ejercicio, ecuaciones diferenciales ordinarias
TRANSCRIPT
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8/17/2019 EJercicio, Ecuaciones diferenciales ordinarias
1/2
I 1 =∫1
3
1 − z dz+∫1
3 z+
2
3
z2+ z+1
dz
I 1 =− 1
3 ln (1 − z)+1
3 ∫ z+2
z2 + z+1 dz
I 1 = −1
3 ln (1 − z)+
1
3 I 2
I 2 =∫ z+2 z
2+ z+1dz
t = z2 + z+1 ; dt =( 2 z+1 )dz
I 2 =1
2 ∫(2 z+1 )+3 z
2 + z+1dz
I 2 =1
2 ∫(2 z+1 ) z
2 + z+1dz+
3
2 ∫ dz
( z+1
2)
2
+(√ 32
)2
I 2 =1
2 ln ( z2 + z+1 )+ 3
√ 3acrtan (
2 z+1√ 3
)
I 1 = −1
3 ln (1 − z)+
1
6 ln ( z2 + z+1 )+ 1
√ 3acrtan (
2 z+1√ 3
)
1
2
(1
6 ln ( z2 + z+1 )− 1
3 ln (1 − z)+ 1√ 3 acrtan
(2 z+1
√ 3
))= x+c
( 112 ln ( z2+ z+1 )− 16 ln (1 − z)+ 12 √ 3 acrtan (2 z+1√ 3 ))= x+c
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8/17/2019 EJercicio, Ecuaciones diferenciales ordinarias
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((2 x− 2 y+2 )2+(2 x− 2 y+2 )+1 )1
12 ln ¿
(¿−1
6 ln (1 −( 2 x− 2 y+2 ))+ 1
2 √ 3 acrtan (2 (2 x− 2 y+2 )+1√ 3 ))= x+c
( 112 ln (4 x2 +4 y2 − 8 xy+10 x− 10 y+7 )− 16 ln (2 y− 2 x− 1 )+ 12 √ 3 acrtan (4 x− 4 y+5√ 3 ))= x+c