Texto 2

Ecuaciones Diferenciales Ordinarias
de Orden Superior
Pedro Maldonado Tapia, pdmaldonado1@espe.edu.ec

Ejercicio 0.1.

yy y + 2y 2 y = 3yy

2

Soluci´
on

y

y =p

(1)

y = pp

(2)

= p(p 2 + pp )

(3)

(1),(2),(3), en ecuaci´
on principal
ypp(p 2 + pp ) + 2p2 pp = 3yp2 p 2
yp2 p 2 + yp3 p + 2p3 p = 3yp2 p 2
yp3 p + 2p3 p = 2yp2 p 2
Ejercicio 0.2.

y 2 (y y − 2y 2 ) = 1
Soluci´
on
(1),(2),(3), enecuaci´
on principal
y 2 (pp(p 2 + pp ) − 2p2 p 2 ) = 1
y 2 p2 p 2 + y 2 p3 p − 2y 2 p2 p 2 = 1
y 2 p3 p − y 2 p2 p 2 = 1
y 2 p2 (pp − p 2 ) = 1

Page 1 of 24

Ejercicio 0.3.

(yy − 3 2 )y = y 5
Soluci´
on
(1),(2),(3), en ecuaci´
on principal
yp(p 2 + pp ) − 3p2 p 2 )y = p5
y 2 pp 2 + y 2 p2 p − 3yp2 p 2 = p5
yp(yp 2 + y 2 p − 3pp 2 ) = p5
Ejercicio 0.4.

yy 2 + y

3

+y

4

=0

Soluci´
on(1),(2),(3), en ecuaci´
on principal
yp2 + p3 p 3 + p4 (p + pp )4 = 0
p2 (y + pp 3 + p42 (p + pp )4 ) = 0
Ejercicio 0.5.

xy −

y 2
= ey
y

Soluci´
on
(1),(2),(3), en ecuaci´
on principal
xp −

p2 p 2
= epp
p

xp − pp 2 = epp
p(x − p 2 ) = epp
Ejercicio 0.6.

yy − y 2 − √

yy
=0
1 + x2

Soluci´
on

Page 2 of 24

y=1

(4)

y =z

(5)

y =z +z

(6)

(4),(5),(6),en ecuaci´
on principal
z + z2 − z2 − √
z=√

z
=0
1 + x2

z
1 + x2

dz
=
z

√

dx
1 + x2
1 + x2 )

lnz = ln(x +
C1 z = x +
zdx =

(x +

= x2 + x

1 + x2

1 + x2 + C1 )dx
1 + x2 + ln(x +

1 + x2 ) + C 1 x + C 2

y = ezdx
2

y = ex
Para:
C1 = 1
C1 = 1
C1 = 1
C1 = 1

C2
C2
C2
C2

√
+x 1+x2 +C1 x+C2

(x +

1 + x2 )

= 1; negro
= 2; azul
= 3; verde
= 4; coral

Page 3 of 24

Ejercicio 0.7.

yy − y 2 −

xyy
=0
1 + x2

Soluci´
on
(4),(5),(6),enecuaci´
on principal

xz
=0
1 + x2
xz
z =
1 + x2
dz
xdx
=
z
1 + x2

z + z2 − z2 −

lnz =

1
ln(1 + x2 )
2

z = ( 1 + x2 ) + C 1
zdx =

( 1 + x2 ) + C1 dx
y=e

y = C1 (x +
Para:
C1 = 1
C1 = 2
C1 = 3
C1 = 4

C2
C2
C2
C2

zdx

1 + x2 )C2

= 1; negro
= 2; azul
= 3; verde
= 4; coral

Page 4 of 24

Ejercicio 0.8.

x2 yy − (y − xy )2 = 0
Soluci´
on
(4),(5),(6),en ecuaci´
on principal
x2 (z + z 2 ) − (1 −xz)2 = 0
x2 z + x2 z 2 − 1 + 2xz − x2 z 2 = 0
x2 z − 1 + 2xz = 0
2z
1
=0
z − 2+
x
x
2dx
−
x = e−2lnx = x−2
v=e
u=

dx = x + C1

1 C1
+ 2
x
x
C1
1 C1
+ 2 dx = lnx −
+ C2
x
x
x

z = vu = (x + C1 )x−2 =
zdx =

y=e

zdx

y = C2 xe
Para:
C1 = 1
C1 = 2
C1 = 3
C1 = 4

C2
C2
C2
C2

C1
x

= 1; negro
= 2; azul
= 3; verde
= 4; coral

Page 5 of 24

Ejercicio 0.9.

xyy + y (y + xy ) = 0
Soluci´
on(4),(5),(6),en ecuaci´
on principal
x(z + z 2 ) − z(1 + xz) = 0
xz + z 2 x − z − xz 2 = 0
xz − z = 0
dx
dz
=
z
x
dz
=
z

dx
x

lnz = lnx + C1
z = C1 x
zdx =

C 1 x2
+ C2
2

y=e

zdx

2

y = C2 eC1 x
Para:
C1 = 1
C1 = 2
C1 = 3
C1 = 4

C2
C2
C2
C2

= 1; negro
= 2; azul
= 3; verde
= 4; coral

Page 6 of 24

Ejercicio 0.10.

x2 yy − (y + xy )2 = 0
Soluci´
on
(4),(5),(6),en ecuaci´
on principal
x2 (z + z 2 ) − (1+ xz)2 = 0
x2 z + x2 z 2 − 1 − 2xz − x2 z 2 = 0
x2 z − 1 − 2xz = 0
2z
1
z −
= 2
x
x
v=e
u=

2dx

= e2lnx = x2
−1
1
dx = 3 + C1
4
x
3x

z = uv =

1
+ C 1 x2
3x

y=e
zdx =

zdx

C1 x3
1
lnx +
+ C2
3
3
3

y=
Para:
C1 = 1
C1 = 2
C1 = 3
C1 = 4

C2
C2
C2
C2

C2 eC1 x
x1/3

= 1; negro
= 2; azul
= 3; verde
= 4; coral

Page 7 of 24

Ejercicio 0.11.

xyy − xy 2 + yy = 0
Soluci´
on
(4),(5),(6),en ecuaci´on principal
x(z + z 2 ) − xz 2 + z = 0
xz + z 2 x − xz 2 + z = 0
dx
dz
=
−
z
x
C1
=z
x
zdx = C1 lnx + C2
y=e

zdx

y = C1 xC2
Para:
C1 = 1
C1 = 2
C1 = 3
C1 = 4

C2
C2
C2
C2

= 1; negro
= 2; azul
= 3; verde
= 4; coral

Page 8 of 24

Ejercicio 0.12.

x2 (yy − y 2 ) + xyy − y

y 2 + x2 y 2 = 0

Soluci´
on
(4),(5),(6),en ecuaci´
on principal
x(z + z 2 ) + xz −

z=

x2

1 + x2 z 2 = 0

xz + xz −

1 +x2 z 2 = 0

u
;
x

x2 u − u
x2

z =

(xu − u) ux
+
−
x2
x
ux=
√

; en(7)
x2 u2
=0
x2

1 + u2

du
=
1 + u2

ln(u +

dx
x

1 + u2 ) = lnx
1 + u2 = x + C1

u+
zx +

1+

(7)

1 + z 2 x2 = x + C 1

1 + z 2 x2 = x2 C1 2 − 2zx2 C1 + z 2 x2
1 − x2 C 1 2
=
−2x2 C1
zdx =

1
2

C1 x +
y=e

zdx
1
xC1 + C2

zdx

C1 x
1
+
2
2xC
1 (C3 )
y=e
Para:
C1 = 1
C1 = 2
C1 = 3
C1 = 4

C2
C2
C2
C2

= 1; negro
= 2;...