Ecuaciones diferenciales ordinarias

1. UNIVERSIDAD DE LAS FUERZAS
ARMADAS 'ESPE'
May 23, 2016
ECUACIONES DIFERENCIALES ORDINALES
TEMA: RICCATI
DEBER N°10
NOMBRE:
Asqui guillermo
nrc: 2110
FECHA: MAYO, 23 DE 2016
Resolver las siguientes ecuaciones diferenciales de Riccati
1

2. 1.y = y2
− y
x + 1 − 1
4x2 ;
y, = 1
2x + tgx
y , = −1
2x2 + sec2
x
−1
2x2 + sec2
x = ( 1
2x + tgx)2
− 1
x ( 1
2x + tgx) + 1 − 1
4x2
−1
2x2 + sec2
x = 1
4x2 + tgx
x + tg2
x − 1
2x2 + tgx
x + 1 − 1
4x2
sec2
x = tg2
x+1
sec2
x = sec2
x; entonces y,es solucion
y = z + y,
y = z + y ,
z + y, = (z + y, )2
− z+y,
x + 1 − 1
4x2
z + y, = z2
+ 2zy, +y2
, −z
x − y
x + 1 − 1
4x2
z = z2
+ 2zy, −z
x
z = z2
+ z(2y, −1
x )
z = z2
+ z(1
x − 1
x + 2tgx)
z + z(−2tgx) = z2
.............n=2
v = z1−n
= z1−2
= z−1
v = −z−2
z
−z−2
z + (−z−2
)z(−2tgx) = (−z−2
)z2
−z−2
z + z−1
(2tgx) = −1
u(x) =e
´
2tgxdx
= e2lnsecx
= sec2
x
vsec2
x = −
´
sec2
xdx
vsec2
x = −tgx + c
v = (−tgx + c)/sec2
x
1
z = −tgxcos2
x + ccos2
x
z = 1
−tgxcos2x+ccos2x
SOL:
y = 1
−tgxcos2x+ccos2x + 1
2x + tgx
2.y = xy2
− 2y + 4 − 4x
y, = 2
y , = 0
0 = x(2)2
− 2(2) + 4 − 4x
0 = 4x − 4 + 4 − 4x
0 = 0; entonces y,es solución
y = 1
z + y,
y = −z−2
+ y ,
−z−2
+ y , = x(1
z + y)2
− 2(1
z + y) + 4 − 4x
−z−2
+ y , = x
z2 + 2y,
z + xy, −2
z − 2y, +4 − 4x
−z
z2 = x
z2 + 2y,
z − 2
z
z = −x − 4z + 2z
z = −x − 2z
z + 2z = −x
u(x) = e
´
2dx
= e2x
z = −1
e2x
´
e2x
xdx + c
e2x´
e2x
xdx :
2

3. a = x.................da = dx
db = e2x
dx........b = e2x
/2
´
e2x
xdx = xe2x
2 − 1
2
´
e2x
dx
´
e2x
xdx = xe2x
2 − e2x
4
z = − 1
e2x (xe2x
2 − e2x
4 ) + c
e2x
z = e2x
−2xe2x
+c
4e2x
SOL:
y = e2x
−2xe2x
+c
4e2x + 2
3.y = y2
x−1 − xy
x−1 + 1
y, = 1
y, = 0
y = y2
−xy
x−1 + 1
0 = 1−x
x−1 + 1
0 = −1 + 1
0 = 0
y = 1
z + y,
y = −z−2
z + y ,
−z−2
z + y , =
( 1
z +y)2
−x( 1
z +y)
x−1 + 1
−z−2
z + y , =
1
z2 + 2y,
z +y,2
− x
z +yx)
x−1 + 1
−z−2
z = 1
z2(x−1) + 2y,
z(x−1) − x
z(x−1)
z = −1
(x−1) − 2zy,
(x−1) − xz
(x−1)
z = z(− 2y,
(x−1) − x
(x−1) ) + −1
(x−1)
z = z(− 2
(x−1) − x
(x−1) ) + −1
(x−1)
z = z(− x−2
(x−1) ) −1
+(x−1)
z − z( x−2
(x−1) ) = −1
(x−1)
u(x) = e
´
− x−2
x−1 dx
= e−x+ln(x−1)
= (x − 1)e−x
z = 1
(x−1)e−x
´
(x − 1)e−x −1
(x−1) dx + cex
x−1
z = 1
(x−1)e−x
´
−e−x
dx + cex
x−1
z = 1
(x−1)e−x (−ex
) + cex
x−1
z = 1
(x−1) + cex
x−1
z = cex
+1
x−1
SOL:
y = cex
+1
x−1 + 1
4. y = (1 + x + 2x2
cos) − (1 + 4xcosx)y + 2y2
cosx
y, (x) = x
y, ´(x) = 1
1 = 1 + x + 2x2
cosx − x − 4x2
cosx + 2x2
cosx
1 = 1; entonces y, (x)si es solución
3

4. y = z + y,
y´=z´+y, ´
z + y, = 1 + x + 2x2
cosx − (1 + 4xcosx)(z + y, ) + 2(z + y, )2
cosx
z +y, = 1+x+2x2
cosx−z−y, −z4xcosx−4xy, cosx+2z2
cosx+4zy, cosx+
2y, cosx
z = −4xzcosx − z + 2z2
cosx + 4zy, cosx
z = −4xzcosx − z + 2z2
cosx + 4zx, cosx
z = −z + 2z2
cosx
z +z = 2z2
cosx......................n=2
v = z1−n
= z1−2
= z−1
v = −z−2
z
−z−2
z +(-z−2
)z = 2(−z−2
)z2
cosx
−z−2
z + z−1
= −2cosx
v − v = −2cosx
u(x) = e
´
−dx
= e−x
v = 1
e−x
´
−2e−x
cosxdx + c
e−x´
e−x
cosxdx :
a = cosx.............da = senxdx
db = e−x
dx.........b = −e−x
´
e−x
cosxdx = −e−x
cosx + e−x
senx −
´
e−x
cosxdx´
e−x
cosxdx = (−e−x
cosx + e−x
senx)/2
v = 1
e−x − 2(−e−x
cosx + e−x
senx)/2 + c
e−x
v = −e−x
ex
senx + e−x
ex
cosx + cex
v = −senx + cosx + cex
1
z = 1
cosx−senx+cex
SOL:
y = cosx − senx + cex
+ x
5.y − sen2
xy2
+ 1
senxcosx y + cos2
x = 0
y, = cosx
senx
y , = −csc2
x
−csc2
x − sen2
x( cosx
senx )2
+
cosx
senx
senxcosx + cos2
x = 0
-
1
sen2x − cos2
x + 1
sen2x + cos2
x= 0
0 = 0,entonces y,es solución
y = 1
z + y,
y = −z−2
z + y ,
−z−2
z + y , −sen2
x(1
z + y, )2
+
1
z +y,
senxcosx + cos2
x = 0
−z−2
z + y , −2y,sen2
x
z − sen2
x
z2 − sen2
xy,2
+ 1
zcosxsenx + y,
senxcosx + cos2
x = 0
−z−2
z − 2y,sen2
x
z − sen2
x
z2 + 1
zcosxsenx = 0
−z−2
z −
2 cosx
senx sen2
x
z − sen2
x
z2 + 1
zcosxsenx = 0
−z−2
z − 2senxcosx
z − sen2
x
z2 + 1
zcosxsenx = 0
z + 2zsenxcosx + sen2
x − z
cosxsenx = 0
z + z(2senxcosx − 1
cosxsenx ) = −sen2
x
u(x) = e
´
(2senxcosx− 1
cosxsenx )dx
= esen2
x
e−2
´ dx
sen2x
4

5. ´ dx
sen2x :
v = 2x...........dv = 2dx´ dx
sen2x = 1
2
´ dv
senv
v = tg(t/2).........dv = 2dt
1+t2 .....................senv = 2t
1+t2
´ dx
sen2x = 1
2
´ 2dt
1+t2
1+t2
2t = 1
2
´ dt
t = 1
2 lnt = 1
2 ln(tg 2x
2 )=
1
2 ln(tgx)
u(x) = esen2
x
e−2 1
2 ln(tgx)
= esen2
x
e−ln(tgx)
= tg−1
x.esen2
x
ztg−1
x.esen2
x
=
´
tg−1
x.esen2
x
. − sen2
xdx
´
tg−1
x.esen2
x
. − sen2
xdx=
´
esen2
x
.senxcosxxdx
m = sen2
x...........dm = 2senxcosdx´
tg−1
x.esen2
x
. − sen2
xdx=
´
em
.senxcosx 1
2senxcosx dm =
´ −1
2 em
dm
ztg−1
x.esen2
x
=−em
2 + c
z = −senxesen2x
2cosxesen2x
+ csenx
cosesenˆe
z = 2csenx−esen2x
senx
2cosxesen2x
SOL:
y = 2csenx−esen2x
senx
2cosxesen2x
+ cosx
senx
Resolver las ecuaciones diferenciales siguientes
1.x = (y )5
+ y
sea y = p
x = p5
+ p
1 = 5p4
p + p
1 = p (5p4
+ 1)
1
5p4+1 = dp
dx = dp
dy
dy
x
1
5p4+1 = dp
dy p
´
dy =
´
(5p4
+ 1)pdp
SOL:
y = 5p6
6 + p2
2 + c
x = p5
+ p
2. y = (y )−3
− y
y = p
y = p−3
− p
y = −3p−4
p − p
y = p (−3p−4
− 1)
p = dp
dx (−3p−4
− 1)
dx = (−3
p5 − 1
p )dp
´
dx =
´ −3
p5 dp-
´ dp
d
x = 3p−4
4 − lnp + c
SOL:
y = p−3
− p
x = 3
4p4 − lnp + c
5

6. 3.2y + y − 2y log(y ) = 0
y = p
2p + y − 2plogp = 0
y = 2plogp − 2p
y = 2p logp + 2logep − 2p
y = p (2logp + 2loge − 2)
p = dp
dx (2logp + 2loge − 2)
dxp =(2logp + 2loge − 2)dp/p´
dx =
´ 2lnpdp
pln10 +
´ 2lne
pln10 dp −
´ 2dp
p
´
dx = 2
ln10
´ lnpdp
p + 2
ln10
´ 1
p dp −
´ 2dp
p
´ lnpdp
p :
m = lnp........................dm = dp
p
´ mpdm
p =
´
mdm = m2
2
´
dx =
´ 2lnpdp
pln10 +
´ 2lne
pln10 dp −
´ 2dp
p
x = 2
ln10 (ln2
p
2 ) + 2
ln10 lnp − 2lnp + c
SOL:
y = 2plogp − 2p
x =ln2
p
lnqo + 2lnp( 1
ln10 − 1) + c
4.x = y√
1+(y )2
y = p
x = p√
1+(p)2
1 =
√
1+(p)2p − ppp
√
1+(p)2
1+p2 =
√
1+(p)2p − p2p
√
1+(p)2
1+p2
1 = p (1+p2
−p2
)
√
(1+(p)2)3
(1 + (p)2)3 = dp
dx = dp
dy
dy
dx
(1 + (p)2)3 = dp
dy p
´ dpd√
(1+(p)2)3
=
´
dy
´ dpd√
(1+(p)2)3
:
u = 1 + p2
........................du = 2pdp´ dpd√
(1+(p)2)3
=
´ dpu
2p
√
(u)3
= 1
2
´ du
u3/2
´ dpd√
(1+(p)2)3
= u−1/2
−1 = −1√
1+p2
+c
SOL:
y = −1√
1+p2
+c
x = p√
1+(p)2
6

7. 5.y = xy − 1
y
y = p
y = xp − 1
p
y = p + xp + p
p2
p = p + xp + p
p2
0 = xp + p
p2
0 = p (x + 1
p2 )
Si p = 0........p = c
Si x + 1
p2 = 0.....x = − 1
p2
SOL:
y = xp − 1
p = − 1
p2 p − 1
p = −2
p
x = − 1
p2
6. y = xy + y +
√
y
y = p
y = xp + p +
√
p
y = p + xp + p + p
2
√
p
p = p + xp + p + p
2
√
p
0 = xp + p + p
2
√
p
0 = p (x + 1 + 1
2
√
p )
Si p = 0........p = 0
Si x + 1+ 1
2
√
p ..................x = −(1 + 1
2
√
p )
SOL:
y = xp + p +
√
p=−(1 + 1
2
√
p )p + p +
√
p =
√
p
2
x = −(1 + 1
2
√
p )
7.x = y e2y
y = p
x = pe2p
1 =p e2p
+ p2e2p
p
1 =p (e2p
+ p2e2p
)
1 =(e2p
+ p2e2p
)dp
dx = e2p
+ p2e2p
)dp
dy
dy
dx
1 =(e2p
+ p2e2p
)pdp
dy´
dy =
´
(e2p
+ p2e2p
)pdp´
(e2p
+ p2e2p
)pdp=
´
pe2p
dp +
´
2e2p
p2
dp´
pe2p
dp :
u = p..............du = dp
dv = e2p
dp.........p = e2p
/2
´
pe2p
dp = pe2p
2 − e2p
4´
2e2p
p2
dp :
u = p2
..............du = 2pdp
7

8. dv = e2p
dp.........p = e2p
/2
´
2e2p
p2
dp=
p2
e2p
2 −
´
pe2p
dp=
p2
e2p
2 − pe2p
2 + e2p
4
y =pe2p
2 − e2p
4 + p2
e2p
2 − pe2p
2 + e2p
4 + c
SOL:
y =p2e2p
− pe2p
2 + e2p
4 + c
x = pe2p
8.(y )2
− 5xy − 6x2
= 0
y = p
(p)2
− 5xp − 6x2
= 0
(p − 6x)(p + x) = 0
p − 6x = 0...................p = 6x
p + x = 0.....................p = −x
1)p = 6x
y = 6x´
dy =
´
6xdx
y = 3x2
+ c
y − 3x2
+ c
2)p = −x
y = −x´
dy =
´
−xdx
y = −x2
2 + c
y + x2
2 + c = 0
SOL:
(y − 3x2
+ c)(y + x2
2 + c) = 0
9.y = x(1+(y )2
2y )
y = p
y = x(1+(p )2
2p )
dy
dx = 1+p2
2p + x(2p2pp −(1−p2
)2p
4p2
p = 1+p2
2p + 4xp2
p −2xp −2xp2
p
4p2
p − 1+p2
2p = p (4xp2
−2x−2xp2
)
4p2
2p2−
1−p2
2p = p x(2p2
−1−p2
)
2p2
p2
− 1 = p x(p2
−1)
2p2
1 = xdp
2p2dx
´ dp
p2 =
´ dx
2x
−2p−1
= lnx + lnc
lnxc= −2p−1
SOL:
x = e−2p−1
y = x(1+(p )2
2p ) = e(1+(p )2
2p )
8