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

8/17/2019 EJercicio, Ecuaciones diferenciales ordinarias

2/2

((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