𝒅𝒆 𝒐𝒃𝒕𝒆𝒏𝒆𝒓 𝒔𝒖 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
𝒎𝒆𝒅𝒊𝒂𝒏𝒕𝒆 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒄𝒊ó𝒏 𝒅𝒆 𝒍𝒂𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆𝒔 𝒐 𝒑𝒂𝒓á𝒎𝒆𝒕𝒓𝒐𝒔 , 𝒆𝒔𝒕𝒐 𝒔𝒆 𝒐𝒃𝒕𝒊𝒆𝒏𝒆 𝒂𝒊𝒔𝒍𝒂𝒏𝒅𝒐 𝒍𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆
𝒆𝒏 𝒖𝒏 𝒎𝒊𝒆𝒎𝒃𝒓𝒐 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒚 𝒅𝒆𝒓𝒊𝒗𝒂𝒏𝒅𝒐. 𝑻𝒂𝒎𝒃𝒊é𝒏 𝒔𝒆 𝒑𝒖𝒆𝒅𝒆 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒓 𝒍𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆 𝒅𝒆𝒓𝒊𝒗𝒂𝒏𝒅𝒐
𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒂𝒅𝒂 𝒕𝒂𝒏𝒕𝒂𝒔 𝒗𝒆𝒄𝒆𝒔 𝒄𝒐𝒎𝒐 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆𝒔 𝒂𝒓𝒃𝒊𝒕𝒓𝒂𝒓𝒊𝒂𝒔 𝒕𝒆𝒏𝒈𝒂, 𝒚 𝒍𝒖𝒆𝒈𝒐 𝒔𝒆 𝒓𝒆𝒔𝒖𝒆𝒍𝒗𝒆 𝒆𝒍 𝒔𝒊𝒔𝒕𝒆𝒎𝒂
𝒅𝒆 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒇𝒐𝒓𝒎𝒂𝒅𝒐 𝒄𝒐𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍.
𝑺𝒖𝒎𝒂𝒏𝒅𝒐: 𝒚′′ + 𝒚 𝒐𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
𝟔) 𝑬𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒂 𝒑𝒂𝒓𝒕𝒊𝒓 𝒅𝒆 𝒖𝒏𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒖𝒓𝒗𝒂𝒔
𝑺𝒊 𝒚 = 𝑨 𝒔𝒆𝒏 𝒙 + 𝑩 𝒄𝒐𝒔 𝑿 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔: ቊ
𝒚′
= 𝑨𝒄𝒐𝒔𝒙 − 𝑩𝒔𝒆𝒏𝒙
y′′ = −A sen x − B cos X
ቊ
𝒚′′
= −𝑨𝒄𝒐𝒔𝒙 − 𝑩𝒔𝒆𝒏𝒙
y = A sen x + B cos X 𝒚′′
+ 𝒚 = 𝟎
 𝑶𝒕𝒓𝒂 𝒎𝒂𝒏𝒆𝒓𝒂 𝒅𝒆 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒓 𝒍𝒂𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆𝒔 𝒆𝒔, 𝒄𝒐𝒏𝒔𝒊𝒅𝒆𝒓𝒂𝒏𝒅𝒐 𝒆𝒍 𝒔𝒊𝒔𝒕𝒆𝒎𝒂 𝒔𝒊𝒈𝒖𝒊𝒆𝒏𝒕𝒆:
y = Asenx + Bcosx
𝑦′ = Acos x − Bsenx
y′′
= −Asenx − Bcosx
−y + Asenx + Bcosx = 0
−y′
+ Acos x − Bsenx = 0
−y′′ − Asen x − Bcos x = 0
𝑬𝒔𝒕𝒆 𝒔𝒊𝒔𝒕𝒆𝒎𝒂 𝒅𝒆 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒆𝒏 𝒅𝒐𝒔 𝒊𝒏𝒄ó𝒈𝒏𝒊𝒕𝒂𝒔 𝑨 𝒚 𝑩, 𝒕𝒊𝒆𝒏𝒆 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒔𝒊 𝒚 𝒔ó𝒍𝒐 𝒔𝒊:
−𝒚 𝒔𝒆𝒏 𝒙 𝒄𝒐𝒔 𝒙
−𝒚′ 𝒄𝒐𝒔 𝒙 − 𝒔𝒆𝒏 𝒙
−𝒚′′
−𝒔𝒆𝒏 𝒙 − 𝒄𝒐𝒔 𝒙
= 𝟎, 𝒓𝒆𝒔𝒐𝒍𝒗𝒊𝒆𝒏𝒅𝒐 𝒆𝒔𝒕𝒆 𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒂𝒏𝒕𝒆, 𝒔𝒆 𝒕𝒊𝒆𝒏𝒆
𝑦′′ + 𝑦 = 0
𝐿𝑎 𝑓𝑜𝑟𝑚𝑎 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑑𝑒 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑟 𝑢𝑛𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑜𝑟𝑑𝑖𝑛𝑎𝑟𝑖𝑎 𝑑𝑒 𝑝𝑟𝑖𝑚𝑒𝑟 𝑜𝑟𝑑𝑒𝑛 𝑦 𝑝𝑟𝑖𝑚𝑒𝑟
𝑔𝑟𝑎𝑑𝑜 𝑒𝑠:
𝐿𝑎𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙𝑒𝑠 𝑑𝑒 𝑙𝑎 𝑓𝑜𝑟𝑚𝑎:
𝑆𝑒 𝑙𝑙𝑎𝑚𝑎𝑛 𝐸𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝐷𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙𝑒𝑠 𝑑𝑒 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑆𝑒𝑝𝑎𝑟𝑎𝑏𝑙𝑒𝑠 𝑠𝑖 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒𝑛 𝑒𝑠𝑐𝑟𝑖𝑏𝑖𝑟 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒
𝑚𝑎𝑛𝑒𝑟𝑎 𝑴 𝒙 𝒅𝒙 + 𝑵 𝒚 𝒅𝒚 = 𝟎 𝑑𝑜𝑛𝑑𝑒 𝑠𝑢𝑠 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑒𝑠𝑡á𝑛 𝑠𝑒𝑝𝑎𝑟𝑎𝑑𝑎𝑠
𝑃𝑎𝑟𝑎 ℎ𝑎𝑙𝑙𝑎𝑟 𝑙𝑎 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝑑𝑒 𝑒𝑠𝑡𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑠𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎 𝑐𝑎𝑑𝑎 𝑢𝑛𝑜 𝑑𝑒 𝑙𝑜𝑠 𝑡é𝑟𝑚𝑖𝑛𝑜𝑠
𝑴 𝒙, 𝒚 𝒅𝒙 + 𝑵 𝒙, 𝒚 𝒅𝒚 = 𝟎
𝑴 𝒙, 𝒚 + 𝑵 𝒙, 𝒚 = 𝟎
න 𝑴 𝒙 𝒅𝒙 + න 𝑵 𝒚 𝒅𝒚 = 𝑪
𝑬𝒔𝒕𝒂𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒔𝒆 𝒄𝒍𝒂𝒔𝒊𝒇𝒊𝒄𝒂𝒏 𝒆𝒏: 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒆 𝑽𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝑺𝒆𝒑𝒂𝒓𝒂𝒃𝒍𝒆𝒔, 𝑯𝒐𝒎𝒐𝒈é𝒏𝒆𝒂𝒔,
𝑬𝒙𝒂𝒄𝒕𝒂𝒔,𝑳𝒊𝒏𝒆𝒂𝒍𝒆𝒔 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓 𝒐𝒓𝒅𝒆𝒏, 𝒅𝒆 𝑩𝒆𝒓𝒏𝒐𝒖𝒍𝒍𝒊 𝒚 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝒅𝒆 𝑹𝒊𝒄𝒄𝒂𝒕𝒊
𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑶𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂𝒔 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓 𝒐𝒓𝒅𝒆𝒏 𝒚 𝒑𝒓𝒊𝒎𝒆𝒓 𝒈𝒓𝒂𝒅𝒐
𝟏. 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑶𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂𝒔 𝒅𝒆 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒔𝒆𝒑𝒂𝒓𝒂𝒃𝒍𝒆𝒔
𝑬𝒋𝒆𝒎𝒑𝒍𝒐 𝟏.
𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏:
𝑨𝒒𝒖í 𝑴 𝒙, 𝒚 = 𝒆𝒙𝒔𝒆𝒄𝒚 𝒚 𝑵 𝒙, 𝒚 = 𝟏 + 𝒆𝒙 𝒔𝒆𝒄 𝒚 𝒕𝒈 𝒚
𝑺𝒆𝒑𝒂𝒓𝒂𝒏𝒅𝒐 𝒍𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒕𝒆𝒏𝒆𝒎𝒐𝒔:
𝒆𝒙
𝒅𝒙
𝟏 + 𝒆𝒙 + 𝒕𝒈 𝒚 𝒅𝒚 = 𝟎
𝑄𝑢𝑒 𝑒𝑠 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑑𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑎𝑑𝑎, 𝑝𝑒𝑟𝑜, 𝑐𝑜𝑚𝑜 𝑛𝑜𝑠 𝑑𝑎𝑛 𝑐𝑜𝑛𝑑𝑖𝑐𝑖𝑜𝑛𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑙𝑒𝑠
𝑑𝑒𝑏𝑒𝑚𝑜𝑠 ℎ𝑎𝑙𝑙𝑎𝑟 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑙𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑖𝑎 𝑘 𝑦 𝑙𝑢𝑒𝑔𝑜 𝑒𝑠𝑐𝑟𝑖𝑏𝑖𝑟 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟
𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒏𝒅𝒐 න
𝒆𝒙
𝒅𝒙
𝟏 + 𝒆𝒙
+ න 𝒕𝒈 𝒚 𝒅𝒚 = 𝑪 , 𝑶𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝑳𝒏
𝟏 + 𝒆𝒙
𝒄𝒐𝒔 𝒚
= 𝒍𝒏 𝒌
𝑪𝒖𝒂𝒏𝒅𝒐 𝒙 = 𝟑 , 𝒚 = 𝟔𝟎°; 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 1 + e3 =
k
2
𝒅𝒆 𝒅𝒐𝒏𝒅𝒆 𝒌 = 𝟐(𝟏 + 𝒆𝟑)
𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐, 𝒍𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒑𝒂𝒓𝒕𝒊𝒄𝒖𝒍𝒂𝒓 𝒆𝒔: 𝟏 + 𝒆𝒙 = 𝟐 𝟏 + 𝒆𝟑 𝒄𝒐𝒔 𝒚
𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒆𝒙𝒔𝒆𝒄𝒚 𝒅𝒙 + 𝟏 + 𝒆𝒙 𝒔𝒆𝒄𝒚 𝒕𝒈𝒚𝒅𝒚 = 𝟎, 𝒄𝒖𝒂𝒏𝒅𝒐 𝒚 = 𝟔𝟎° 𝒔𝒊 𝒙 = 𝟑
𝑸𝒖𝒆 𝒍𝒐 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒎𝒐𝒔 𝒄𝒐𝒎𝒐: 𝟏 + 𝒆𝒙
= 𝒌 𝒄𝒐𝒔 𝒚
𝟐. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒙𝒚𝟐 − 𝒚𝟐 + 𝒙 − 𝟏 𝒅𝒙 + (𝒙𝟐𝒚 − 𝟐𝒙𝒚 + 𝒙𝟐 + 𝟐𝒚 − 𝟐𝒙 + 𝟐)𝒅𝒚 = 𝟎
𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏
𝑬𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒙𝒚𝟐 − 𝒚𝟐 + 𝒙 − 𝟏 𝒅𝒙 + 𝒙𝟐𝒚 − 𝟐𝒙𝒚 + 𝒙𝟐 + 𝟐𝒚 − 𝟐𝒙 + 𝟐 𝒅𝒚 = 𝟎
𝑨𝒈𝒓𝒖𝒑𝒂𝒎𝒐𝒔 𝒚 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒎𝒐𝒔 𝒄𝒐𝒎𝒐
𝒚𝟐
𝒙 − 𝟏 + (𝒙 − 𝟏) 𝒅𝒙 + 𝒙𝟐
𝒚 + 𝟏 − 𝟐𝒙 𝒚 + 𝟏 + 𝟐(𝒚 + 𝟏) 𝒅𝒚 = 𝟎
𝑬𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝒕𝒆𝒏𝒆𝒎𝒐𝒔: (𝒚𝟐+𝟏) 𝒙 − 𝟏 𝒅𝒙 + 𝒙𝟐 − 𝟐𝒙 + 𝟐 𝒚 + 𝟏 𝒅𝒚 = 𝟎
𝑺𝒆𝒑𝒂𝒓𝒂𝒏𝒅𝒐 𝒍𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒐𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔:
𝒙 + 𝟏 𝒅𝒙
𝒙𝟐 − 𝟐𝒙 + 𝟐
+
𝒚 + 𝟏 𝒅𝒚
𝒚𝟐 + 𝟏
= 𝟎
𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒎𝒐𝒔 𝒂𝒎𝒃𝒐𝒔 𝒎𝒊𝒆𝒎𝒃𝒓𝒐𝒔 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏
න
𝐱 − 𝟏 𝐝𝐱
𝐱𝟐 − 𝟐𝐱 + 𝟐
+ න
𝐲 + 𝟏 𝐝𝐲
𝐲𝟐 + 𝟏
= න 𝟎
𝟏
𝟐
𝑳𝒏(𝒙𝟐
−𝟐𝒙 + 𝟐) +
𝟏
𝟐
𝒍𝒏(𝒚𝟐
+𝟏) + 𝒂𝒓𝒕𝒈 𝒚 = 𝑪
𝑻𝒂𝒎𝒃𝒊é𝒏 𝒑𝒐𝒅𝒆𝒎𝒐𝒔 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒓𝒍𝒐 𝒄𝒐𝒎𝒐
𝑳𝒏(𝒙𝟐
− 𝟐𝒙 + 𝟐)(𝒚𝟐
+𝟏) = 𝒌 − 𝟐𝒂𝒓𝒕𝒂𝒈 𝒚
(𝐱𝟐
−𝟐𝐱 + 𝟐)(𝐲𝟐
+𝟏) = 𝑨𝒆−𝟐𝒂𝒓𝒕𝒈𝒚
(𝒙𝟐
−𝟐𝒙 + 𝟐)(𝒚𝟐
+ 𝟏)𝒆𝟐𝒂𝒓𝒂𝒈 𝒚
= 𝑨
𝑨𝒑𝒍𝒊𝒄𝒂𝒎𝒐𝒔 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒄𝒊𝒂𝒍 𝒂 𝒄𝒂𝒅𝒂 𝒎𝒊𝒆𝒎𝒃𝒓𝒐 𝒅𝒆 𝒆𝒔𝒕𝒂 𝒊𝒈𝒖𝒂𝒍𝒅𝒂𝒅 𝒚 𝒐𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔
𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐:
𝟑. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒙 𝟏 + 𝒚𝟐 + 𝒚 𝟏 + 𝒙𝟐 𝒚′ = 𝟎
𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒏𝒅𝒐 𝒑𝒐𝒓 𝒅𝒙 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒑𝒂𝒓𝒂 𝒑𝒐𝒏𝒆𝒓𝒍𝒐 𝒂 𝒍𝒂 𝒇𝒐𝒓𝒎𝒂 𝒈𝒆𝒏𝒆𝒓𝒂𝒍,
𝒔𝒆𝒑𝒂𝒓𝒂𝒏𝒅𝒐 𝒍𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔, 𝒕𝒆𝒏𝒆𝒎𝒐𝒔:
𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒏𝒅𝒐 න
𝒙𝒅𝒙
𝟏 + 𝒙𝟐
+ න
𝒚𝒅𝒚
𝟏 + 𝒚𝟐
= 𝒄
𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐 𝟏 + 𝒚𝟐 + 𝟏 + 𝒙𝟐 = 𝑪
𝒙 𝟏 + 𝒚𝟐 𝒅𝒙 + 𝒚 𝟏 + 𝒙𝟐𝒅𝒚 = 𝟎
𝒙𝒅𝒙
𝟏 + 𝒙𝟐
+
𝒚𝒅𝒚
𝟏 + 𝒚𝟐
= 𝟎
𝑬𝑱𝑬𝑹𝑪𝑰𝑪𝑰𝑶𝑺 𝑵º 𝟎𝟏
𝟏. 𝑽𝒆𝒓𝒊𝒇𝒊𝒄𝒂𝒓 𝒒𝒖𝒆 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏: 𝒚 = 𝒙 න
𝟎
𝒙
𝒔𝒆𝒏𝒕
𝒕
𝒅𝒕 , 𝒔𝒂𝒕𝒊𝒔𝒇𝒂𝒄𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒙𝒚′ = 𝒚 + 𝒙𝒔𝒆𝒏𝒙
𝟒. 𝑬𝒏𝒄𝒐𝒏𝒕𝒓𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒄𝒖𝒚𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒆𝒔: 𝒚 = 𝑨𝒆−𝒙 + 𝑩𝒆−𝟐𝒙 + 𝑪𝒆−𝟑𝒙
𝟔. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏: 𝒚′ =
𝒙𝟐𝒚 − 𝒙𝟐 + 𝒚 − 𝟏
𝒙𝒚 + 𝟐𝒙 − 𝟑𝒚 − 𝟔
𝟕. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒅𝒆: 𝒅𝒚 = 𝟏 + 𝒙 + 𝒚𝟐
+ 𝒙𝒚𝟐
𝒅𝒙
𝟖. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒑𝒂𝒓𝒕𝒊𝒄𝒖𝒍𝒂𝒓 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏: 𝒙 𝒚𝟔
+ 𝟏 𝒅𝒙 + 𝒚𝟐
𝒙𝟒
+ 𝟏 𝒅𝒚 = 𝟎; 𝒅𝒐𝒏𝒅𝒆 𝒚 𝟎 = 𝟏
2. 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑞𝑢𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛: 𝑓(𝑥) = න
0
1
𝑐𝑜𝑠𝑥𝑡
1 − 𝑡2
𝑑𝑡 , 𝑥 ≠ 0, 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑒 𝑎 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑓′′ 𝑥 +
1
𝑥
𝑓′ 𝑥 + 𝑓 𝑥 = 0
3. 𝐶𝑜𝑚𝑝𝑟𝑜𝑏𝑎𝑟 𝑠𝑖 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛: 𝑦 = 𝑐1𝑥 + 𝑐2𝑥 න
0
𝑥
𝑠𝑒𝑛𝑡
𝑡
𝑑𝑡, 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑦′′
. 𝑥𝑠𝑒𝑛𝑥 − 𝑦′
. 𝑥𝑐𝑜𝑠𝑥 + 𝑦𝑐𝑜𝑠𝑥 = 0
𝟓. 𝑫𝒆𝒎𝒐𝒔𝒕𝒓𝒂𝒓 𝒒𝒖𝒆 𝒒𝒖𝒆 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏: 𝒚 = 𝒆𝒙𝟐
(𝑨 + 𝑩 න 𝒆−𝒙𝟐
𝒅𝒙) 𝒔𝒂𝒕𝒊𝒔𝒇𝒂𝒄𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒚′′
− 𝟐𝒙𝒚′
− 𝟐𝒚 = 𝟎
𝑬𝑱𝑬𝑹𝑪𝑰𝑪𝑰𝑶𝑺 𝑵º 𝟎𝟐
𝟏. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟏 − 𝒙𝒚𝒄𝒐𝒔𝒙𝒚 𝒅𝒙 − 𝒙𝟐
𝒄𝒐𝒔𝒙𝒚𝒅𝒚 = 𝟎
𝟒. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟔𝒙 + 𝟑𝒚 − 𝟓 𝒅𝒙 − 𝟐𝒙 + 𝒚 𝒅𝒚 = 𝟎.
𝟔. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏: 𝟏 − 𝒙𝒚 + 𝒙𝟐𝒚𝟐 𝒅𝒙 + 𝒙𝟑𝒚 − 𝒙𝟐 𝒅𝒚 = 𝟎
𝟕. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒅𝒆 𝒍𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒊𝒓𝒄𝒖𝒏𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒒𝒖𝒆 𝒑𝒂𝒔𝒂𝒏 𝒑𝒐𝒓 𝒆𝒍 𝒐𝒓𝒊𝒈𝒆𝒏 𝒚
𝒄𝒖𝒚𝒐𝒔 𝒄𝒆𝒏𝒕𝒓𝒐𝒔 𝒆𝒔𝒕𝒂𝒏 𝒔𝒐𝒃𝒓𝒆 𝒆𝒍 𝒆𝒋𝒆 𝒀.
𝟖. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒅𝒆 𝒍𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒊𝒓𝒄𝒖𝒏𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒒𝒖𝒆 𝒕𝒊𝒆𝒏𝒆𝒏 𝒄𝒆𝒏𝒕𝒓𝒐 𝒔𝒐𝒃𝒓𝒆 𝒍𝒂 𝒓𝒆𝒄𝒕𝒂
𝒚 = 𝒙 , 𝒚 𝒔𝒐𝒏 𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒆𝒔 𝒂 𝒍𝒐𝒔 𝒆𝒋𝒆𝒔 𝒄𝒐𝒐𝒓𝒅𝒆𝒏𝒂𝒅𝒐𝒔.
𝟐. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟐𝒚𝒅𝒙 + 𝒙𝟐
𝒅𝒚 = −𝒅𝒙, 𝒄𝒖𝒂𝒏𝒅𝒐 𝒚
𝟏
𝑳𝒏𝟐
=
𝟕
𝟐
𝟑. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟏 + 𝒆𝒙
𝒚𝒚′
= 𝒆𝒚
, 𝒄𝒖𝒂𝒏𝒅𝒐 𝒚 𝟎 = 𝟎
𝟓. 𝑯𝒂𝒄𝒊𝒆𝒏𝒅𝒐 𝒖 = 𝒙 + 𝒚 , 𝒗 = 𝒙𝒚 , 𝒓𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟏 + 𝒙𝒚𝟑
𝒅𝒙 + 𝟏 + 𝒙𝟑
𝒚 𝒅𝒚 = 𝟎.
5. 𝐻𝑎𝑐𝑖𝑒𝑛𝑑𝑜 𝑢 = 𝑥 + 𝑦 , 𝑣 = 𝑥𝑦 , 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 diferencial 1 + x𝑦3
𝑑𝑥 + 1 + 𝑥3
𝑦 𝑑𝑦 = 0.
1 + x𝑦3
+ 𝑥2
𝑦2
− 𝑥2
𝑦2
𝑑𝑥 + 1 + 𝑥3
𝑦 + 𝑥2
𝑦2
− 𝑥2
𝑦2
𝑑𝑦 = 0
𝑑𝑥 + 𝑑𝑦 + 𝑥𝑦2 𝑦 + 𝑥 𝑑𝑥 − 𝑥2𝑦2𝑑𝑥 + 𝑥2𝑦 𝑥 + 𝑦 𝑑𝑦 − 𝑥2𝑦2𝑑𝑦 = 0
𝑑𝑥 + 𝑑𝑦 + 𝑥𝑦 y + x 𝑦𝑑𝑥 + 𝑥𝑦 𝑥 + 𝑦 𝑥𝑑𝑦 − 𝑥2
𝑦2
(𝑑𝑥 + 𝑑𝑦) = 0
𝐴ℎ𝑜𝑟𝑎 𝑐𝑜𝑚𝑜 𝒖 = 𝒙 + 𝒚 , 𝒗 = 𝒙𝒚 , 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠: 𝒅𝒖 = 𝒅𝒙 + 𝒅𝒚; 𝒅𝒗 = 𝒙𝒅𝒚 + 𝒚𝒅𝒙
𝑑𝑢 + 𝑣𝑢𝑑𝑣 − 𝑣2
𝑑𝑢 = 0
𝑟𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑡𝑒𝑛𝑒𝑚𝑜𝑠:
𝑑𝑥 + 𝑑𝑦 + 𝑥𝑦 𝑥 + 𝑦 𝑦𝑑𝑥 + 𝑥𝑑𝑦 − 𝑥2
𝑦2
(𝑑𝑥 + 𝑑𝑦) = 0
1 − 𝑣2
𝑑𝑢 + 𝑢𝑣𝑑𝑣 = 0

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Clase 1- Mat IV - Ecuaciones deif. Ordin. .pdf

  • 1. ESCUELA DE INGENIERÍA CIVIL MATEMÁTICA IV Dr. FRANCISCO BAUTISTA LOYOLA fbautista@unjfsc.edu.pe SEMESTRE ACADEMICO 2022- I
  • 2. 𝑪𝑶𝑵𝑪𝑬𝑷𝑻𝑶𝑺 𝑩𝑨𝑺𝑰𝑪𝑶𝑺 𝐸𝑠 𝑎𝑞𝑢𝑒𝑙𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑞𝑢𝑒 𝑒𝑛 𝑒𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑎 𝑐𝑜𝑛𝑡𝑖𝑒𝑛𝑒 𝑙𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑜 𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑢𝑛𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑑𝑒𝑠𝑐𝑜𝑛𝑜𝑐𝑖𝑑𝑎 𝑙𝑙𝑎𝑚𝑎𝑑𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝒂) 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑶𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂𝒔: 𝐸𝑗𝑒𝑚𝑝𝑙𝑜𝑠 𝑑𝑒 𝐸𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙𝑒𝑠: 1) dy dx = 4X + 7 , 2) y2 dx − x2 dy = 0, 3) m d2y dt2 = mg − K dy dx , 4) d2y dx2 3 − cos x dy dx = sen x , 5) 𝛿2 𝜔 𝛿𝑥2 + 𝛿2 𝜔 𝛿𝑦2 + 𝛿2 𝜔 𝛿𝑧2 = 0, 𝑑𝑜𝑛𝑑𝑒 𝜔 = 𝑓 𝑥, 𝑦, 𝑧 , 6)𝑥2 𝛿2 𝜔 𝛿𝑥2 + 𝑦2 𝛿2 𝜔 𝑑𝑦2 + 𝑧2 𝛿2 𝜔 𝑑𝑧2 = 0, 𝑑𝑜𝑛𝑑𝑒 𝜔 = 𝑓(𝑥, 𝑦, 𝑧) 𝑳𝒂𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝒔𝒆 𝒄𝒍𝒂𝒔𝒊𝒇𝒊𝒄𝒂𝒏 𝒆𝒏 𝒅𝒐𝒔 𝒕𝒊𝒑𝒐𝒔: 1) Ecuación Diferencial. 2) Clases de ecuaciones diferenciales Es aquella ecuación cuya función incógnita depende de una sola variable independiente donde sólo aparecen derivadas ordinarias. Son ecuaciones ordinarias 1, 2, 3 y 4 de los ejemplos anteriores.
  • 3. 𝒃) 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑷𝒂𝒓𝒄𝒊𝒂𝒍𝒆𝒔: 𝑬𝒍 𝒐𝒓𝒅𝒆𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒆𝒔𝒕á 𝒅𝒂𝒅𝒐 𝒑𝒐𝒓 𝒍𝒂 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒎𝒂𝒚𝒐𝒓 𝒒𝒖𝒆 𝒂𝒑𝒂𝒓𝒆𝒄𝒆 𝒆𝒏 𝒍𝒂 𝒆𝒄. 𝒅𝒊𝒇 𝑬𝒍 𝒈𝒓𝒂𝒅𝒐 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒆𝒔𝒕á 𝒅𝒂𝒅𝒐 𝒑𝒐𝒓 𝒆𝒍 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕𝒆 𝒅𝒆 𝒍𝒂 𝒎𝒂𝒚𝒐𝒓 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂 𝒒𝒖𝒆 𝒂𝒑𝒂𝒓𝒆𝒄𝒆 𝒆𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍. 𝑬𝒋𝒆𝒎𝒑𝒍𝒐𝒔: 𝑪𝒖𝒂𝒏𝒅𝒐 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏 𝒊𝒏𝒄ó𝒈𝒏𝒊𝒕𝒂 𝒅𝒆𝒑𝒆𝒏𝒅𝒆 𝒅𝒆 𝒗𝒂𝒓𝒊𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒊𝒆𝒏𝒕𝒆𝒔 𝒚 𝒍𝒂𝒔 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂𝒔 𝒔𝒐𝒏 𝒅𝒆𝒓𝒊𝒗𝒂𝒅𝒂𝒔 𝒑𝒂𝒓𝒄𝒊𝒂𝒍𝒆𝒔. 𝑳𝒂𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝟓 𝒚 𝟔 𝒂𝒏𝒕𝒆𝒓𝒊𝒐𝒓𝒆𝒔, 𝒔𝒐𝒏 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑷𝒂𝒓𝒄𝒊𝒂𝒍𝒆𝒔 𝟑) 𝑶𝒓𝒅𝒆𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟒) 𝑮𝒓𝒂𝒅𝒐 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 1) 𝑒𝑥 𝑑2𝑦 𝑑𝑥2 + 𝑠𝑒𝑛 𝑥 𝑑𝑦 𝑑𝑥 = 𝑥, 2𝑑𝑜 𝑜𝑟𝑑𝑒𝑛 𝑦 1𝑒𝑟 𝑔𝑟𝑎𝑑𝑜. 2) 𝑑3 𝑦 𝑑𝑥3 + 2 𝑑2 𝑦 𝑑𝑥2 3 + 𝑑𝑦 𝑑𝑥 = 𝑡𝑔 𝑥, 3𝑒𝑟 𝑜𝑟𝑑𝑒𝑛 𝑦 1𝑒𝑟 𝑔𝑟𝑎𝑑𝑜 3) 𝑑𝑦 𝑑𝑥 + 𝑝 𝑥 𝑦 = 𝑄 𝑥 , 1𝑒𝑟 𝑜𝑟𝑑𝑒𝑛 𝑦 1𝑒𝑟 𝑔𝑟𝑎𝑑𝑜. 4) 𝑑3𝑦 𝑑𝑥3 2 − 2 𝑑𝑥 𝑑𝑥 4 + 𝑥𝑦 = 0, 𝑂𝑟𝑑𝑒𝑛 3 𝑦 2𝑑𝑜 𝑔𝑟𝑎𝑑𝑜.
  • 4. 𝑺𝒖𝒑𝒐𝒏𝒈𝒂𝒎𝒐𝒔, 𝒒𝒖𝒆 𝑮 𝒙 = 𝑭 𝒙 + 𝑪, 𝒅𝒐𝒏𝒅𝒆 𝑪 𝒆𝒔 𝒖𝒏𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆 𝒂𝒓𝒃𝒊𝒕𝒓𝒂𝒓𝒊𝒂 𝒚 𝒒𝒖𝒆 𝒍𝒂 𝒅 𝑮 𝒙 = 𝒅 𝑭 𝒙 + 𝒄 = 𝑭′ (𝒙)𝒅𝒙 = 𝒇 𝒙 𝒅𝒙 𝑺𝒊 𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒐𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂 𝒅𝒆 𝒍𝒂 𝒇𝒐𝒓𝒎𝒂 𝒅𝒚 𝒅𝒙 = 𝒇 𝒙 𝑳𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒅𝒆 𝒆𝒔𝒕𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒆𝒔 𝒖𝒏𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏 𝒅𝒆 𝒍𝒂 𝒇𝒐𝒓𝒎𝒂 𝒚 = 𝑮 𝒙 𝒅𝒆 𝒕𝒂𝒍 𝒎𝒂𝒏𝒆𝒓𝒂 𝒒𝒖𝒆 𝒗𝒆𝒓𝒊𝒇𝒊𝒒𝒖𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒂𝒅𝒂. 𝑬𝒏𝒕𝒐𝒏𝒄𝒆𝒔: 𝒚 = 𝑮 𝒙 = 𝑭 𝒙 + 𝒄, 𝒔𝒆 𝒍𝒍𝒂𝒎𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝑳𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍, 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒂 𝒖𝒏𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒖𝒓𝒗𝒂𝒔 𝒒𝒖𝒆 𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏 𝒅𝒆 𝒍𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆 𝒂𝒓𝒃𝒊𝒕𝒓𝒂𝒓𝒊𝒂. 𝟓) 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍
  • 5. 𝑬𝒋𝒆𝒎𝒑𝒍𝒐𝒔: 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏: 𝑺𝒊 𝒚 = 𝒆𝒙𝟐 න 𝟎 𝒙 𝒆−𝒕𝟐 𝒅𝒕 + 𝒆𝒙𝟐 𝑳𝒖𝒆𝒈𝒐, 𝒓𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒏𝒅𝒐, 𝒕𝒆𝒏𝒆𝒎𝒐𝒔: 𝒚′ − 𝟐𝒙𝒚 = 𝟐𝒙𝒆𝒙𝟐 න 𝟎 𝒙 𝒆−𝒕𝟐 𝒅𝒕 + 𝟏 + 𝟐𝒙𝒆𝒙𝟐 − 𝟐𝒙 𝒆𝒙𝟐 න 𝟎 𝒙 𝒆−𝒕𝟐 𝒅𝒕 + 𝒆𝒙𝟐 𝑬𝒏𝒕𝒐𝒏𝒄𝒆𝒔: 𝒚′ − 𝟐𝒙𝒚 = 𝟐𝒙𝒆𝒙𝟐 න 𝟎 𝒙 𝒆−𝒕𝟐 𝒅𝒕 + 𝟏 + 𝟐𝒙𝒆𝒙𝟐 − 𝟐𝒙𝒆𝒙𝟐 න 𝟎 𝒙 𝒆−𝒕𝟐 𝒅𝒕 − 𝟐𝒙𝒆𝒙𝟐 = 𝟏 𝒚′ = 𝟐𝒙𝒆𝒙𝟐 න 𝟎 𝒙 𝒆−𝒕𝟐 + 𝟏 + 𝟐𝒙𝒆𝒙𝟐 𝟏. 𝑽𝒆𝒓𝒊𝒇𝒊𝒄𝒂𝒓 𝒒𝒖𝒆 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏 𝒚 = 𝒆𝒙𝟐 න 𝟎 𝒙 𝒆−𝒕𝟐 𝒅𝒕 + 𝒆𝒙𝟐 , 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒚′ − 𝟐𝒙𝒚 = 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐: 𝒚′ − 𝟐𝒙𝒚 = 𝟏
  • 6. 𝑹𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒏𝒅𝒐 𝒆𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒂𝒅𝒂 𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝟐. 𝑽𝒆𝒓𝒊𝒇𝒊𝒄𝒂𝒓 𝒔𝒊 𝒇 𝒙 = 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒅𝜽 , 𝒔𝒂𝒕𝒊𝒔𝒇𝒂𝒄𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒇′′ 𝒙 + 𝒇′(𝒙) 𝒙 + 𝒇 𝒙 = 𝟎 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏: 𝑺𝒊 𝒇 𝒙 = 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒅𝜽 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑡𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒: 𝒇′ 𝒙 = − 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒅𝜽 𝒅𝒆 𝒅𝒐𝒏𝒅𝒆 𝒇′′ 𝒙 = − 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏𝟐 𝜽 𝒅𝜽 𝑓′′ 𝑥 + 𝑓′ 𝑥 𝑥 + 𝑓 𝑥 = − 2 𝜋 න 0 𝜋 2 𝑐𝑜𝑠 𝑥𝑠𝑒𝑛 𝜃 𝑠𝑒𝑛2 𝜃 𝑑𝜃 − 2 𝜋 න 0 𝜋 2 𝑠𝑒𝑛 𝑥 𝑠𝑒𝑛 𝜃 𝑠𝑒𝑛 𝜃 𝑥 𝑑𝜃 + 2 𝜋 න 0 𝜋 2 𝑐𝑜𝑠(𝑥 𝑠𝑒𝑛 𝜃) 𝑑𝜃
  • 7. 𝒇′′ 𝒙 + 𝒇′ 𝒙 𝒙 + 𝒇 𝒙 = 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝟏 − 𝒔𝒆𝒏𝟐 𝜽 𝒅𝜽 − 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒔𝒆𝒏 𝒙𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒅𝜽 𝒙 = 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐𝜽 𝒅𝜽 − 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒅𝜽 𝒙 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒏𝒅𝒐 𝒑𝒐𝒓 𝒑𝒂𝒓𝒕𝒆𝒔 𝒍𝒂 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒍 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐𝜽 𝒅𝜽 𝒉𝒂𝒄𝒆𝒎𝒐𝒔: 𝒖 = 𝒄𝒐𝒔 𝜽 → d𝒖 = − 𝒔𝒆𝒏 𝜽𝒅𝜽 𝒅𝒗 = 𝒄𝒐𝒔 𝒙𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔 𝜽 𝒅𝜽 → 𝒗 = 𝒔𝒆𝒏 (𝒙 𝒔𝒆𝒏𝜽) 𝒙 𝑟𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜: න 0 𝜋 2 𝑐𝑜𝑠 𝑥 𝑠𝑒𝑛 𝜃 𝑐𝑜𝑠2 𝜃 𝑑𝜃 = 𝑐𝑜𝑠𝜃. 𝑠𝑒𝑛(𝑥 𝑠𝑒𝑛𝜃) 𝑥 /0 𝜋 2 + න 0 𝜋 2 𝑠𝑒𝑛 𝑥 𝑠𝑒𝑛 𝜃 𝑠𝑒𝑛 𝜃 𝑥 𝑑𝜃 𝒑𝒓𝒊𝒎𝒆𝒓𝒐 𝒂 𝒍𝒂 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒍𝒐 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒎𝒐𝒔 𝒄𝒐𝒎𝒐 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝜽. 𝒄𝒐𝒔𝜽 𝒅𝜽
  • 8. 𝑹𝒆𝒎𝒑𝒍𝒂𝒛𝒂𝒏𝒅𝒐 𝒆𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝒒𝒖𝒆: 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐 𝜽 𝒅𝜽 = 𝟎 − 𝟎 + න 𝟎 𝝅 𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒙 𝒅𝜽 න 𝟎 𝝅 𝟐 𝒄𝒐𝒔 𝒙 𝒔𝒆𝒏 𝜽 𝒄𝒐𝒔𝟐 𝜽 𝒅𝜽 = න 𝟎 𝝅 𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒙 𝒅𝜽 𝒇′′ 𝒙 + 𝒇′(𝒙) 𝒙 + 𝒇 𝒙 = 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽 𝒙 𝒅𝜽 − 𝟐 𝝅 න 𝟎 𝝅 𝟐 𝒔𝒆𝒏 𝒙 𝒔𝒆𝒏 𝜽 𝒔𝒆𝒏 𝜽𝒅𝜽 𝒙 = 𝟎 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐 𝒇′′ 𝒙 + 𝒇′(𝒙) 𝒙 + 𝒇 𝒙 = 𝟎
  • 9. 𝑬𝒋𝒆𝒎𝒑𝒍𝒐𝒔 𝟏. 𝑬𝒏𝒄𝒐𝒏𝒕𝒓𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍, 𝒄𝒖𝒚𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒆𝒔: 𝒚 = 𝑨𝒔𝒆𝒏𝒙 + 𝑩𝒄𝒐𝒔𝒙 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏: 𝑺𝒊 𝒔𝒆 𝒕𝒊𝒆𝒏𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒆 𝒖𝒏𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒖𝒓𝒗𝒂𝒔, 𝒔𝒆 𝒑𝒖𝒆𝒅𝒆 𝒐𝒃𝒕𝒆𝒏𝒆𝒓 𝒔𝒖 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒎𝒆𝒅𝒊𝒂𝒏𝒕𝒆 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒄𝒊ó𝒏 𝒅𝒆 𝒍𝒂𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆𝒔 𝒐 𝒑𝒂𝒓á𝒎𝒆𝒕𝒓𝒐𝒔 , 𝒆𝒔𝒕𝒐 𝒔𝒆 𝒐𝒃𝒕𝒊𝒆𝒏𝒆 𝒂𝒊𝒔𝒍𝒂𝒏𝒅𝒐 𝒍𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆 𝒆𝒏 𝒖𝒏 𝒎𝒊𝒆𝒎𝒃𝒓𝒐 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒚 𝒅𝒆𝒓𝒊𝒗𝒂𝒏𝒅𝒐. 𝑻𝒂𝒎𝒃𝒊é𝒏 𝒔𝒆 𝒑𝒖𝒆𝒅𝒆 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒓 𝒍𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆 𝒅𝒆𝒓𝒊𝒗𝒂𝒏𝒅𝒐 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒂𝒅𝒂 𝒕𝒂𝒏𝒕𝒂𝒔 𝒗𝒆𝒄𝒆𝒔 𝒄𝒐𝒎𝒐 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆𝒔 𝒂𝒓𝒃𝒊𝒕𝒓𝒂𝒓𝒊𝒂𝒔 𝒕𝒆𝒏𝒈𝒂, 𝒚 𝒍𝒖𝒆𝒈𝒐 𝒔𝒆 𝒓𝒆𝒔𝒖𝒆𝒍𝒗𝒆 𝒆𝒍 𝒔𝒊𝒔𝒕𝒆𝒎𝒂 𝒅𝒆 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒇𝒐𝒓𝒎𝒂𝒅𝒐 𝒄𝒐𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍. 𝑺𝒖𝒎𝒂𝒏𝒅𝒐: 𝒚′′ + 𝒚 𝒐𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟔) 𝑬𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒂 𝒑𝒂𝒓𝒕𝒊𝒓 𝒅𝒆 𝒖𝒏𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒖𝒓𝒗𝒂𝒔 𝑺𝒊 𝒚 = 𝑨 𝒔𝒆𝒏 𝒙 + 𝑩 𝒄𝒐𝒔 𝑿 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔: ቊ 𝒚′ = 𝑨𝒄𝒐𝒔𝒙 − 𝑩𝒔𝒆𝒏𝒙 y′′ = −A sen x − B cos X ቊ 𝒚′′ = −𝑨𝒄𝒐𝒔𝒙 − 𝑩𝒔𝒆𝒏𝒙 y = A sen x + B cos X 𝒚′′ + 𝒚 = 𝟎
  • 10.  𝑶𝒕𝒓𝒂 𝒎𝒂𝒏𝒆𝒓𝒂 𝒅𝒆 𝒆𝒍𝒊𝒎𝒊𝒏𝒂𝒓 𝒍𝒂𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒆𝒔 𝒆𝒔, 𝒄𝒐𝒏𝒔𝒊𝒅𝒆𝒓𝒂𝒏𝒅𝒐 𝒆𝒍 𝒔𝒊𝒔𝒕𝒆𝒎𝒂 𝒔𝒊𝒈𝒖𝒊𝒆𝒏𝒕𝒆: y = Asenx + Bcosx 𝑦′ = Acos x − Bsenx y′′ = −Asenx − Bcosx −y + Asenx + Bcosx = 0 −y′ + Acos x − Bsenx = 0 −y′′ − Asen x − Bcos x = 0 𝑬𝒔𝒕𝒆 𝒔𝒊𝒔𝒕𝒆𝒎𝒂 𝒅𝒆 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒆𝒏 𝒅𝒐𝒔 𝒊𝒏𝒄ó𝒈𝒏𝒊𝒕𝒂𝒔 𝑨 𝒚 𝑩, 𝒕𝒊𝒆𝒏𝒆 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒔𝒊 𝒚 𝒔ó𝒍𝒐 𝒔𝒊: −𝒚 𝒔𝒆𝒏 𝒙 𝒄𝒐𝒔 𝒙 −𝒚′ 𝒄𝒐𝒔 𝒙 − 𝒔𝒆𝒏 𝒙 −𝒚′′ −𝒔𝒆𝒏 𝒙 − 𝒄𝒐𝒔 𝒙 = 𝟎, 𝒓𝒆𝒔𝒐𝒍𝒗𝒊𝒆𝒏𝒅𝒐 𝒆𝒔𝒕𝒆 𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒂𝒏𝒕𝒆, 𝒔𝒆 𝒕𝒊𝒆𝒏𝒆 𝑦′′ + 𝑦 = 0
  • 11. 𝐿𝑎 𝑓𝑜𝑟𝑚𝑎 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑑𝑒 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑟 𝑢𝑛𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑜𝑟𝑑𝑖𝑛𝑎𝑟𝑖𝑎 𝑑𝑒 𝑝𝑟𝑖𝑚𝑒𝑟 𝑜𝑟𝑑𝑒𝑛 𝑦 𝑝𝑟𝑖𝑚𝑒𝑟 𝑔𝑟𝑎𝑑𝑜 𝑒𝑠: 𝐿𝑎𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙𝑒𝑠 𝑑𝑒 𝑙𝑎 𝑓𝑜𝑟𝑚𝑎: 𝑆𝑒 𝑙𝑙𝑎𝑚𝑎𝑛 𝐸𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝐷𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙𝑒𝑠 𝑑𝑒 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑆𝑒𝑝𝑎𝑟𝑎𝑏𝑙𝑒𝑠 𝑠𝑖 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒𝑛 𝑒𝑠𝑐𝑟𝑖𝑏𝑖𝑟 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑚𝑎𝑛𝑒𝑟𝑎 𝑴 𝒙 𝒅𝒙 + 𝑵 𝒚 𝒅𝒚 = 𝟎 𝑑𝑜𝑛𝑑𝑒 𝑠𝑢𝑠 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑒𝑠𝑡á𝑛 𝑠𝑒𝑝𝑎𝑟𝑎𝑑𝑎𝑠 𝑃𝑎𝑟𝑎 ℎ𝑎𝑙𝑙𝑎𝑟 𝑙𝑎 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝑑𝑒 𝑒𝑠𝑡𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑠𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎 𝑐𝑎𝑑𝑎 𝑢𝑛𝑜 𝑑𝑒 𝑙𝑜𝑠 𝑡é𝑟𝑚𝑖𝑛𝑜𝑠 𝑴 𝒙, 𝒚 𝒅𝒙 + 𝑵 𝒙, 𝒚 𝒅𝒚 = 𝟎 𝑴 𝒙, 𝒚 + 𝑵 𝒙, 𝒚 = 𝟎 න 𝑴 𝒙 𝒅𝒙 + න 𝑵 𝒚 𝒅𝒚 = 𝑪 𝑬𝒔𝒕𝒂𝒔 𝒆𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒔𝒆 𝒄𝒍𝒂𝒔𝒊𝒇𝒊𝒄𝒂𝒏 𝒆𝒏: 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒆 𝑽𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝑺𝒆𝒑𝒂𝒓𝒂𝒃𝒍𝒆𝒔, 𝑯𝒐𝒎𝒐𝒈é𝒏𝒆𝒂𝒔, 𝑬𝒙𝒂𝒄𝒕𝒂𝒔,𝑳𝒊𝒏𝒆𝒂𝒍𝒆𝒔 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓 𝒐𝒓𝒅𝒆𝒏, 𝒅𝒆 𝑩𝒆𝒓𝒏𝒐𝒖𝒍𝒍𝒊 𝒚 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝒅𝒆 𝑹𝒊𝒄𝒄𝒂𝒕𝒊 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑶𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂𝒔 𝒅𝒆 𝒑𝒓𝒊𝒎𝒆𝒓 𝒐𝒓𝒅𝒆𝒏 𝒚 𝒑𝒓𝒊𝒎𝒆𝒓 𝒈𝒓𝒂𝒅𝒐 𝟏. 𝑬𝒄𝒖𝒂𝒄𝒊𝒐𝒏𝒆𝒔 𝑫𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍𝒆𝒔 𝑶𝒓𝒅𝒊𝒏𝒂𝒓𝒊𝒂𝒔 𝒅𝒆 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒔𝒆𝒑𝒂𝒓𝒂𝒃𝒍𝒆𝒔
  • 12. 𝑬𝒋𝒆𝒎𝒑𝒍𝒐 𝟏. 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏: 𝑨𝒒𝒖í 𝑴 𝒙, 𝒚 = 𝒆𝒙𝒔𝒆𝒄𝒚 𝒚 𝑵 𝒙, 𝒚 = 𝟏 + 𝒆𝒙 𝒔𝒆𝒄 𝒚 𝒕𝒈 𝒚 𝑺𝒆𝒑𝒂𝒓𝒂𝒏𝒅𝒐 𝒍𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒕𝒆𝒏𝒆𝒎𝒐𝒔: 𝒆𝒙 𝒅𝒙 𝟏 + 𝒆𝒙 + 𝒕𝒈 𝒚 𝒅𝒚 = 𝟎 𝑄𝑢𝑒 𝑒𝑠 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑑𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑎𝑑𝑎, 𝑝𝑒𝑟𝑜, 𝑐𝑜𝑚𝑜 𝑛𝑜𝑠 𝑑𝑎𝑛 𝑐𝑜𝑛𝑑𝑖𝑐𝑖𝑜𝑛𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑙𝑒𝑠 𝑑𝑒𝑏𝑒𝑚𝑜𝑠 ℎ𝑎𝑙𝑙𝑎𝑟 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑙𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑖𝑎 𝑘 𝑦 𝑙𝑢𝑒𝑔𝑜 𝑒𝑠𝑐𝑟𝑖𝑏𝑖𝑟 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒏𝒅𝒐 න 𝒆𝒙 𝒅𝒙 𝟏 + 𝒆𝒙 + න 𝒕𝒈 𝒚 𝒅𝒚 = 𝑪 , 𝑶𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝑳𝒏 𝟏 + 𝒆𝒙 𝒄𝒐𝒔 𝒚 = 𝒍𝒏 𝒌 𝑪𝒖𝒂𝒏𝒅𝒐 𝒙 = 𝟑 , 𝒚 = 𝟔𝟎°; 𝒆𝒏𝒕𝒐𝒏𝒄𝒆𝒔 1 + e3 = k 2 𝒅𝒆 𝒅𝒐𝒏𝒅𝒆 𝒌 = 𝟐(𝟏 + 𝒆𝟑) 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐, 𝒍𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒑𝒂𝒓𝒕𝒊𝒄𝒖𝒍𝒂𝒓 𝒆𝒔: 𝟏 + 𝒆𝒙 = 𝟐 𝟏 + 𝒆𝟑 𝒄𝒐𝒔 𝒚 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒆𝒙𝒔𝒆𝒄𝒚 𝒅𝒙 + 𝟏 + 𝒆𝒙 𝒔𝒆𝒄𝒚 𝒕𝒈𝒚𝒅𝒚 = 𝟎, 𝒄𝒖𝒂𝒏𝒅𝒐 𝒚 = 𝟔𝟎° 𝒔𝒊 𝒙 = 𝟑 𝑸𝒖𝒆 𝒍𝒐 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒎𝒐𝒔 𝒄𝒐𝒎𝒐: 𝟏 + 𝒆𝒙 = 𝒌 𝒄𝒐𝒔 𝒚
  • 13. 𝟐. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒙𝒚𝟐 − 𝒚𝟐 + 𝒙 − 𝟏 𝒅𝒙 + (𝒙𝟐𝒚 − 𝟐𝒙𝒚 + 𝒙𝟐 + 𝟐𝒚 − 𝟐𝒙 + 𝟐)𝒅𝒚 = 𝟎 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏 𝑬𝒏 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒙𝒚𝟐 − 𝒚𝟐 + 𝒙 − 𝟏 𝒅𝒙 + 𝒙𝟐𝒚 − 𝟐𝒙𝒚 + 𝒙𝟐 + 𝟐𝒚 − 𝟐𝒙 + 𝟐 𝒅𝒚 = 𝟎 𝑨𝒈𝒓𝒖𝒑𝒂𝒎𝒐𝒔 𝒚 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒎𝒐𝒔 𝒄𝒐𝒎𝒐 𝒚𝟐 𝒙 − 𝟏 + (𝒙 − 𝟏) 𝒅𝒙 + 𝒙𝟐 𝒚 + 𝟏 − 𝟐𝒙 𝒚 + 𝟏 + 𝟐(𝒚 + 𝟏) 𝒅𝒚 = 𝟎 𝑬𝒏𝒕𝒐𝒏𝒄𝒆𝒔 𝒕𝒆𝒏𝒆𝒎𝒐𝒔: (𝒚𝟐+𝟏) 𝒙 − 𝟏 𝒅𝒙 + 𝒙𝟐 − 𝟐𝒙 + 𝟐 𝒚 + 𝟏 𝒅𝒚 = 𝟎 𝑺𝒆𝒑𝒂𝒓𝒂𝒏𝒅𝒐 𝒍𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒐𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔: 𝒙 + 𝟏 𝒅𝒙 𝒙𝟐 − 𝟐𝒙 + 𝟐 + 𝒚 + 𝟏 𝒅𝒚 𝒚𝟐 + 𝟏 = 𝟎 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒎𝒐𝒔 𝒂𝒎𝒃𝒐𝒔 𝒎𝒊𝒆𝒎𝒃𝒓𝒐𝒔 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 න 𝐱 − 𝟏 𝐝𝐱 𝐱𝟐 − 𝟐𝐱 + 𝟐 + න 𝐲 + 𝟏 𝐝𝐲 𝐲𝟐 + 𝟏 = න 𝟎
  • 14. 𝟏 𝟐 𝑳𝒏(𝒙𝟐 −𝟐𝒙 + 𝟐) + 𝟏 𝟐 𝒍𝒏(𝒚𝟐 +𝟏) + 𝒂𝒓𝒕𝒈 𝒚 = 𝑪 𝑻𝒂𝒎𝒃𝒊é𝒏 𝒑𝒐𝒅𝒆𝒎𝒐𝒔 𝒆𝒔𝒄𝒓𝒊𝒃𝒊𝒓𝒍𝒐 𝒄𝒐𝒎𝒐 𝑳𝒏(𝒙𝟐 − 𝟐𝒙 + 𝟐)(𝒚𝟐 +𝟏) = 𝒌 − 𝟐𝒂𝒓𝒕𝒂𝒈 𝒚 (𝐱𝟐 −𝟐𝐱 + 𝟐)(𝐲𝟐 +𝟏) = 𝑨𝒆−𝟐𝒂𝒓𝒕𝒈𝒚 (𝒙𝟐 −𝟐𝒙 + 𝟐)(𝒚𝟐 + 𝟏)𝒆𝟐𝒂𝒓𝒂𝒈 𝒚 = 𝑨 𝑨𝒑𝒍𝒊𝒄𝒂𝒎𝒐𝒔 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒄𝒊𝒂𝒍 𝒂 𝒄𝒂𝒅𝒂 𝒎𝒊𝒆𝒎𝒃𝒓𝒐 𝒅𝒆 𝒆𝒔𝒕𝒂 𝒊𝒈𝒖𝒂𝒍𝒅𝒂𝒅 𝒚 𝒐𝒃𝒕𝒆𝒏𝒆𝒎𝒐𝒔 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐:
  • 15. 𝟑. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒙 𝟏 + 𝒚𝟐 + 𝒚 𝟏 + 𝒙𝟐 𝒚′ = 𝟎 𝑺𝒐𝒍𝒖𝒄𝒊ó𝒏 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒏𝒅𝒐 𝒑𝒐𝒓 𝒅𝒙 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒑𝒂𝒓𝒂 𝒑𝒐𝒏𝒆𝒓𝒍𝒐 𝒂 𝒍𝒂 𝒇𝒐𝒓𝒎𝒂 𝒈𝒆𝒏𝒆𝒓𝒂𝒍, 𝒔𝒆𝒑𝒂𝒓𝒂𝒏𝒅𝒐 𝒍𝒂𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔, 𝒕𝒆𝒏𝒆𝒎𝒐𝒔: 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒏𝒅𝒐 න 𝒙𝒅𝒙 𝟏 + 𝒙𝟐 + න 𝒚𝒅𝒚 𝟏 + 𝒚𝟐 = 𝒄 𝑷𝒐𝒓 𝒍𝒐 𝒕𝒂𝒏𝒕𝒐 𝟏 + 𝒚𝟐 + 𝟏 + 𝒙𝟐 = 𝑪 𝒙 𝟏 + 𝒚𝟐 𝒅𝒙 + 𝒚 𝟏 + 𝒙𝟐𝒅𝒚 = 𝟎 𝒙𝒅𝒙 𝟏 + 𝒙𝟐 + 𝒚𝒅𝒚 𝟏 + 𝒚𝟐 = 𝟎
  • 16. 𝑬𝑱𝑬𝑹𝑪𝑰𝑪𝑰𝑶𝑺 𝑵º 𝟎𝟏 𝟏. 𝑽𝒆𝒓𝒊𝒇𝒊𝒄𝒂𝒓 𝒒𝒖𝒆 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏: 𝒚 = 𝒙 න 𝟎 𝒙 𝒔𝒆𝒏𝒕 𝒕 𝒅𝒕 , 𝒔𝒂𝒕𝒊𝒔𝒇𝒂𝒄𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒙𝒚′ = 𝒚 + 𝒙𝒔𝒆𝒏𝒙 𝟒. 𝑬𝒏𝒄𝒐𝒏𝒕𝒓𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒄𝒖𝒚𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒆𝒔: 𝒚 = 𝑨𝒆−𝒙 + 𝑩𝒆−𝟐𝒙 + 𝑪𝒆−𝟑𝒙 𝟔. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏: 𝒚′ = 𝒙𝟐𝒚 − 𝒙𝟐 + 𝒚 − 𝟏 𝒙𝒚 + 𝟐𝒙 − 𝟑𝒚 − 𝟔 𝟕. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒈𝒆𝒏𝒆𝒓𝒂𝒍 𝒅𝒆: 𝒅𝒚 = 𝟏 + 𝒙 + 𝒚𝟐 + 𝒙𝒚𝟐 𝒅𝒙 𝟖. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒔𝒐𝒍𝒖𝒄𝒊ó𝒏 𝒑𝒂𝒓𝒕𝒊𝒄𝒖𝒍𝒂𝒓 𝒅𝒆 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏: 𝒙 𝒚𝟔 + 𝟏 𝒅𝒙 + 𝒚𝟐 𝒙𝟒 + 𝟏 𝒅𝒚 = 𝟎; 𝒅𝒐𝒏𝒅𝒆 𝒚 𝟎 = 𝟏 2. 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑞𝑢𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛: 𝑓(𝑥) = න 0 1 𝑐𝑜𝑠𝑥𝑡 1 − 𝑡2 𝑑𝑡 , 𝑥 ≠ 0, 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑒 𝑎 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑓′′ 𝑥 + 1 𝑥 𝑓′ 𝑥 + 𝑓 𝑥 = 0 3. 𝐶𝑜𝑚𝑝𝑟𝑜𝑏𝑎𝑟 𝑠𝑖 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛: 𝑦 = 𝑐1𝑥 + 𝑐2𝑥 න 0 𝑥 𝑠𝑒𝑛𝑡 𝑡 𝑑𝑡, 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑦′′ . 𝑥𝑠𝑒𝑛𝑥 − 𝑦′ . 𝑥𝑐𝑜𝑠𝑥 + 𝑦𝑐𝑜𝑠𝑥 = 0 𝟓. 𝑫𝒆𝒎𝒐𝒔𝒕𝒓𝒂𝒓 𝒒𝒖𝒆 𝒒𝒖𝒆 𝒍𝒂 𝒇𝒖𝒏𝒄𝒊ó𝒏: 𝒚 = 𝒆𝒙𝟐 (𝑨 + 𝑩 න 𝒆−𝒙𝟐 𝒅𝒙) 𝒔𝒂𝒕𝒊𝒔𝒇𝒂𝒄𝒆 𝒂 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒚′′ − 𝟐𝒙𝒚′ − 𝟐𝒚 = 𝟎
  • 17. 𝑬𝑱𝑬𝑹𝑪𝑰𝑪𝑰𝑶𝑺 𝑵º 𝟎𝟐 𝟏. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟏 − 𝒙𝒚𝒄𝒐𝒔𝒙𝒚 𝒅𝒙 − 𝒙𝟐 𝒄𝒐𝒔𝒙𝒚𝒅𝒚 = 𝟎 𝟒. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟔𝒙 + 𝟑𝒚 − 𝟓 𝒅𝒙 − 𝟐𝒙 + 𝒚 𝒅𝒚 = 𝟎. 𝟔. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏: 𝟏 − 𝒙𝒚 + 𝒙𝟐𝒚𝟐 𝒅𝒙 + 𝒙𝟑𝒚 − 𝒙𝟐 𝒅𝒚 = 𝟎 𝟕. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒅𝒆 𝒍𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒊𝒓𝒄𝒖𝒏𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒒𝒖𝒆 𝒑𝒂𝒔𝒂𝒏 𝒑𝒐𝒓 𝒆𝒍 𝒐𝒓𝒊𝒈𝒆𝒏 𝒚 𝒄𝒖𝒚𝒐𝒔 𝒄𝒆𝒏𝒕𝒓𝒐𝒔 𝒆𝒔𝒕𝒂𝒏 𝒔𝒐𝒃𝒓𝒆 𝒆𝒍 𝒆𝒋𝒆 𝒀. 𝟖. 𝑯𝒂𝒍𝒍𝒂𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝒅𝒆 𝒍𝒂 𝒇𝒂𝒎𝒊𝒍𝒊𝒂 𝒅𝒆 𝒄𝒊𝒓𝒄𝒖𝒏𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒔 𝒒𝒖𝒆 𝒕𝒊𝒆𝒏𝒆𝒏 𝒄𝒆𝒏𝒕𝒓𝒐 𝒔𝒐𝒃𝒓𝒆 𝒍𝒂 𝒓𝒆𝒄𝒕𝒂 𝒚 = 𝒙 , 𝒚 𝒔𝒐𝒏 𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒆𝒔 𝒂 𝒍𝒐𝒔 𝒆𝒋𝒆𝒔 𝒄𝒐𝒐𝒓𝒅𝒆𝒏𝒂𝒅𝒐𝒔. 𝟐. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝟐𝒚𝒅𝒙 + 𝒙𝟐 𝒅𝒚 = −𝒅𝒙, 𝒄𝒖𝒂𝒏𝒅𝒐 𝒚 𝟏 𝑳𝒏𝟐 = 𝟕 𝟐 𝟑. 𝑹𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟏 + 𝒆𝒙 𝒚𝒚′ = 𝒆𝒚 , 𝒄𝒖𝒂𝒏𝒅𝒐 𝒚 𝟎 = 𝟎 𝟓. 𝑯𝒂𝒄𝒊𝒆𝒏𝒅𝒐 𝒖 = 𝒙 + 𝒚 , 𝒗 = 𝒙𝒚 , 𝒓𝒆𝒔𝒐𝒍𝒗𝒆𝒓 𝒍𝒂 𝒆𝒄𝒖𝒂𝒄𝒊ó𝒏 𝒅𝒊𝒇𝒆𝒓𝒆𝒏𝒄𝒊𝒂𝒍 𝟏 + 𝒙𝒚𝟑 𝒅𝒙 + 𝟏 + 𝒙𝟑 𝒚 𝒅𝒚 = 𝟎.
  • 18.
  • 19.
  • 20. 5. 𝐻𝑎𝑐𝑖𝑒𝑛𝑑𝑜 𝑢 = 𝑥 + 𝑦 , 𝑣 = 𝑥𝑦 , 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 diferencial 1 + x𝑦3 𝑑𝑥 + 1 + 𝑥3 𝑦 𝑑𝑦 = 0. 1 + x𝑦3 + 𝑥2 𝑦2 − 𝑥2 𝑦2 𝑑𝑥 + 1 + 𝑥3 𝑦 + 𝑥2 𝑦2 − 𝑥2 𝑦2 𝑑𝑦 = 0 𝑑𝑥 + 𝑑𝑦 + 𝑥𝑦2 𝑦 + 𝑥 𝑑𝑥 − 𝑥2𝑦2𝑑𝑥 + 𝑥2𝑦 𝑥 + 𝑦 𝑑𝑦 − 𝑥2𝑦2𝑑𝑦 = 0 𝑑𝑥 + 𝑑𝑦 + 𝑥𝑦 y + x 𝑦𝑑𝑥 + 𝑥𝑦 𝑥 + 𝑦 𝑥𝑑𝑦 − 𝑥2 𝑦2 (𝑑𝑥 + 𝑑𝑦) = 0 𝐴ℎ𝑜𝑟𝑎 𝑐𝑜𝑚𝑜 𝒖 = 𝒙 + 𝒚 , 𝒗 = 𝒙𝒚 , 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠: 𝒅𝒖 = 𝒅𝒙 + 𝒅𝒚; 𝒅𝒗 = 𝒙𝒅𝒚 + 𝒚𝒅𝒙 𝑑𝑢 + 𝑣𝑢𝑑𝑣 − 𝑣2 𝑑𝑢 = 0 𝑟𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑡𝑒𝑛𝑒𝑚𝑜𝑠: 𝑑𝑥 + 𝑑𝑦 + 𝑥𝑦 𝑥 + 𝑦 𝑦𝑑𝑥 + 𝑥𝑑𝑦 − 𝑥2 𝑦2 (𝑑𝑥 + 𝑑𝑦) = 0 1 − 𝑣2 𝑑𝑢 + 𝑢𝑣𝑑𝑣 = 0