Problemario mate
Enviado por Danny Perez • 5 de Marzo de 2020 • Apuntes • 5.498 Palabras (22 Páginas) • 180 Visitas
𝒂) (𝒙𝒚 − 𝒙)𝒅𝒙 + (𝒙𝒚 + 𝒚)𝒅𝒚 = 𝟎 ; 𝑺𝒐𝒍. (𝒚 − 𝟏)𝒆 ⁄ = 𝑪(𝒙 + 𝟏)[pic 1]
(𝑦 − 1)𝑑𝑥 + (𝑥 + 1)𝑦 = 0
(𝑦 − 1)𝑑𝑥 = −𝑦(𝑥 + 1)𝑑𝑦
𝑥 𝑦
(𝑥 + 1) 𝑑𝑥 = − (𝑦 − 1) 𝑑𝑦[pic 2][pic 3]
𝑥 𝑦 1 1
∫ 𝑑𝑥 = − ∫ 𝑑𝑦 = ∫ (1 + ) 𝑑𝑥 = − ∫ (1 + ) 𝑑𝑦 (𝑥 + 1) (𝑦 − 1) (𝑥 + 1) (𝑦 + 1)[pic 4][pic 5][pic 6][pic 7]
𝑥 − ln|𝑥 + 1| = −𝑦 − ln|𝑦 − 1| + 𝐶
𝑥 − ln|𝑥 + 1| + 𝑦 + ln|𝑦 − 1| + 𝐶
𝑦 − 1
𝑥 + 𝑦 + ln ([pic 8]
𝑥 + 1
) = 𝐶
𝑥+𝑦+ln(𝑦−1)
𝑒 𝑥+1
= 𝑒𝑐
𝑒𝑥+𝑦
(𝐼𝑛)(𝑦−1)
𝑒 𝑥+1
= 𝑒𝑐
𝑒𝑥+𝑦
𝑦 − 1
( ) = 𝑒𝑐[pic 9]
𝑥 + 1
(𝑦 − 1)𝑒𝑥+𝑦 = (𝑥 + 1) (𝑦 − 1)𝑥+𝑦 = 𝐶(𝑥 + 1)
𝒃) 𝒙𝟐𝒅𝒙 + 𝒚(𝒙 − 𝟏)𝒅𝒚 = 𝟎
𝑥2𝑑𝑥 = −𝑦(𝑥 − 1)𝑑𝑦
𝑥2[pic 10]
(𝑥 − 1)
𝑥2
∫
𝑑𝑥 = −𝑦𝑑𝑦
𝑑𝑥 = − ∫ 𝑦 𝑑𝑦[pic 11]
(𝑥 − 1)
𝑥 + 1 𝑥 + 1 + 1 𝑥 − 1 𝑦2 𝑥−1 −𝑥2 + 𝑥 𝑥 −𝑥 + 1 [pic 12] 1 | 𝑥2 1 ∫ 𝑑𝑥 = ∫ 𝑥 + 1 + 𝑑𝑥 (𝑥 − 1) 𝑥 − 1 |
∫ (𝑥 + 1 +
𝑥2[pic 13]
1[pic 14]
𝑥 − 1[pic 15]
) 𝑑𝑥 =
𝑦2[pic 16]
𝑦2 2[pic 17][pic 18][pic 19]
+ 𝐶[pic 20]
2
𝑥2 2[pic 21]
+ 𝑥 + ln|𝑥 − 1| =
+ 𝑥 + ln|𝑥 − 1| −
2
𝑦2 2[pic 22]
+ 𝐶
= 𝐶
𝒄) (𝒙𝒚 + 𝒙)𝒅𝒙 = (𝒙𝟐𝒚𝟐 + 𝒙𝟐 + 𝒚𝟐 + 𝟏)𝒅𝒚 𝑺𝒐𝒍. 𝐥𝐧(𝒙𝟐 + 𝟏) = 𝒚𝟐 − 𝟐𝒚 + 𝟒 𝐥𝐧[𝒄(𝒚 + 𝟏)]
(𝑦 + 1)𝑑𝑥 = (𝑥2 + 1)(𝑦2 + 1)
𝑥 (𝑦2 + 1) 𝑥 (𝑦2 + 1)[pic 23][pic 24][pic 25]
(𝑥2 + 1) 𝑑𝑥 =[pic 26][pic 27][pic 28][pic 29]
→ ∫ 𝑑𝑥 = ∫ 𝑑𝑦 (𝑦 + 1) (𝑥2 + 1) (𝑦 + 1)
[pic 30][pic 31][pic 32][pic 33][pic 34][pic 35][pic 36][pic 37][pic 38][pic 39][pic 40][pic 41][pic 42]
𝑥 1 1∫ 𝑑𝑥 = ln|(𝑥2 + 1)| + 𝐶 (𝑥2+1) 2 | 𝑢 = 𝑥2 + 1 𝑑𝑢 = 𝑥𝑑𝑥 2 |
(𝑦2+1) 𝑦2 1 1 1 2∫ (𝑦+1) 𝑑𝑦 = ∫ 𝑦+1 + 𝑦+1 𝑑𝑦 = ∫ 𝑦 − 1 − 𝑦+1 + 𝑦+1 𝑑𝑦 | 𝑦2 − 𝑦 + ln|𝑦 + 1||𝐼𝑛||𝑦 + 1| 2 + 𝐶 |
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