E
j
e r
c i
c i
o
s
r
e s
u
e l
t
H
o
s
ug
d
o
E
1.
1
c
1
ua c
E
i
c
o ne
uaci
s
di
ones
l
i
f
e
e
r
br
i
nc
es
a c i
ba
A
r
neal
e c u
Lom
13
1
e
i
y
r
do
l
F
20
al
o n
l
o
r
e s
e
i
ni
m
os
el
f
actor
e
s
de
educi
pr
bl
e
s
i
a
e
i
ntegra
.
p(x) = 2 ´
f
actor
i
m
ul
t
i
n
pl
tegr
i
an
cam
e
te:
os
2dx
l
a
ec
e2x
=
uaci
on
p
or
el
f
act
or
i
nte
gr
ante
e2x ddxy + 2e2x = 0 e
l
l
ado
i
z
qui
d 2x dx [e y] s
epar
am
e
r
do
de
l
a
e
c
uac
i
on
s
e
r
e
duce
a:
=0 os
var
i
abl
es
e
i
nt
egr
am
os
.
d 2x dx [e y]
´
=0
´
dx + c
e2x y = c y = ce−2x 2
.
dy = 3y dx f
orm
a
l
i
n
eal
.
dy dx
3y = 0
− p(x) = −3 ´
F
m
act
or
ul
t
i
i
pl
ntegr
i
an
cam
os
t
p
e −3dx e−3x
e:
or
=
f
act
or
i
nte
gr
ante
.
1
f
e r
e n
c i
a l
1
dy + 2y = 0 dx ef
i
s
.
D
d
.
m
s
e
t
as
r
.
or
de
n
e s
e−3x ddyx
− 3e−3x y = 0
dy 3x y dx [e
−
´
´
= 0 dx + c
e−3x y = c y = ce3x 3
.
3 pas
am
os
l
a
e
c
uac
i
on
a
l
a
f
or
m
dy + 12y = 4 dx a
l
dy dx
i
nea
l
.
+ 4y =
4 3
p(x) = 4 ´
F
act
or
i
ntegr
an
t
e
e:
4dx
e4x
=
e4x ddxy + 4e4x y = 43 e4x
´
d 4x dx [e y]
=
´
e4x dx + c
e4x y = 14 e4x + c y= 4
1 4
4x
+ ce−
.
y = 2y + x2 + 5
f
or
m
a
l
i
neal
y
− 2y = x2 + 5
´
F
act
or
i
ntegr
an
t
e −2dx = e−2x
e:
e−2x y
´
− 2e−2x y = e−2x x2 + 5e−2x
d 2x y] dx [e
−
=
´ −2x 2 ´ e x + 5 e−2x + c
− 52 e−2x − 14 e−2x (2x2 + 2x + 1) + C y = − x2 − x2 − 14 + 52 + ce2x
e−2x y =
2
5
.
ydx
− 4(x + y6)dy = 0
ydx = 4(x + y6 )dy dx dy
=
4(x+y6 ) y
dx dy
;
2
=
4x y
+
4y6 y
deni
m
os
l
a
f
or
m
a
l
i
neal
.
dx dy
F
act
or
i
ntegr
an
t
e−4
e:
´
1 y dy
− 4yx = 4y5
e−4 log(y) elog(y )
4
−
;
;
1 dx y4 dy
1 4x 4 y
−y
d 1 dy [ y 4 x] d 1 [ x] dy y4
´
1 y4 x
;
y −4 =
1 5 y4 4y
=
= 4y
´
=4
ydy
= 2y2 + C
x = 2y6 + cy4 6
.
xy + y = ex
ex x
y + x1 y =
F
act
or
i
ntegr
an
t
e:
1 x dx
´
e
= elog x = x x
xy + xx y =
xe x
d [xy] dx I
nte
gr
am
os
= ex
:
´
d dx [xy]
=
´
ex dx + c
xy = ex + c y = ex x−1 + cx−1 7
.
x dy dx ha
ce
m
os
l
a
s
ust
i
t
uci
e
r
i
vam
os
e
s
t
a
ul
t
i
m
+
y x
u = y 1−n
on:
u= D
dy 2 +y = 2 dx y =
2 xy2
.
.(1
n=
done
)
2
− − = y 3 u1/3−= y
y1 ( 2)
;
a.
1 2/3 du 3u dx
−
3
=
dy dx
1 y4
Su
s
t
i
t
ui
m
os
en
l
a
ec
ua
ci
on
di
f
er
enci
1 2/3 du 3u dx
−
A
c
om
odam
os
a
l
a
f
or
m
a
l
i
nea
l
,
al
u
+
m
t
i
pl
+3
dx E
s
t
a
es
un
a
ec
uaci
on
l
i
neal
ul
t
i
pl
i
cam
os
p
or
f
act
D
1 x dx
´
e3 M
.
or
i
i
u
c
m
ando
t
oda
l
a
e
c
uac
x
os
elf
act
or
i
nt
egr
ant
3
= e3log x = elog x = x3
nte
gr
ante
.
3u 36 x3 du dx + 3x x = x x d 3 dx [x u] i
ntegr
am
= 6x2
os.
d 3 dx [x u]
´
=6
´
x2 + c
x3 u = 2x3 + c u = 2 + cx−3 S
u
s
t
i
t
u
i
m
u = y3
os
y 3 = 2 + cx−3 8
dy y 1/2 dx + y 3/2 = 1
.
dy dx
y 3/2 y 1/2
+
u = y 1−n n =
d
i
ci
on
y(0) = 4
u = y 1−(−1/2) = y 3/2
−1/2
;
con
↔ dydx + y = y−1/2
1 y1/2
=
;
;
u2/3 = y 2 1/3 du dx 3u
−
S
u
st
i
t
u
i
m
−
M
ul
t
i
pl
i
ca
m
os
l
a
ec
uaci
on
a
e
c
uac
i
on
s
e
r
e
t
e:
duj
o
3
F
act
or
i
dy dx
+ u2/3 = (u2/3 )−1/2 2 1/3 3u
por
du dx L
=
os.
2 1/3 du dx 3u
ntegr
an
e2
a
´
una
dx
l
i
nea
+ 32 u = l
.
3
= e2x
4
i
6
=
x
eni
2(u1/3 )2 x
=
x
ul
du
1.
1/3
3 2
e.
on
por
1 2/3 3u
.
3
3
3
3 3 2x u = e2x e 2 x du dx + e 2 2 3 d x 2 u] dx [e 3 d 2 x u] dx [e
´
=
3
= 32 e 2 x 3 32 x dx 2e
´
3 2x
+c
3 2x
e u=e +c 3 u = 1 + ce− 2 x S
u
s
t
i
t
u
i
m
u = y3/2
os
3
y 3/2 = 1 + ce− 2 x A
h
ora
ap
l
i
cam
os
l
as
cond
i
ci
on
es
i
S
ni
ci
al
ol
u
es
ci
on
genral
.
y(0) = 4
.
3
43/2 = 1 + ce− 2 0 8 1=c c=7
−
Sus
t
ut
ui
m
os
e
l
va
l
or
de
9
c
e
n
l
a
e
.
y +
u = y 1−n ento
ces
c
3 = 1 + 7 e− 2 x
y 3/2
uac
i
on
S
ol
2 y= x
ge
u
ner
ci
on
al
p
.
art
i
cu
−2xy2
n =2
;done
:
u = y 1−2 u = y −1 u−1 = y ;
;
−u−2 ddux = ddxy s
ust
i
t
ui
m
os
en
l
a
ec
uaci
on.
−u−2 ddux + x2 u−1 = −2x(u−1)2 m
ul
t
i
pl
i
cam
os
p
−u2
or
du dx es
t
ob
a
t
e
s
un
enm
a
os
e
elf
cua
i
act
on
or
l
i
nte
i
− x2 u = 2x p(x) = − x2
nealco
gr
ante
e−2
.
´
1 x dx
x−2 ddux
= elog x
−
ntegr
am
= x−2
− x−2 x2 u = x−22x
d 2 dx [x u] i
2
−
os.
5
= 2x−1
l
ar.
d 2 dx [x u]
−
´
=
2x−1 dx + c
´
x−2 u = 2 log x + c u = 2x2 log x + cx2 sut
i
y
l
t
a
u
s
i
m
ol
u = y−1
os
uci
ón
es
ent
onces
:
1 2x2 log x+cx2
y= 1
0
,
y + xy = xy −1/2
n=
sea.
−1/2 u = y 1−n u = y 1−(−1/2) u = y 3/2 y = u2/3 ;
;
dy dx s
ust
i
t
ui
m
os
en
l
a
ec
uaci
+ xu2/3 = x(u2/3 )−1/2
−
ul
t
i
pl
i
cam
du dx F
act
or
i
os
p
2 1/3 3u
or
+ 32 xu = 32 x
ntegr
an
t
= 23 u−1/3
on.
2 u 1/3 3 m
;
que
e
s
una
e
c
uac
i
on
l
3
3
e4x
´
2
du dx
´
xdx 3
2
+ e4x
2
3
3
3 2 xu
2
d 34 x2 u dx e
=
3
3 2
2
´ 3
2
3
xe 4 x dx + c 2
3
i
m
os
3 2x
= 32 xe 4 x dx + c
2
u = 1 + ce− 4 x u
2
= e4x
d 34 x2 u dx e
3
t
l
= e4x
e4x u = e4x + c
i
nea
e:
e2
sut
i
u = y3/2 y 3/2 = 1 + ce− 34 x2
6
c
on
p(x) = 32 x
1.
2
E
c
uac
i
ones
e
C
om
pr
=
∂N ∂x
oba
m
t
as
y
r
os
que
l
a
e
c
uac
i
on
s
∂M ∂y s
on
i
A
ho
gual
es
r
a
t
,
om
e
du
c
i
por
l
am
o
os
t
ant
o
una
f
l
unc
a
e
i
e
a
e
=0
cua
i
xac
t
∂N ∂x
;
nt
eg
r
am
os
r
es
pec
t
o
a
x,
y
l
∂M ∂x
´
on
e
a
c
ons
t
´
s
=2
ant
t
a
f
unci
on
l
a
der
i
va
m
gual
am
os
c
on
N
(
x,
e
a
s
t
e
xac
t
as
.
o
e
s
s
i
s
e
c
um
pl
e
ct
nt
eg
r
am
os
r
e
s
pec
o
a
os
c
on
r
e
de
−
i
´
nt
eg
s
pec
t
o
r
ac
i
on
s
e
r
a
un
a
dx + g(y)
de
y
.
.
(1)
.
= g (y)
y
´ ´ ´ g (y) = 3 ydy + dy + c g(y) = 32 y2 + y + c s
ust
i
t
ui
m
os
l
a
f
un
ci
on
en
(
1)
.
− x + 32 y2 + y = c
x2 es
2
t
a
e
s
un
a
s
ol
uci
on
e
n
f
or
c
ondi
c
i
on
−1
y)
t
a
a.
g (y) = 3y + 1 i
l
=0
.
∂f ∂y i
s
− x + g(y)
f (x, y) = x s
a,
exa
e
xdx 2
E
e
fx (x, y) = M (x, y)
on
fx (x, y) = 2x i
bl
1.(2x − 1)dx + (3y + 1)dy = 0 − 1; N (x, y) = 3y + 1
M (x, y) = 2x ∂M ∂y
xac
m
a
i
m
pl
i
ci
t
a
de
l
a
ec
uaci
on.
.
(seny
− ysenx)dx + (cosx + xcosy − y)dy = 0 M (x, y) = seny − ysenx N (x, y) = cosx + xcosy − y ∂M ∂y = cosy − senx ∂N ∂x = −senx + cosy ;
7
f
un
ci
on
g(y)
∂M ∂y t
i
o
nt
∂N ∂x
=
m
a
e
m
o
gr
por
l
o
t
ant
o
e
s
una
fx (x, y) = seny
s
am
os
c
on
r
e
s
pec
t
´
o
e
c
uac
i
on
e
xac
t
a.
− ysenx
a
x
´
fx (x, y)dx = (seny
f (x, y) = xseny
− ysenx)dx
y( cosx) + g(y)...
(
1
)
− − der
i
va
m
os
e
s
t
a
e
c
uac
i
on
r
e
s
pec
t
o
a
y
,
e
i
gual
am
os
c
on
N
(
x,
y)
fy (x, y) = cosx + xcosy + g (y) = cosx + xcosy − y g (y) = i
nt
eg
r
am
os
r
e
s
pec
t
o
de
−y
y
´ ´ g (y) = − ydy + c
− 12 y2 + c
g(y) = s
ust
i
t
ui
m
os
en
(
1)
f (x, y) = xseny + ycosx nos
queda
l
a
s
ol
uci
on
i
m
pl
i
ci
t
a.
xseny + ycosx 3
− 12 y2
− 12 y2 = c
.
(3x2 y + ey )dx =
−(x3 + xey − 2y)dy
M (x, y) = 3x2 y + ey N (x, y) = x3 + xey
− 2y
;
My (x, y) = 3x2 + ey Nx (x, y) = 3x2 + ey My (x, y) = Nx (x, y) fx (x, y) I
nte
gr
const
an
am
t
e
ent
os
d
e
i
ntegr
aci
on
c
on
o
ce
n
r
e
s
s
es
un
pe
c
t
a
o
ec
de
ua
ci
x,
on
y
di
o
bt
f
e
er
enci
ne
al
m
o
s
exact
una
.
f (x, y) = (3x2 y + ey )dx
´
f (x, y) = x3 y + xey + g(y) .
D
e
r
i
v
am
os
c
on
r
e
s
pe
c
t
o
fy (x, y) =
de
y
x3
(
1)
e
i
gua
l
am
y
os
c
.
on
.
(1)
N
(
x,
y)
+ xe + g (y) = x3 + xey − 2y
g (y) =
8
−2y
a.
f
unc
i
o
n
g(y)
d
e
I
nt
e
gr
am
os
r
es
pec
t
o
de
y
−2 ´ ydy + c g(y) = −y 2 + c
g(y) =
s
ust
i
t
ui
m
os
en
(
1)
x3 y + xey 4
− y2 = c
.
.
.
sol
u
ci
on
i
m
p
l
i
ci
ta.
.
− 2y2)dx + (3x2 − 4xy)dy = 0 My (x, y) = 6x − 4y Nx (x, y) = 6x − 4y (6xy
,
l
a
i
nte
ec
uaci
on
gr
am
es
exac
t
a.
fx (x, y)
os
r
e
s
pec
t
o
a
x.
´
f (x, y) = (6xy f (x, y) = 3x2 y der
i
va
m
os
r
e
s
pct
o
de
− 2xy2 + g(y) .
gual
am
os
c
on
N
(
x,
nt
eg
r
am
os
r
e
s
pec
)
− 4xy + g(y)
y)
3x2 i
.(1
y
fy (x, y) = 3x2 i
− 2y2)dx
t
− 4xy + g(y) = 3x2 − 4xy o
de
g (y) = 0
y
g(y) = c s
ut
i
t
ui
m
os
e
n
l
a
ec
uaci
on
(
1)
3x2 y 5
− 2xy2 = c
.
− 2xy3 + 4x + 6)dx + (2x − 3x2y2 − 1)dy = 0 y(−1) = 0 My = 2 − 6xy 2 = NX
(2y con
l
U
i
a
condi
ci
na
nte
v
gr
e
am
z
on
c
os
om
pr
ob
ad
a
fx (x, y)
que
r
e
s
s
pec
e
t
a
o
e
a
´
xa
f (x, y) = (2y f (x, y) = 2xy
c
t
a.
x
− 2xy3 + 4x + 6)dx
− 3x2y3 + 2x2 + 6x + g(y) .
9
.(1
)
der
i
va
m
os
r
e
s
pec
t
o
a
y:
− 3x2y2 + g(y)
fx (x, y) = 2x i
gual
am
o
co
N (x, y)
n
3x2 y 2 + g (y) = 2x
2x i
ntegr
am
−
os:
3x2 y 2
g(y) =
s
ust
i
t
ui
m
os
en
(
2xy
onces
l
a
s
.
uci
on
part
i
cul
r
−
ar
.
.
sol
u
ci
on
i
m
p
l
i
ci
ta.
a
al
cas
x2 y3
−
2xy 6
−y + c
− x2y3 + 2x2 + 6x − y = c y(−1) = 0 2(−1)2 + 6(−1) = c c = −4
ol
1
−
1)
pa
ent
1 g (y) =
−
o
y(
+ 2x2
-
1)
=
+ 6x
0
es
:
− y = −4
.
( xy sin x + 2y cos x)dx + 2x cos xdy = 0;
−
U
s
e
el
f
act
or
i
nte
gr
µ(x, y) = xy
ante
−x sin x + 2 cos x Nx (x, y) = −2x sin x + 2 cos x NX = My My (x, y) =
l
l
o
t
a
ec
ant
uaci
o
on
pr
e
oce
s
no
exa
dem
os
ct
a
m
a,en
ul
e
t
i
pl
i
s
ca
t
e
r
t
ej
e
m
pl
oda
l
o
a
e
s
e
cua
nos
di
i
on
o
elf
ac
por
t
elf
ac
or
t
i
nt
or
i
eg
r
nt
eg
ant
r
e
ant
,
por
e.
xy( xy sin x + 2y cos x)dx + xy(2x cos x)dy = 0
−
( x2 y2 sin x + 2xy 2 cos x)dx + (2x2 y cos x)dy = 0
−
c
om
pr
oba
m
os
que
e
s
t
a
e
c
uac
i
on
s
My (x, y) =
−
e
a
e
xac
t
por
l
o
t
ant
o
e
s
t
a
e
c
uac
i
on
e
1
0
s
+ 4xy cos x
− 2x2y sin x
NX (x, y) = 4xy cos x MY = NX
a.
2yx 2 sin x
e
xac
t
a
y
l
a
r
e
s
ol
ve
m
os
c
om
o
t
al
.
fx (x, y) = i
nt
eg
r
am
os
r
e
s
pec
t
o
a
−x2y2 sin x + 2xy2 cos x
x:
f (x, y) = ( x2 y 2 sin x + 2xy 2 cos x)dx
´
−
f (x, y) = x2 y 2 cos x + g(y) .
der
i
va
m
os
r
e
s
pec
t
o
a
.(1
)
y:
fy (x, y) = 2x2 y cos x + g (y)
i
gual
am
os
c
Nx
on
2x2 y cos x + g (y) = 2x2 y cos x
g (y) = 0 i
nt
eg
r
am
os
r
e
s
pec
t
o
a
y:
g(y) = c s
ust
i
t
ui
m
os
en
(
1)
f (x, y) = x2 y 2 cos x + c
2
2.
E
c
1
ua
E
c
a
1
.
c
u
i
ac
pr
i
i
o
ne
s
de
ones
m
o
di
e
r
or
f
de
er
r
de
n
enci
al
s
es
up e
r
de
or
n.
y = 2x2
I
nt
e
´
gr
am
os
y =2
am
bos
l
ados
de
l
a
e
c
uac
i
x2 dx + c
´
y = 23 x3 + c1
V
ol
ve
m
os
a
i
nt
S
ol
u
´
2 3
ci
on
(x3
eg
r
ar
:
y = + + c2 y = ( 23 )( 14 )x4 + xc1 + c2
´
c1 )dx
:
y = 16 x4 + c1 x + c2 1
1
on:
i
d
o
en
r
s
up
er
i
or
r
edu
c
i
b
l
es
2
I
nt
´
y
.
e
gr
= sen(kx)
am
os
am
bos
l
ados
de
l
= ´ sen(kx)dx + c1 y = −kcos(kx) + c1 ´ ´
a
e
c
uac
i
on:
y
y =
k
cos(kx)dx + c1
´
dx + c2
−k−2sen(kx) + xc1 + c2 ´ ´ ´ ´ y = −k 2 sen(kx)dx + c1 xdx + c2 dx + c3 y =
y = k 3 cos(kx) + 12 c1 x2 + c2 x + c3 3
I
y
.
= x1
ntegr
an
´
y
do
:
= ´ x1 dx + c1
y = log x + c1
´
y =
´
´
log xdx + c1
´
dx + c2
− x + c1x´ + c2 ´ ´ ´ y = x log xdx − xdx + c1 xdx + c2 dx + c3 y = x2 (log x − 12 ) − 12 x2 + c1 12 x2 + c2 x + c3 y = x + sin x y = x log x 2
4
I
.
ntegr
an
´
y =
do
´
y = 12 x2
´
y= 5
R
´
y
.
1 2 1 3 6x
y =
es
ol
:
xdx +
−
´
´
sin xdx + c1
cos x + c1
x2 dx
− ´ cos xdx + c1 ´ dx + c2
− sin x + c1x + c2 = x sin x, y(0) = 0 y(0) = 0 y(0) = 2 vem
os
l
a
ec
ua
ci
on
di
f
er
enci
al
i
nte
gr
and
o
t
r
es
vec
es
:
= ´ x sin xdx + c! y = sin x − x cos x + c1 ´ ´ ´ ´ y = sin xdx − x cos xdx + c1 dx + c2 y = − cos x − (cos x + x sin x) + c1 x + c2 ´ ´ ´ ´ ´ ´ y = − cos xdx − cos xdx − x sin xdx + c1 xdx + c2 dx + c3 y = − sin x − sin x − (−x cos x + sin x) + 12 c1 x2 + c2 x + c3 1 2 y = −3sin x + x cos x + 2 c1 x + c2 x + c3 y
1
2
2
.
2
R
1
e
duc
i
bl
e
s
a
pr
i
m
e
r
o
r
de
n
xy + y = 0
.
D
eni
e
nd
p(x) =
o:
dy dx
dp dx
d2 y dx2
=
xp + p = 0
nos
que
da
sepa
una
rab
1 x dx
l
e
c
uac
i
on
l
i
ne
al
hom
og
e
ne
a
de
or
de
n
1
de
v
ar
i
abl
e
s
es.
− p1 dp ´ = − 1p dp + c1 log x = − log p + log c1 =
1 x dx
´
log x = log( cp1 ) A
pl
i
c
ando
hac
e
xponec
i
al
am
bos
l
ados
de
l
a
e
c
uac
i
on.
c1 p
x= e
m
dy dx
p(x) =
os
c! dy/dx
x=
x = c1 dx dy i
ntegrad
o:
1 x dx
1
= c1 dy + c2 log x = c11 y + c2
´
´
y = c1 log x + c2 val
2
e
p
ni
m
os
p(x) = (x D
i
ti
vo
.
L
di
− −
m
a
c
ons
t
an
t
e
de
i
gr
ac
=
0
:
dy dx
dp dx
d2 y dx2
=
os
ent
r
(x
e
− x−1 1 p = 0
queda
dp dx
− 1)
una
e
c
uac
i
on
l
i
nea
l
hom
oge
− x−1 1 p = 0
dp = x 1 1 p dx 1 1 p dp = x 1 dx
− −
ntegrad
´
e
− x−1 1 p = 0
nos
i
nt
.
− 1)p − p = 0 vi
x 1 x 1p
p
osi
− 1)y − y
(x
.
D
or
o:
1 p dp
=
´
1 x 1 dx
−
+ c1
1
3
nea
.
i
on
c
on
vi
e
ne
que
t
om
e
log(p) = log(x
− 1) + log(c1) − 1)]
log(p) = log[c1 (x p = c1 (x h
aci
end
dy dx
− 1) p=
o
= c1 (x
1)
− 1)dx
dy = c1 (x i
ntegrad
o:
´
´
− 1)dx + c2 − x + c2
dy = c1 (x
y = c1 12 x2 3
2.
.
3
E
1
dy dx
c
uac
i
y +y
.
R
es
ol
ones
l
i
nea
l
e
s
hom
oge
nea
s
.
− 2y = 0
vem
os
l
a
ec
ua
ci
on
car
act
er
i
s
t
i
ca
as
oci
ad
.
m2 + m
−2= 0 (m + 2)(m − 1) = 0 m1 = −2 m2 = 1 Sup
one
m
os
una
s
ol
uci
y = emx
on
y1 = e−2x y2 = ex y(x) = c1 e−2x + c2 ex 2
.
y E
cua
ci
(m
on
cart
er
i
st
i
ca
− 1)2 = 0
asoi
ad
− 2y + y = 0 m2 − 2m + 1 = 0 a
m1,2 = 1 sol
u
ci
y = emx
on
x
y1 = e
y2 = y1
´
´ x
y2 = e
´
e
p(y)dy
y12
dx
e2x e2x dx
y2 = ex x sol
u
ci
on
.
y(x) = c1 ex + c2 xex 3
.
4y
− 8y + 5y = 0 1
4
E
cua
ci
4m2
on
cart
er
m1,2 = 8±
u
st
i
ca.
ci
on
64 80 8
−
± 12 i
m1,2 = 1 sol
i
− 8m√+ 5 = 0 .
y = c1 ex ei 12 x + c2 ex e−i 12 x 1
1
y = ex (c1 ei 2 x + c2 e−i 2 x ) y = ex (c1 cos 12 x + c2 sen 12 x) 4
3y
.
E
− 2y − 8y = 0
cua
ci
on
cart
er
i
st
i
ca:
3m2
− 2y − 8 = 0 (3m + 4)(m − 2) m1 = 2
− 43
m2 = Sol
uci
on
pr
opues
t
a
de
l
a
f
or
m
y = emx
a,
2x
y1 = e
−e−
y2 = S
ol
u
ci
on
4 3x
:
y(x) = c1 e2x + c2 e 43 x 5
yv
.
E
− 10y + 9y = 0
cua
ci
on
cart
er
i
st
i
ca.
m5
− 10m3 + 9m = 0 m(m4 − 10m2 + 9) = 0 m1 = 0 (m2 − 9)(m2 − 1) m2,3 = ±3 m4,5 = ±1 E
nt
onces
t
enm
os
l
as
s
ol
uci
ones
:
y1 = e0 = 1 y2 = e3x y3 = e−3x y4 = ex y5 = e−x S
ol
u
ci
on
:
y(x) = c1 + c2 e3x + c3 e−3x + c4 ex + c5 ex 6
E
y + 4y + 3y = 0 y(0) = 2 y (0) =
.
cua
ci
on
cart
er
i
st
i
ca.
m2 + 4m + 3 = 0
1
5
−3
√
m1,2 = −4±2 −36 m1,2 = S
ol
u
ci
−2 ± 3i
on
:
y(x) = e−2x (c1 cos3 x + c2 sin3 x) y (x) = e−2x ( 3c1 sin3 x + 3c2 cos3 x) R
P
ar
es
ol
ve
r
em
−
os
par
a
l
os
c
as
y(0) = 2
os
2e−2x (c1 cos3 x + c2 sin3 x)
−y(0) = −3
y
y(0) = 2
a
p
art
i
cul
arm
ent.
2 = e0 (c1 cos 0 + c2 sin 0) 2 = c1 y (0) =
−3 −3 = e0(−3c1 sin 0 + 3 c2 cos0) − 2e0(c1 cos 0 + c2 sin 0) −3 = 3c2 − 2c1 −3 = 3c2 − 2(2) −3 + 4 = 3 c2 P
ar
a
1 3
c2 = P
or
l
o
t
ant
o
l
a
s
ol
uci
on
par
d4 y 7
a
1 3
y(x) = e−2x (2cos3 x +
e
l
c
as
o
e
n
ge
ner
ales
:
sin3 x)
d4 y
.
− 7 dx − 18y = 0 m4 − 7m2 − 18 = 0 dx4
E
2
2.
.
cua
4
4.
E
c
1
on
cart
ua
C
1
2
ci
c
er
i
o
oeci
ne
i
s
ent
st
i
ca:
no
es
i
ho
nd
m
et
o
er
m
g
i
e
ne
nad
a
os
s
de
s
e
g
undo
o
r
de
n
.
.
y + 3y + 2y = 6
R
e
s
ol
ve
m
os
l
a
e
c
uac
i
on
hom
oge
nea
as
oc
i
ad
yh = y + 3y + 2y = 0 E
cua
ci
on
cart
er
i
st
i
ca:
m2 + 3m + 2 = 0 (m
− 1)(m − 2)
m1
= 1 m2 = 2 yh = c1 ex + c2 e2x A
hor
a
p
a
r
rt
e
i
s
cu
ol
ve
l
a
m
os
l
a
par
t
e
no
hom
og
r.
1
6
e
na
s
uponi
e
ndo
una
s
ol
uc
i
on
e
n
e
s
t
e
c
as
una
s
o
ol
l
uc
a
par
i
t
on
e
no
de
l
hom
a
f
og
or
m
e
ne
a
e
s
6,l
o
que
nos
s
ugi
e
r
e
us
e
m
os
m
os
A
a
yp = A y
p = 0 p = 0
y Su
s
t
i
t
ui
m
os
en
l
a
ec
ua
ci
on
or
i
gi
nal
.
0 + 3(0) + 2A = 6 A=3 E
nt
onces
l
a
s
ol
uci
on
y(x) = yh + yp
es
y(x) = c1 ex + c2 e2x + 3 2
y + y = sin x
.
R
e
s
ol
ve
m
os
pr
i
m
e
r
l
a
e
c
uac
i
on
hom
oge
nea
as
oc
i
ad.
y +y = 0
L
a
ec
uaci
on
car
act
er
i
s
t
i
ca
de
es
t
a
ec
uaci
on
es
.
m2 + 1 = 0
−1 m1,2 = ±√−1 m1,2 = α ± βi m1,2 = ±i m2 =
α=0
done
β=1
y
yh = c1 eαx cos βx + c2 eαx sin βx yh = c1 cos x + c2 sin x A
hor
a
bus
c
una
s
obs
e
ol
r
v
asoci
a
de
os
ar
e
s
s
de
que
l
e
s
t
a
a
f
ol
uci
or
ya
t
e
ca
s
os
o,d
p
os
eb
i
t
i
m
vos
s
,
t
i
c
ul
ar
,
par
sin x
a
,s
una
s
ent
ol
uc
onces
os
qu
par
A sin x + B cos x
a
e
er
on
m
y +y = 0
a
ent
una
on
ad
par
am
uci
m
e
i
s
ul
el
t
i
m
i
i
pl
i
na
on
de
egun
ca
r
l
l
l
a
a
a
r
pl
i
ca
n
e
e
c
egl
xn
por
du
i
m
uac
on
n
pode
hom
m
n
opne
go
on
de
nde
i
pr
r
i
a
do
nos
ba
ul
e
og
t
s
i
e
pl
lnum
i
e
ca
ne
a
i
e
on
r
o
.
yp = Ax sin x + Bx cos x y y
p = A sin x + Ax cos x + B cos x − Bx sin x p = A cos x + A cos x − Ax sin x − B sin x − Bx cos x − B sin x = 2A cos x − 2B sin x − Ax sin x − Bx cos x
Su
s
t
i
t
ui
m
os
en
l
a
ec
uaci
on
or
i
gi
nal
− 2B sin x − Ax sin x − Bx cos x + Ax sin x + Bx cos x = sin x 2A cos x − 2B sin x = sin x 2A cos x
ent
2A = 0 2B = 1
− S
u
s
ti
yp =
t
u
yen
on
ces
ent
on
d
ces
A=0 B=
− 12
o
− 12 x cos x 1
7
y(x) = yh + y
p
− 12 x cos x − 10y + 25y = 30x + 3
y(x) = c1 cos x + c2 sin x 3
R
e
s
y
.
ol
ve
m
m2
os
l
a
e
c
uac
ui
on
hom
oge
nea
as
oc
i
ad.
10m + 25 = 0
−
m1,2 = 5 yh = c1 e5x + c2 xe5x L
a
s
ol
uci
on
part
i
cul
ar
proues
t
a
30x + 3
par
e
Ax + B
s
yp = Ax + B
p = A p = 0
y y s
ust
i
t
ui
m
os
en
l
a
ec
uaci
on
−10(A) + 25(Ax + B) = 30 x + 3 25A = 30
.
.
.
(1)
en
ton
− 10A = 3 − 10( 65 ) = 3
25B
.
25B
.(2
6 5
A=
ces
)
25B = 3 + 12 3 5 6 5x
B= yp =
+
3 5
y(x) = yh + yp y(x) = c1 e5x + c2 xe5x + 65 x + 4
R
1 4y
.
e
s
ol
3 5
2
+ y + y = x − 2x ve
m
os
l
a
e
c
uac
i
on
hom
oge
nea
as
oc
i
ad.
1 4y + y + y = 0 1 2 4m + m + 1 = 0
m1,2 =
−2
yh = c1 e−2x + c2 xe−2x A
hor
a
s
x2
upone
m
os
una
s
ol
uc
i
on
par
t
i
c
ul
ar
par
− 2x
yp = Ax2 + Bx + C yp = 2Ax + B yp = 2A Su
s
t
i
t
ui
m
os
en
l
a
ec
ua
ci
on
or
i
gi
nal
.
1 2 4 (2A) + 2Ax + B + Ax + Bx + C = 1 2 2 2 A + B + Ax + 2Ax + Bx + C = x
1
8
x2
− 2x − 2x
a
e
l
c
as
o
de
f (x) =
A=1 2A + B = 2 B =2 1 2A 1 2A
−2 = 0
+B+C = 0 +C =0
− 12 A = − 12 yp = x2 − 12 C=
y(x) = yh + yp
− 12
y(x) = c1 e−2x + c2 xe−2x + x2 5
y + 3y =
.
Se
r
y
e
s
+
uel
3
ve
y
=
l
−
a
48x2 e3x
par
t
e
hom
oge
nea
.
0
m2 + 3 = 0
√−3 m = √3i √ 1,2 √ = c cos 3x + c sen 3x
m1,2 = yh s
1
up
2
onem
os
e3x (Ax2
yp = y
un
a
s
ol
uci
on
par
t
i
cul
ar
par
+ Bx + C )
a
−48x2e3x
= 3e3x (Ax2 + Bx + C ) + e3x (2Ax + B)
p
p 3=x 9e3x (Ax23+x Bx2+ C ) + 3e3x (2Ax3x+ B) + 3e3x (2Ax3x + B) +
y
e (2A) = 9e (Ax + Bx + C ) + 3e (4Ax + 2B) + e (2A)
Su
s
i
t
ui
m
os
en
l
a
e
cua
i
on.
9e3x (Ax2 + Bx + C ) + 3e3x (4Ax + 2B) + e3x (2A) + 9e3x (Ax2 + Bx + C ) + 3e3x (2Ax + B) = 48x2 e3x
−
9e3x Ax2 + 9e3x Bx + 9e3x C + 12e3x Ax + 6e3x B + 2e3x A + 9e3x Ax2 + 9e3x Bx + 9e3x C + 6e3x Ax + 3e3x B = 48x2 e3x 9A + 9A = 18A = A=
−48 − 83
−
−48
B =0
C=0
− y = −3 y2 m −m= 0 m(m − 1) = 0 6
.
y -
y
=
0
m1 = 0 m2 = 1
1
9
yh = c1 e0x + c2 ex = c1 + c2 ex E
n
e
s
e
ent
t
e
c
as
c1
s
on
ce
i
o
podem
gua
s
os
l
p
or
l
ve
r
c
l
ar
am
e
nt
e
que
e
xi
s
t
e
ya
una
s
ol
uc
i
−3
con
a
r
egl
a
de
m
ul
t
i
pl
i
ci
da
d.
l
a
s
ol
uci
on
prou
es
t
on
que
yp = Ax
a
yp = Ax y y
p = A p = 0
Su
s
0
t
i
t
uyend
o
en
− A = −3
l
a
ec
ento
uaci
ces
on.
A=3
,
yp = 3x
y(x) = yh + yp y(x) = c1 + c2 ex + 3x 7
y
.
E
c
− 6y = 3 − cosx
uac
i
on
hom
oge
nea
as
oc
i
yh = y
ad
m3
− 6m2 = 0 2 m (m − 6) = 0
− 6y = 0
m1,2 = 0 m3 = 6 yh = c1 + c2 x + c3 e6x L
a
s
ol
uci
on
par
t
i
c
ul
ar
Bcosx + Csenx s
t
an
m
t
ul
t
e
i
,
e
pl
i
Ax2
yp =
nt
onc
ci
e
da
s
pr
i
n
s
l
nos
e
a
opues
m
t
bar
m
ul
a
par
go
t
i
e
pl
i
c
3
a
n
l
am
a
os
s
por
ol
uci
− cosxy on
x
de
e
ac
yp 1 = A yp 2 =
s
p1 ue
s
e
r
r
e
do
yp1 = Ax2
queda.
+ Bcosx + Csenx
y
p = 2Ax − Bsenx + Ccosx yp = 2A − Bcosx − Csenx yp = Bsenx − Ccosx Su
s
i
t
uyen
do
en
l
a
ec
ua
ci
on
or
i
gi
na
l
.
Bsenx
− Ccosx − 12A + 6Bcosx + 6Csenx = 3 − Cosx −12A = 3 A = − 14 6B − C = 1 ;
.
.(1
6C + B = 0 .
I
gu
a
l
B= C= yp =
an
do
6 37 1 37 1 3 x 2
y(x) =
1
)
.(2
y
)
2
6 1 cosx + 37 senx 37 6 c1 + c2 x + c3 e x 14 x2
+
−
6 1 37 cosx + 37 senx
+
2
0
pi
t
a
e
l
l
a
a
co
l
e
y
n-
de
9
y + 2y + y = senx + 3cos2x
.
yh = y + 2y + y = 0 m2 + 2m + 1 = 0 (m + 1)2 = 0 m1,2 =
1
yh = c1 ex + c2 xex
−
S
ol
u
ci
on
p
art
i
cul
ar
yp = Acosx + Bsenx + Ccos2x + Dsen2x yp =
−Asenx + Bcosx − 2Csen2x + 2Dcos2x −Acosx − Bsenx − 4Ccos2x − 4Dsen2x
yp = s
u
st
i
tuyen
d
o.
−Acosx−Bsenx−4Ccos2x−4Dsen2x−2Asenx+2Bcosx−4Csen2x+ 4Dcos2x + Acosx + Bsenx + Ccos2x + Dsen2x = senx + 3cos2x
−3Ccos2x − 3Dsen2x − 2Asenx + 2Bcosx − 4Csen2x + 4Dcos2x = senx + 3Cos2x
−3C + 4D = 3 −3D − 4C = 0 .
.(1
)
.
.(2
)
9 25
C=
25 D = 12 2A = 1
−
A=
;
2B = 0 B = 0
− 12
;
y(x) = c1 ex + c2 xex
2.
1
5
.
V
a
r
i
a
c
i
on
de
pa
− 12 cosx + 259 cos2x + 1225 sen2x
r
a
m
e
t
r
o.
y + y = sec x R
P
e
ar
s
ol
a
ve
l
m
a
os
ec
l
uaci
a
par
on
t
ho
e
hom
m
og
e
ne
ogena
a
as
de
l
oci
a
ad
e
a,
c
r
uac
es
i
ol
on
vem
e
os
s
t
a
l
e
a
m2 + 1 = 0 m2 = m1,2 m1,2 = α
√ = −1
± βi
;
−1 m1,2 =
±i
α=0β=1
;done
2
1
yh = y + y = 0
s
ec
uaci
on
car
act
er
i
s
t
i
ca.
yh = c1 cosx + c2 senx A
hor
a
i
dent
i
cam
y1 = cosx
os
y2 = senx
y
y
l
as
der
i
vam
os
.
y1 = cosx y 2 = senx y1 = A
c
ont
i
nuaci
on
y1 y1
W =
ca
y2 = y2
l
cul
am
os
elW
cosx
senx y 2 = cosx
−
r
ons
ki
senx = [(cosx)(cosx)] cosx 2 cos x + sen2 x = 1
−senx
−
−
y1 y1
W1 W
u1 =
− [(senx)(−senx)] =
0 y2 0 senx = = [(0)(cosx)] [(senx)(secx)] = f (x) y2 secx cosx senxsecx = senx cosx = tanx
W1 =
W2 =
ano:
0 = f (x)
= −tanx = 1
−
−
cosx
0 = [(cosx)(secx) secx cosxsecx = cosx cosx = 1
−senx
−tanx u2 =
u1 =
;
W2 W
=
1 1
− (0)(−senx)] =
− ´ tanxdx = −[−ln(cosx)] = ln(cosx) = 1 u2 = ;
´
dx = x
yp = u1 y1 + u2 y2 yp = ln(cosx)cosx + xsenx y(x) = yh + yp y(x) = c1 cosx + c2 senxi + cosxln(cosx) + xsenx 2
R
e
s
ol
ve
m
os
.
y + y = senx
yh = y + y = 0 m2 + 1 = 0 m2 = m1,2
D
on
de
√ = ± −1
−1 ;
m1,2 =
±i
:
α=0
β =1
y
yh = eαx (c1 cosβx + c2 senβx) yh = e0x (c1 cosβx + c2 senβx)
2
2
yh = c1 cosx + c2 senx D
e
ni
m
y1 y2
os
,
y1 = cosx y1 = ;
−senx
y2 = senx y2 = cosx ;
C
al
c
ul
am
os
e
l
W
r
ons
ki
ano.
cosx senx = cos2 x + sen2 x = 1 senx cosx
W =
−
0 senx = senx cosx
W1 =
cosx
W2 = A
hor
a
c
al
c
ul
am
,
0 = senxcosx senx
−senx
u1 u2
os
−sen2x
.
2
− sen1 x = −sen2x ´ u1 = − sen2 xdx = x2 − 14 sen2x u1 =
senxcosx 1
u2 = u2 =
= senxcosx
senxcosxdx = 12 sen2 x
´
yp = u1 y1 + u2 y2 = ( x2
− 14 sen2x)cosx + 12 sen2x(senx) yp = 12 xcosx − 14 cosxsen2x + 12 sen3 x y(x) = yp + yh
y(x) = c1 cosx + c2 senx + 12 xcosx 3
E
c
E
s
uac
i
t
a
e
on
hom
cua
oge
i
on
nea
t
i
e
as
ne
s
oc
ol
i
uci
− 14 cosxsen2x + 12 sen3x
y + y = cos2 x
.
yh = y + y = 0
ad
on
de
l
a
f
or
m
a:
yh = c1 cosx + c2 senx D
e
ni
m
y1 y2
os
,
y1 = cosx y1 = ;
−senx
y2 = senx y2 = cosx ;
C
al
cul
am
os
l
os
W
r
ons
ki
anos
:
2
3
cosx
W =
senx = cos2 x + sen2 x = 1 cosx
−senx
0 senx = cos2 x cosx
W1 =
cosx W2 = D
e
ni
m
0 cos 2 x = cos3 x
−senx
u1 u2
os
,
−senxcos2x = −senxcos2x
u1 = u1 =
−
1
senxcos2 xdx =
´
u2 = u2 =
− −
cos3 x 3
cos4 x + sen2 x 3
y(x) = c1 cosx + c2 senx +
uac
i
on
hom
oge
nea
as
oc
i
.
y
ad
yh = c1 ex + c2 e−x e
ni
m
y1 y2
os
,
y1 = ex y1 = ex ;
y2 = e−x y2 = −e−x ;
C
al
cul
am
os
l
os
W
r
ons
4
ki
anos
:
2
4
−
cos4 x + sen2 x 3
;
D
sen3 x
− sen3 x
− y = cosh x y − y = 0 m2 − 1 = 0 √ m2 = 1 m1,2 = ± 1 = ±1 4
c
3
− sen3 x
(cosx) + senx
yp =
cos3 x cos3 x = 3 3
cos3 x = cos3 x 1
cos3 xdx = senx
´
yp = u1 y1 + u2 y2 =
E
−senxcos2x
3
(senx)
4
− sen3 x
ex ex
W =
e−x
=
−e−x
e−x
0 coshx
W1 =
−e−x (ex ) − ex (e−x ) = −1 − 1 = −2
ex W2 = ex C
al
cul
am
u1
os
−e−x (coshx) = −e−x coshx
=
−e−x
0 coshx = ex coshx
u2
y
x u1 = −e −coshx = 12 e−x coshx 2 −
1 2
u1 =
´ −x e coshxdx = 18 e−2x (2e2x x − 1) ex coshx = 2
u2 =
−
− 12 ex coshx 2x
u2 =
− 12 ´ ex coshxdx = − 12 [ x2 + e4
yp = ex [ 18 e−2x (2e2x x y = 1e p
x
2x
− 1)] + ( −e−x )(− x4 − e8
(2e2x x
−
8
]
1) +
xe−x
−
4
+
)
ex 8
xe−x ex y(x) = c1 e + c2 e−x + 18 e−x (2e2x x − 1) + + x
4
4
E
c
uac
i
on
hom
oge
nea
as
1 y + 3y + 2y = 1 + ex
.
oc
i
ad
yh = y + 3y + 2y = 0 m2 + 3m + 2 = 0
(m + 2)(m + 1) = 0 m1 =
−2 m2 = −1
yh = c1 e−2x + c2 e−x D
e
ni
m
y1 y2
os
,
.
y1 = e−2x y1 =
−2e−2x y2 = −e−x
;
y2 = e−x C
al
cul
am
os
W
r
ons
ki
anos
;
:
2
5
8
e−2x
W =
e−x
= (e−2x )( e−x )
−
−2e−2x −e−x
W1 = W2 = C
al
cul
am
os
e−x
0 1 1+ex
− (e−x )(−2e−2x ) = −e−3x + 2e−3x =
e−3x
=
−e−x
e−2x
0
1 1+ex
−2e−2x
−x
− 1 e+ ex −2x = e x 1+e
u1 ,u2
e−x x 1 u1 = − e+3xe = −
u1 =
1 e−x = −2x = (e−3x )(1 + ex ) e (1 + ex )
− e−2x 1+ e−x
− ´ e−2x 1+ e−x dx = −ex + ln(ex + 1) − 1
e−2x 1 e−2x 1 ex = u2 = + = −x e−3x (e−3x )(1 + ex ) e +1
´
u2 =
1 dx = x + ln(e−x + 1) e−x + 1
yp = (e−2x )[ ex + ln(ex + 1)
− 1] + [x + ln(e−x + 1)](e−x )
−
−e−x + e−2x ln(ex + 1) − e−2x + xe−x + e−x ln(e−x + 1) y(x) = c1 e−2x + c2 e−x − e−x + e−2x ln(ex + 1) − e−2x + xe−x + e−x ln(e−x + 1) 3y − 6y + 6y = ex secx yh = 3y − 6y + 6y = 0 3m2 − 6m + 6 = 0 a = 3 b = −6 c = 6 yp =
5
.
,
D
e
ni
m
m1,2 =
os
y1 y2 ,
−(−6) ± (−6)2 − 4(3)(6) = 6 ± √36 − 72 = 1 ± √−36 = 1 ± i 2(3)
D
e
n
i
endo
,
6
6
α=1 β=1 ,
x
yh = e (c1 cosx + c2 senx)
y1 y2 ,
2
6
6
y1 = ex cosx y1 = ex cosx ;
− ex senx
y2 = ex senx y2 = ex senx + ex cosx ;
C
al
cul
am
os
l
os
W
r
ons
ki
anos
ex cosx
x
e
senx
x
x
x
W = ex cosx ex senx e x senx + ex cosx = (e cosx)(e senx + e cosx) (ex senx)(ex cosx ex senx) = ex (cosxsenx + cos2 x cosxsenx + sen2 x)
−
−
−
−
W = ex (cos2 x + sen2 x) = ex 0 ex senx = (ex senx)(ex secx) = ex secx e x senx + ex cosx ex tanx
W1 =
−
−
−ex ( senx )= cosx
x
cosx e cosx 0 = (ex cosx)(ex secx) = ex ( ) = ex ex cosx ex senx e x secx cosx
W2 = C
al
cul
−
am
u1 u2
os
,
x
= −tanx − e tanx ex ´ u1 = − tanxdx = −(−lncosx) = lncosx u1 =
ex =1 ex
u2 = u2 =
´
dx = x
x
yp = lncosx(e cosx) + x(ex senx) y(x) = ex (c1 cosx + c2 senx) + ex cosxlncosx + xex senx 2.
6
E
c
ua
c
i
one
s
de
C
a
uc
hy-
E
2
x y Supone
m
os
una
s
ol
uc
i
ón
de
l
a
f
or
ul
e
r
− 2y = 0 m
y = xm
a
.
y = xm y = mxm−1 y = (m Su
s
t
i
t
ui
m
os
en
l
a
ec
ua
ci
ón
2 [(m
x
or
−
i
gi
na
− 1)mxm−2 l
.
1)mxm 2 ]
−
2
7
− 2(xm) = 0
x2 [(m
− 1)mxm x−2] − 2(xm) = 0 (m − 1)mxm − 2xm = 0 xm [(m − 1)m − 2] = 0 xm (m2
as
i
obte
nem
os
l
a
ec
uaci
on
aux
i
l
i
m
2) = 0
− −
ar
m2
−m−2 = 0 (m + 1)(m − 2) = 0 m1 = −1 m2 = 2 ;
So
n
r
aí
ce
s
r
eal
es
y
di
s
t
i
ntas
,
as
i
qu
e
l
a
s
ol
uci
ón
es
:
y = c1 x−1 + c2 x2 2
.
x2 y + y = 0 m
Su
pon
e
m
os
l
a
s
ol
uci
on
y=x
y = xm y = mxm−1 y = (m Su
s
t
i
t
ui
m
os
en
l
a
ec
uaci
ón
x2 [(m (m
E
cuai
ón
au
xi
l
i
− 1)mxm−2
− 1)mxm−2] + xm = 0
− 1)mx2xm x−2 + xm = 0 (m2 − m)xm + xm = 0 xm (m2 − m + 1) = 0
ar
m2
don
e
:
α=
1 2
β=
1 2
√3
−m+1=0 m1,2 = 12 ± 12 √3i 1
1
y = c1 x 2 + 2
√3i
2
8
1
1
+ c2 x 2 − 2
√3i
U
s
ando
l
a
i
dent
i
da,
xiβ = (elnx )iβ = eiβlnx c
on
l
a
f
or
m
ul
a
de
E
ul
e
r
,es
l
o
m
i
s
m
o
que
xiβ = cos(βlnx) + isen(βlnx) x−iβ = cos(βlnx)
− isen(βlnx)
entocs
xiβ + x−iβ = cos(βlnx) + isen(βlnx) + cos(βlnx)
− isen(βlnx) = 2cos(βlnx)
iβ
x − x−iβ = cos(βlnx) + isen(βlnx) − cos(βlnx) + isen(βlnx) = 2isen(βlnx) y = C1 xα+iβ + C2 xα−iβ s
i
y1 = xα (xiβ + x−iβ ) = 2xα cos(βlnx) y2 = xα (xiβ s
e
c
onc
l
uye
− x−iβ ) = 2xα isen(βlnx)
que
y1 = xα cos(βlnx) A
s
í
l
a
s
ol
uci
on
genr
al
y
= xα sen(βlnx)
es
y = xα [c1 cos(βlnx) + c2 sen(βlnx)]
√
1
√
y = x 2 [c1 cos( 12 3lnx) + c2 sen( 12 3lnx)] 3
.
x2 y + xy + 4y = 0 Su
pon
e
m
os
l
a
s
ol
uci
ón:
y = xm y = mxm−1 y = (m Su
s
t
i
t
ui
m
os
en
l
a
ec
uaci
− 1)mxm−2
ón.
1)mxm−2 ] + x(mxm−1 ) + 4(xm ) = 0
x2 [(m
−
xm (m2
− m + m + 4) = 0
m
x (m2 + 4) = 0 m2 =
2
9
−4
± √− 4 m1,2 = ±2i
m1,2 =
α=0β=2 y=
x0 (c1 cos2lnx
+ c2 sen2lnx)
y = c1 cos2lnx + c2 sen2lnx 4
.
x2 y So
l
uci
on
prou
es
t
− 3xy − 2y = 0
a.
y = xm y = mxm−1
− 1)mxm−2
y = (m S
u
st
i
t
u
i
m
os.
x2 [(m
− 1)mxm−2] − 3x(mxm−1) − 2(xm) = 0 xm [(m2
m)
3m
2] = 0
− − − xm (m2 − 4m − 2) = 0 √ m =2± 6 1,2
√6
y = c2 x2+ 5
√6
+ c1 x2−
.
25x2 y + 25xy + y = 0 So
l
uci
ón
prou
es
t
a.
y = xm y = mxm−1 y = (m S
u
st
i
t
u
i
m
− 1)mxm−2
os.
25x2 [(m
− 1)mxm−2] + 25x(mxm−1) + xm = 0
xm [25m2
− 25m + 25m + 1] = 0 3
0
25m2 + 1 = 0
m1,2 =
± − 251 = ± 15 i
α = 0, β = 1 5 1 1 0 y = x (c1 cos lnx + c2 sen lnx) 5 5 1 1 y = c1 cos lnx + c2 sen lnx 5 5 6
.
x2 y + 5xy + y = 0 So
l
uci
on
prou
es
t
a.
y = xm y = mxm−1 y = (m S
u
st
i
t
u
i
m
− 1)mxm−2
os.
x2 [(m 1)mxm−2 ] + 5x(mxm−1 ) + xm = 0 xm (m2 m + 5m + 1) = 0 m2 + 4m + 1 = 0 m1,2 = 2 √ 3 √ y = c1 x2+ 3 + c2 x2− 3
− − ±√
7
.
− 4y = x4
xy So
l
uci
ón
prou
es
t
a.
y = xm y = mxm−1 y = (m H
po
r
ac
e
m
os
l
a
e
c
uac
i
on
de
l
a
f
or
m
a
− 1)mxm−2
de
C
auc
x.
− 4xy = x5 yh = x2 y − 4xy = 0 x2 [(m − 1)mxm−2 ] − 4x(mxm−1 ) = 0 x2 y R
S
e
u
s
s
ol
t
ve
i
t
m
u
i
os
m
l
a
par
t
e
hom
oge
nea
.
os
3
1
hy
E
ul
e
r
,par
a
e
s
t
o
l
a
m
ul
t
i
pl
i
c
am
os
xm (m2 m 4m) = 0 m(m 5) = 0 m1 = 0 m2 = 5 yh = c1 x0 + c2 x5 yh = c1 + c2 x5
−
R
e
P
s
ol
ar
− −
ve
a
m
e
s
os
t
por
o
t
e
var
ne
m
i
os
ac
i
on
que
de
e
s
c
par
r
i
ám
bi
r
l
et
a
e
r
c
os
uac
.
i
on
e
n
Q(x)y = f (x) D
i
vi
di
m
os
l
a
ec
ua
ci
ón
or
− 4 yx = x3
y
i
gi
nal
ent
r
x
e
3
f (x) = x y1 y2 y1 = 1 y1 = 0 y2 = x5 y2 = 5x4 1 x5 W = = 5x4 0 = 5x4 0 5x4 5 0 x W1 = 3 = 0 x8 = x8 x 5x4 1 0 = x3 W2 = 0 x3 u1 u2 x8 u1 = − = 15 x4 5x4 ´ 1 5 u1 = 15 x4 dx = 25 x i
d
enti
D
e
cam
ni
os
m
os
,
,
,
− −
C
al
cul
am
os
−
,
−
−
−
3
u2 = 51xx´4 = 1
51x
u2 = 5 x dx = 15 lnx 1 5 yp = 25 x (1) + 15 lnx(x5 ) 5 1 5 yp = 25 x + x5 lnx y(x) = yh + yp 5 1 5 y(x) = c + c2 x5 25 x + x5 lnx
− −
−
7
.
x2 y So
l
uci
on
prou
es
t
− xy + y = 2x
a.
y = xm y = mxm−1
− 1)mxm−2
y = (m R
e
s
ol
ve
m
os
l
a
e
c
uac
i
on
hom
oge
nea
as
oc
yh = x2 y
i
ad.
− xy + y = 0 x2 [(m − 1)mxm−2 ] − x(mxm−1 ) + xm = 0 m2 − m − m + 1 = 0 m2 − 2m + 1 = 0 Su
s
t
i
t
ui
m
os
en
l
a
ec
ua
ci
ón
or
i
gi
na
l
.
3
2
l
a
f
or
m
a
e
s
t
andr
P (x)y +
(m 1)2 m1,2 = 1 yh = c1 x + c2 xlnx
−
P
one
m
I
os
l
a
e
c
uac
i
ón
e
− x1 y + x1 y = x2
y
n
l
a
f
or
m
a
e
s
t
andr
2
denti
cam
2 x
f (x) =
os
y1 = x y2 = xlnx y1 = 1 y2 = lnx + 1 x lnx W = = (x)(lnx + 1) (lnx)(1) = xlnx + x 1 lnx + 1 x lnx + x = xln x + x = xln(1) + x = x 0 lnx W1 = 2 = (lnx)( x2 ) = x2 lnx lnx + 1 x x 0 W2 = = x2 x 0 = 2 1 x2 u1 u2 2 lnx u1 = x x = 2xlnx 2 ´ lnx+1 u1 = 2 lnx x2 = x u2 = x2 ´ 1 u2 = 2 x = 2lnx yp = y1 u1 + y2 u2 = x( lnxx+1 ) + xlnx(2lnx) = lnx + 1+ I
denti
cam
C
al
os
cul
am
,
os
l
os
W
r
y
ons
ki
,
anos
−
−
C
al
cul
am
os
,
−
−
8
−
.
x y − 2xy + 2y = x4 ex 2
So
l
uci
ón
prou
es
t
a.
y = xm y = mxm−1 y = (m
− 1)mxm−2
− 1)mxm−2] − 2x(mxm−1) + 2xm = x4ex x2 y − 2xy + 2y = 0 xm (m2 − m − 2m + 2) = 0 m2 − 3m + 2 = 0 (m − 2)(m − 1) = 0 x2 [(m Sol
uci
onam
os
l
a
e
c
uac
i
on
hom
oge
nea
m1 = 2 m2 = 1 yh = c1 x2 + c2 x ,
C
onve
r
t
2
y
i
m
os
y +
l
2
a
e
cua
i
on
or
i
gi
nal
a
l
a
f
y = x2 ex
− x yx2 y f (x) = x2 ex 1 2 y1 = x2 y1 = 2x y2 = x y = 1 D
e
ni
m
os
,
,
;
2
;
C
al
c
ul
am
os
e
l
W
r
ons
ki
ano
3
3
or
m
a
es
t
andr
− lnx = xlnx −
x2 x = x2 2x2 = x2 2x 1 0 x W1 = 2 x = 0 x3 ex = x3 ex x e 1 x2 0 W2 = = x4 ex 2x x2 ex
− −
W =
−
−
3 x u1 u2 u1 = −−xxe2 = xex ´ x u1 = xe dx = ex (x 1) 4 x u2 = x−xe2 = x2 ex ´ 2 x u2 = x e dx = ex (x2 2x + 2) yp = u1 y1 + u2 y2 = [ex (x 1)]x2 + [ex (x2 yp = x2 ex (x 1) + xex (x2 2x + 2) y(x) = yp + yh y(x) = c1 x2 + c2 x + x2 ex (x 1) + xex (x2 C
al
cul
am
os
,
−
−
−
− − − −
−
9
− 2x + 2)
.
3
1
− 2x + 2)]x
Sol
uc
i
o ne
s
e
n
s
e
r
i
e
s
de
pot
e
nc
i
as
.
y S
u
t
i
t
u
yend
∞ c xn n=0 n
y=
o
y
l
− xy = 0
a
s
e
gund
a
der
i
∞ (n 1)nc xn−2 n n=2
y =
vad
∞ (n − 1)nc xn−2 x ( ∞ c xn ) = 0 n n=2 n=0 n ∞ (n − 1)nc xn−2 ∞ c xn+1 = 0 n n=2 n=0 n
A
hor
a
s
um
am
os
l
as
2(1)c2 x0 2c2 hac
de
e
m
o
S
u
m
t
−
q
i
t
u
ue
i
m
s
e
r
i
e
s
i
gual
ando
l
os
∞ n(n − 1)c xn−2 − n n=3
∞ n(n − 1)c xn−2 − n n=3
k=n 2 n = k+2 n = k
os
do
dos
− −
par
:
a
,
l
a
pr
i
m
e
−1
r
r
a
s
es
p
e
r
i
ect
e
y
i
i
ndi
c
e
s
de
bas
s
par
a
l
a
ent.
2c2
∞ c xn+1 = 0 n=0 n ∞ (k + 2)(k + 1)c xk − ∞ c xk = 0 k+2 k=1 k=1 k−1
2c2
∞ n(n − 1)c xn−2 − n n=3
∞ k=1 [(k + 2)(k + 1)ck+2 − ck −1 ]xk = 0 (k + 2)(k + 1)ck+2 ck+2 =
− ck−1 = 0
ck−1 (k + 2)(k + 1) 3
4
as
.
∞ c xn+1 = 0 n=0 n
os
2c2
um
∞ c xn+1 = 0 n=0 n
k = n+1
vam
am
−
s
eg
un
da
s
er
i
e
,
E
s
t
elva
a
r
or
A
el
aci
on
k
de
hor
genr
com
a
o
ent
coe
er
ci
os
p
ent
os
i
es
t
i
cons
vos
ec
uti
vos
de
l
a
s
ol
uci
on
proues
t
a,
con
.
a
2c2 = 0 c2 = 0 ;
c0
= 1 c0
3(2)
6
k = 1 c3 = ,
k =2
c1 = 4(3)
c4 =
,
c2 = 5(4)
k = 3 c5 = ,
1 20 c2
1 1 30 ( 6 )c0
=
1 180 c0
k = 5 c7 =
c4 = 7(6)
1 1 42 ( 12 )c1
=
1 504 c1
k =6
c8 =
,
c5 =0 8(7)
c6 = 9(8)
k = 7 c9 = ,
,
,
t
uy
end
o
coei
ent
es
en
l
a
s
up
os
i
ci
=
1 12960 c0
1 10(9)(504) c1
c8 =0 11(10)
k = 9 c11 = i
← c5 = 0
1 1 72 ( 180 )c0
c7 k = 8 c10 = 10(9) =
t
← c2 = 0
c3 = 6(5)
,
s
=0
k = 4 c6 = ,
Su
1 12 c1
on
or
← c8 = 0 i
gi
nal
y= c0 +c1 x+c2 x2 +c3 x3 +c4 x4 +c5 x5 +c6 x6 +c7 x7 +c8 x8 +c9 x9 +c10 x10 +c11 x11 +..., y= 1 1 1 1 1 c0 +c1 x+0+ 16 c0 x3 + 12 c1 x4 +0+ 180 c0 x6 + 504 c1 x7 +0+ 12960 c0 x9 + 90(504) c1 x10 +0 y = c0 (1 + 16 x3 +
1 6 180 x
1 9 12960 x ) + c1 (x
+
+
1 4 12 x
+
1 7 504 x
+
1 10 90(504) x )
2
y + x2 y + xy = 0
S
u
t
i
t
u
yend
o:
n y= ∞ n=0 cn x n−1 y = ∞ c nx n n=1 y = ∞ (n 1)ncn xn−2 n=2
− E
n
l
a
ec
uaci
∞ (n n=2
−
ón
or
i
gi
nal
1)ncn xn−2 + x2
∞ c nxn−1 + x [ n=1 n
3
5
cn xn ] = 0
∞ (n − 1)nc xn−2 + ∞ c nxn+1 + ∞ c xn+1 = 0 n n=2 n=1 n n=0 n n−2 n+1 n+1 2c2 x0 + 6c3 x ∞ + ∞ + c0 x1 ∞ =0 n=4 (n − 1)ncn x n=1 cn nx n=1 cn x
H
seri
a
c
e
m
o
− −
k=n 2
s
es.
pa
r
a
l
a
pri
2c2 x0 + 6c3 x ∞ k =2 (k + 2 k−1+1 c0 x 1 ∞ =0 k=2 ck−1 x
m
er
a
s
er
i
k = n+1
e,y
1)(k + 2)ck+2 xk+2−2 +
par
a
k
l
a
s
e
gund
a
y
t
e
r
c
er
a
∞ c (k − 1)xk−1+1 + k =2 k−1 k
k
∞ k =2 (k + 1)(k + 2)ck +2 x + ck−1 (k 2c2 + 6c3 x + c0 x ∞ 1)x + ck−1 x = 0 2c2 + 6c3 x + c0 x k=2[(k + 1)(k + 2)ck+2 + ck−1 (k 1) + ck−1 ]xk = 0 (k + 1)(k + 2)ck+2 + ck−1 (k 1) + ck−1
E
nt
onc
e
s
t
e
nem
−−
−
os
2c2 = 0 c2 = 0 6c3 + c0 = 0 c3 = 16 c0 ;
−
S
u
st
c4 c5 c6 c7 c8 c9
i
t
−
u
yen
c10
= c11 = c12 = P
or
d
2c1 3(4) 3c2 4(5) 4c3 5(6) 5c4 6(7) 6c5 7(8) 7c6 8(9)
= = = = = =
t
kc k−1 (k + 1)(k + 2) k = 2, 3, 4,... = 16 c1 =0 c2 = 0 2 1 = 15 ( 16 c0 ) = 45 c0 5 1 5 = 42 ( 6 c1 ) = 136 c1 6 = 56 (0) = 0 7 1 7 = 72 ( 45 )c0 = 72(45) c0
[(k 1)+1]ck 1 (k +1)(k +2)
ck+2 =
=
−
o
e
← −
−
−
−
n
l
a
f
or
m
ul
a
s
e
obt
i
e
ne
8c7 = 45 4 ( 136 5 c1 ) = 45(34) 5 c1 9(10) 9c8 9 10(11) = 110 (0) = 0 10c9 5 7 7 = 66 ( 72(45) c0 ) = 66(72)(9) c0 11(12)
−
ant
−
o,
y = c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + c5 x5 + c6 x6 + c7 x7 + c8 x8 + c9 x9 + ... 5 5 1 6 7 7 y = c1 [ 16 x4 + 136 x7 + 9(34) x10 ] c0 [ 45 x + 72(45) x9 + 66(72)(9) x12 ] 3
S
− y − 2xy + y = 0
.
u
t
i
t
u
yend
o:
n y= ∞ n=0 cn x n−1 y = ∞ n=1 cn nx y = ∞ (n 1)ncn xn−2 n=2
− − − − −− E
n
l
a
ec
uaci
ón
or
i
gi
nal
∞ (n 1)nc xn−2 − 2x ∞ c nxn−1 + ∞ c xn = 0 n n n=2 ∞ (n 1)ncn xn−2 − 2 ∞ n=1 ∞ c nx=0 n n =0 n n=2 n=1 cn nx + n=0 n ∞ ∞ ∞ n−2 n 2c2 n=3(n 1)ncn x − 2 n=1 cnnx + c0 n=1 cnxn = 0
−
k = n 21)(k + 2)c xk+2−2 k = ∞ c xk = 0 2c2 ∞ 2 n ∞ c kx k + c k+2 k=1 (k + 2 ∞ ∞ k=1 k ∞ 0 k=1 k 2c2 + c0 k=1(k + 1)(k + 2)ck+2 xk + k=1 ck kx k + k=1 ck xk = 0 k 2c2 + c0 ∞ 2ck kx k + ck xk = 0 k =1 (k + 1)(k + 2)ck +2 x ∞ 2c2 + c0 k=1(k + 1)(k + 2)ck+2 xk 2ck kx k + ck xk = 0 H
a
c
e
m
o
s
par
a
l
a
s
e
r
i
e
uno
y
− −
3
6
par
a
l
as
dos
y
t
r
e
s
.
D
e
e
s
t
a
i
gua
l
da
s
e
c
onc
l
uye
que
2c2 + c0 = 0 c2 = 12 c0 (k + 1)(k + 2)ck+2 xk 2ck kx k + ck xk = 0 [(k + 1)(k + 2)ck+2 2ck k + ck ]xk = 0 (k + 1)(k + 2)ck+2 2ck k + ck
−
− −
ck+2 = S
u
st
i
t
u
yen
−
(2k + 1)ck (k + 1)(k + 2) k = 1, 2, 3, 4,... d
o
3c1 2(3)
c3 = 5c2 3(4)
c4 =
c6 =
9c4 5(6)
c7 =
11c5 6(7)
c0
4
=
11 7 42 ( 40 c1 )
=
c9 =
15c7 8(9)
=
15 11 72 ( 6(40) c1 )
7 5 40 x
5 4 24 x
+
=
11 7 240 x
+
+
161 72(240) c1
161 161 9 11 72(240) x + 55(8)(240) x 17(13) 13 8 10 56(16) x + 90(56)(16) x
−
.
u
t
i
t
u
yend
y =
(x2 + 2)
−y = 0
o:
∞ (n − n=2
∞ c xn n=0 n ∞ c nxn−1
y=
x2
=
1 161 55 ( 8(240) )c1
(x2 + 2)y + 3xy S
11 6(40) c1
17(13) − 9(10)(56)(16) c0
=
18c9 10(11)
1 6 16 x
=
13 − 161 c0) = − 56(16) c0
17c8 9(10)
+
7 40 c1
− 245 c0) = − 161 c0
13 56 (
+
=
9 30 (
=
1 3 2x
7 1 20 ( 2 c1 )
=
13c6 7(8)
c11 =
− 12 c0) = − 245 c0
c8 =
c10 =
y = c1
5 12 (
=
7c3 4(5)
c5 =
= 12 c1
n=1 n
n 2
n=2 (n y = ∞ 1)ncn x − ∞ 1)ncn xn−2 + 3x n=1 cn nxn−1
∞ (n − 1)nc xn−2 +2 n n=2
−
∞ c xn = 0 n=0 n ∞ (n − 1)nc xn−2 + ∞ 3c nxn − ∞ c xn = n n n=2 n=1 n=0 n
0
3
7
−
∞ (n − 1)nc xn + ∞ 2(n − 1)nc xn−2 + ∞ 3c nxn − ∞ c xn = 0 n n n n=2 n=2 n=1 n=0 n ∞ (n − 1)nc xn + ∞ 2(n − 1)nc xn−2 + 3c x ∞ 3c nxn − c + n n n 1 0 n=2 n=2 n=2 c x ∞ c xn = 0
1
n=2 n
n 2−2 3c1 x + c0 + c1 x ∞ + 2(3 − n=2 (n − 1)ncn x + 2(2 − 1)2c2 x ∞ c xn = 0 n 1)3c3 x3−2 ∞ 1)ncn xn−2 + ∞ n=4 2(n n=2 3cn nx n=2 n − ∞− 2(n − 1)nc xn−2 + n 3c1 x + c0 + c1 x + 4c2 + 12c3 x ∞ n n=4 =2 (n − 1)ncn x + ∞ 3c nnx ∞ n n c x =0 −
− − − − n
n=2
H
a
c
e
m
o
k=n
s
2
par
a
l
a
s
e
n=2 n
gunda
s
e
r
i
e
y
k=n
par
a
l
as
de
m
as
3c1 x + c0 + c1 x + 4c2 + 12c3 x ∞ 1)kc k xk + ∞ 2(k + 2 − 1)(k + k=2 (k ∞ c kx=2 k k 2)ck+2 xk+2−2 + ∞ 3c kx =0 k k=2 k=2 k ∞ [(k 1)kc + 2(k +1)( k + 2)c + 3c k − c ]xk = 0 4c x + c + 4c + 12c x 1
0
D
e
e
s
2
t
a
i
3
gual
da
s
k
k=2
e
obt
i
e
k+2
k
k
ne
4c1 + 12c3 = 0 c3 = c31 c0 + 4c2 = 0 c2 = c40 (k 1)kc k + 2(k + 1)(k + 2)ck+2 + 3ck k ck 3kc k + ck (k 1)kc k [ 3k + 1 k 2 k]ck [ 4k + 1 k2 ]ck ck+2 = = = 2(k + 1)(k +k2) 2(k + 1)(k + 2) = 2, 3, 4, 5,...2(k + 1)(k + 2)
− −
−
Su
s
t
i
−
t
uy
− −
end
o
val
or
es
−
−
− −
−
−
de
c2 = c40 c3 = c31 −4−2)c2 = c4 = (−6+1 2(3)(4)
− −
11 1 11 24 ( 4 c0 ) = 96 c0 ( 12+1 9)c3 20 1 1 c5 = = 40 c3 = 2 ( 3 c1 ) = 16 c1 2(4)(5) ( 16+1 16)c4 31(11) 31 11 c6 = = 31 2(5)(6) 60 c4 = 60 ( 96 c0 ) = (60)(96) c0 ( 20+1 25)c5 44 11 1 11 c7 = = 84 c5 = 21 ( 6 c1 ) = 126 c1 2(6)(7) 36)c6 11(31) 59 11(31) c8 = ( 24+1 = 112 ( 60(96) c0 ) = 112(60)(96) c0 2(7)(8) 31(11) 11(31) 4+ 6+ 8 ] + c [ 1 x3 y = c0 [ 14 x2 + 11 x x x 1 90 60(96) 112(60)(96) 3
− − − −
− − − − − − −
− − − −
−
−
4
1
Tr
a
ns
−
f
o
r
m
a
da
de
La
pl
a
c
+ 16 x5 +
e
.
1, 0 f (t) 1,
t<1
t
≥≤ 1 ´ ∞ −st L{f (t)} = 0 e f (t)dt = − ´01 e−st (1) + ´1∞ e−st (1) = − e−s |10 + e−s |∞ 1
−
st
st
−
−
3
8
11 7 126 x ]
s(1)
s(0)
s(1)
s(∞)
− e −s − [− e −s ] + e −s − e −s = e s − 1s + e s + 0s = 2es − 1s −
=
−
−
s
−
s
−
−
s
−
2
.
f (t) =
t, 1,
0
≤t<1 t≥1
L[f (t)] = ´0∞ e−st f (t) = ´01 e−st tdt + ´1∞ e−st (1)dt = − e s (t − 1s )|10 + − e s |∞ 1 = − e s (1 − 1s ) − [− e s (0 − 1s )] + [ − e s − e = − e s + es + s1 − 1s st
st
−
−
s(1)
s(0)
−
s(∞)
−
s
−
s(0)
−
s
s
−
]
−
2
2
f (t) = te4t
L{te4t } = ´0∞ e−st te4tdt = ´0∞ te−(s−4)t dt =
−(s−4)t
− e(s − 4)2 [−s + 3]|∞0 −(s−4)∞ 0
=
(s
1
− 4)2 + (s − 4)2 1
= 3
−(s−4)0
− e(s − 4)2 − [− e(s − 4)2 ]
=
(s
− 4)2
.
y + 3y + 2y = 0
y(0) = 1 A
pl
i
c
am
y (0) = 1
,
os
t
r
ans
f
or
m
ad
de
L
apl
ac
e
a
t
oda
l
a
e
c
uac
i
ón
L[y2 ] + 3L[y] + 2L[y] = 0 [s Y (s) − sy(0) − y (0)] + 3[sY (s) − y(0)] + 2[Y (s)] = 0 s2 Y (s) − sy(0) − y (0) + 3sY (s) − 3y(0) + 2Y (s) = 0 s2 Y (s) − s(1) − 1 + 3sY (s) − 3(1) + 2Y (s) = 0 s2 Y (s) − s − 1 + 3sY (s) − 3 + 2Y (s) = 0 Y (s)(s2 + 3s + 2) − s − 4 S
F
u
st
actor
i
t
u
i
i
zan
m
os
l
d
os
val
ores
i
n
i
ci
al
es
o
3
9
4+s s2 + 3s + 2
Y (s) = Sep
ar
am
os
en
f
r
ac
i
on
es
pa
r
ci
al
es
4+s A B = + (s + 2)(s + 1) s+1 s+2 P
or
e
lm
é
t
od
de
H
e
(4 + s)(s + 1)
A=
(s + 2)(s + 1) (4 + s)(s + 2) B= (s + 1)(s + 2) Su
s
t
i
t
ui
m
os
A
pl
i
en
l
3 s+1
Y (s) = cam
os
l
a
ec
a
vi
s
t
r
de
4 + ( −1) =3 s= 1 | − −4 1−+2 2 |s=−2 = −2 + 1 = −2 =
uaci
on
− s +2 2
a
i
ansf
or
m
t
r
ad
i
ansf
or
nver
m
s
ad.
a
a
cad
t
ér
m
i
no
del
des
ar
r
ol
l
o
ant
er
i
or
.
3 y(t) = −1 [ ] s+1 1 − 1 y(t) = 3 [ ] s+1 y(t) = 3e−t 2e−2t
L L
− L−[ s +2 2 ] − 2L−[ s +1 2 ]
−
y S
u
t
i
t
u
yend
∞ c xn n=0 n
y=
o
y
l
− xy = 0
a
s
e
gund
a
der
i
∞ (n 1)nc xn−2 n n=2
y =
vad
−
∞ (n − 1)nc xn−2 − x ( ∞ c xn ) = 0 n n=2 n=0 n ∞ (n − 1)nc xn−2 − ∞ c xn+1 = 0 n
n=2
A
hor
a
s
um
am
os
l
as
2(1)c2 x0 2c2 hac
de
e
m
o
S
u
m
t
−
q
i
t
u
ue
i
m
dos
s
e
r
i
e
s
i
n=0 n
gual
ando
l
os
∞ n(n − 1)c xn−2 − n n=3
∞ n(n − 1)c xn−2 − n n=3
k=n 2 n = k+2 n = k
os
do
par
:
a
,
l
a
pr
i
m
e
−1
r
r
a
s
es
p
e
r
i
ect
e
y
i
i
ndi
c
e
s
de
bas
s
par
a
l
a
ent.
os
2c2 2c2
∞ c xn+1 = 0 n=0 n ∞ (k + 2)(k + 1)c xk − ∞ c xk = 0 k+2 k=1 k=1 k−1
2c2
∞ n(n − 1)c xn−2 − n n=3
∞ k=1 [(k + 2)(k + 1)ck+2 − ck −1 ]xk = 0 (k + 2)(k + 1)ck+2 ck+2 =
− ck−1 = 0
ck−1 (k + 2)(k + 1) 4
0
um
as
.
∞ c xn+1 = 0 n=0 n
k = n+1
vam
am
∞ c xn+1 = 0 n=0 n
s
eg
un
da
s
er
i
e
,
E
s
t
elva
a
r
or
A
el
aci
on
k
de
hor
genr
com
o
a
ent
coe
er
ci
os
ent
p
os
i
es
t
i
cons
vos
ec
uti
vos
de
l
a
s
ol
uci
on
proues
t
a,
con
.
a
2c2 = 0 c2 = 0 ;
c0
= 1 c0
3(2)
6
k = 1 c3 = ,
k =2
,
c1 = 4(3)
c4 =
c2 = 5(4)
k = 3 c5 = ,
1 c 20 2
1 1 30 ( 6 )c0
=
1 180 c0
k = 5 c7 =
c4 = 7(6)
1 1 42 ( 12 )c1
=
1 504 c1
k =6
c5 =0 8(7)
c8 =
,
c6 = 9(8)
k = 7 c9 = ,
,
,
t
uy
end
o
coei
ent
es
en
l
a
s
up
os
i
ci
=
1 c 12960 0
1 10(9)(504) c1
c8 =0 11(10)
k = 9 c11 = i
← c5 = 0
1 ( 1 )c 72 180 0
c7 k = 8 c10 = 10(9) =
t
← c2 = 0
c3 = 6(5)
,
s
=0
k = 4 c6 = ,
Su
1 12 c1
on
or
← c8 = 0 i
gi
nal
y= c0 +c1 x+c2 x2 +c3 x3 +c4 x4 +c5 x5 +c6 x6 +c7 x7 +c8 x8 +c9 x9 +c10 x10 +c11 x11 +..., y= 1 1 1 1 1 c0 +c1 x+0+ 16 c0 x3 + 12 c1 x4 +0+ 180 c0 x6 + 504 c1 x7 +0+ 12960 c0 x9 + 90(504) c1 x10 +0 y = c0 (1 + 16 x3 + 2
1 6 180 x
1 9 12960 x ) + c1 (x
+
u
1 4 12 x
+
1 7 504 x
+
1 10 90(504) x )
.
y S
+
t
i
t
u
yend
∞
∞ n 1 ∞ (n − 1)nc nx=1 n−c2n nx − n n=2 ∞ (n − 1)nc xn−2 − (x + 1) ∞ c nxn−1 − ∞ c xn = 0 n n n o
y=
− (x + 1)y − y = 0
n=2
n
l
n=0 cn x d
eri
vad
∞ (n − 1)nc xn−2 − n n=2
a
a
pri
m
y =
er
a
der
i
n=1
∞ c nxn − n=1 n 4
1
vad
y
l
a
s
e
gunda
n=0
∞ c nxn−1 − n=1 n
∞ c xn = 0 n=0 n
2c2 x0
hac
par
e
a
l
m
a
e
x0
c0
k=n
os
s
∞ (n − 1)nc xn−2 − n n=3
gunda
y
l
−2 a
c
par
uar
l
t
a
s
a
e
pri
r
i
∞ c nxn − c x0 n 1 ∞n=1 c xn = 0
n=1 n
m
e
er
s
er
i
k=n
e,
−1
∞ c nxn−1 − n=2 n
par
l
a
t
er
ce
r
a,
k=n
.
∞ (n 1)nc xn−2 n
c2 c1 c0 +
a
∞ c nxn n
∞ c nxn−1 n
∞ c xn = n
n=3 − − − − n=10 − n=2 − n=1 ∞ c kx k − ∞ c (k + k c2 − c1 − c0 + ∞ k =1 (k + 2 − 1)(k + 2)ck +2 x − k=1 k k=1 k+1 k 1)xk − ∞ k =1 ck x = 0 k c2 − c1 − c0 + ∞ k=1 [(k + 1)(k + 2)ck+2 − kc k − (k + 1)ck+1 − ck ]x = 0
D
e
ak
is
e
c
onc
l
uye
que
c2 = 0 c1 = 0 c0 = 0 (k + 1)(k + 2)ck+2 ck+2 =
− (k + 1)ck+1 − kck − ck = 0
(k + 1)ck+1 + (k + 1)ck (k + 1)(k + 2)
k = 1, 2, 3, ..., 2 +2c1 k = 1 c3 = 2c2(3) =0 3 3 +3c2 k = 2 c4 = 3c3(4) = 12 c3 = 0 4c4 +4c3 k = 3 c5 = 4(5) = 0 S
u
st
i
t
u
yen
d
o
,
,
,
Ser
1
i
e
de
T
ayl
or
.
y = x + 2y2 y(0) = 0 y (o) = 1 D
er
i
vand
o
y = 1 + 4 yy y = 4y y + 4yy yiv = 4y y + 4y y + 4y y + 4yy = 12y y + 4yy y v = 12y y + 12y y + 4y y + 4yy iv
4
2
y vi = 12y y + 12y y + 12y y + 12y y iv + 4y y + 4y y iv + 4y yiv + 4yy v = 36y y + 12y y iv + 4y y + 4y yiv + 4y y iv + 4yy v y (0) = 1
y (0) = 4
,
x 1!
y(x) =
1
y iv (0) = 12
,
x2 2!
+
4x3 3!
+
+
y v (0) = 76
,
12x4 4!
.
f (t) = 4t
− 10
L[f (t)] = 4 L´ [t]∞ − −10stL[1] ´ ∞ −st L[f (t)] = 4 0 te dt − 10 0 e dt I
nt
e
gr
am
os
por
par
t
e
s
l
a
pr
i
m
er
a
i
nt
eg
r
al
´∞
4 0 te−st dt = u = t du = 1 st dv = e−st v = e s ´ st 1 ∞ −st )]dt = 4[(1)( e s ) ∞ 0 + s 0 (1)(e st = 4s e−s(∞) ( 4s e−s(0)) + 1s e s ∞ 0 ,
= =
−
−
1 s
4 s
+
4
+ 1(
s H
ac
s e
−
m
os
(e
−s a
| −− −e s s(∞)
−
1 l
−
−
,
)= s
e
gunda
−
s(0)
−
s
4
1
s
− s2 i
nt
e
gr
)
|
al
´∞ st 10 0 e−st dt = − e s |∞ 0 =
−e
s(∞)
−
1 = s
s
−
− (− e
s(0)
−
s
)
entocs
L[f (t)] = 4s − s12 − 1s = 3s − s12 2
.
f (t) = et/5 t/5
´∞
L[f (t)] = L[e ] = 0t e e−st dt ´∞ t (1 5s) = 0 e 5 (1−5s) dt = e15(1−5s) |∞ 0 5
t/5
−
=
∞ (1−5s) 5 1 5 (1 5s)
e
−
0 (1−5s) 5 1 5
− e (1−5s) = − 1
5
5s
5
=
5s
− 3
+
1
−
.
f (t) = et−2
L[f (t)] = L
[et 2 ]
−
4
3
76x5 5!
+
,
yvi (0) = 408
408x6 6!
´ ∞ t−2 −st ´ e dt = e−2 et(1−s) dt 0 e t(1 s) −2 )( e( )(1 s) ) − (e−2 )( e(0)(1 s) ) = e−2 [ e 1−s ]|∞ 0 = (e 1−s 1−s − − 2 2 e e =
−
=
4
∞
−
−
−1 − s = s − 1
.
f (t) = et cos t
L[f (t)] = L[et cos t] L[fat(t)] = F (s) L[e f (t)] = F (s − a) P
or
e
l
t
e
or
e
m
a
de
t
r
as
l
ac
i
on
delj
s
e
a=1
L[cos t] = s s+1 =⇒ L[et cos t] = (s − 1)s 2 + 1 2
5
.
f (t) = e−t cos t
L[f (t)] = L[e−t cos t] L[fat(t)] = F (s) P
or
e
l
t
e
or
e
m
a
de
t
r
[e f (t)] = F (s
as
l
ac
i
on
delj
s
e
a)
a = −1 ´ ∞ −st− LL[cos t] = 0 e (cos t)dt I
nt
e
gr
am
os
por
par
t
e
s
−
u = cos t du = sin t ´ dv = e−st v = e−st dt = ,
st
−e s −
,
´ ∞ −st e−st ∞ ´ ∞ −e st e (cos t)dt = cos t(− )| − 0 s (− sin t)dt 0 s 0 e−st ∞ 1 ´ ∞ −st = −(cos t) | − e (sin t)dt −
0
s
E
t
ér
m
v
al
i
n
uam
os
e
l
pr
s
i
m
e
r
0
t
é
r
m
i
no
y
vo
l
ve
m
os
a
i
nt
e
gr
ar
por
par
t
e
s
e
l
s
e
gundo
o
u = sin t du = cos t st dv = e−st v = e s −∞ e e−0 = [(cos ) (cos 0) ] s s ,
−
,
−
∞
= 1+
−
´ ∞ −st e cos tdt
1 s2
0
´ ∞ −st cos tdt = ∴ 0 e A
pl
i
c
ando
e
l
t
e
or
em
1 s
1+ a
de
1 s2 t
r
st
st
− e s )|∞0 − ´0∞(− e s 1 1 1 = 2 st sin t|∞ cos t|∞ 0 − 0 = s e sest s
−
−
sin t(
s s2 + 1
=
as
− 1s
l
ac
i
on
delj
e
L[fat(t)] = F (s) L[e f (t)] = F (s − a) 4
4
s
−
)cos tdt
−1
a=
L[e−t cos t] = 6
s+1 (s + 1)2 + 1
.
1 0 1
f (t) =
0
−
≤
≥
L[f (t)] = ´0∞ e−st f (t)dt = ´02 e−st (−1)dt + ´24 e−st (0)dt + ´4∞ e−st (1)dt ´ 2 −st 1 1 1 1 (−1)dt = −[− e−st ]20 = e−2s − e0 = e−2s − 1 0 e s s s s ´ 4 −st 2
e
(0)dt = 0
´ ∞ −st 1 1 −∞t 1 4s 1 4s (1)dt = − e−st |∞ + e = e 4 =− e 4 e s
L 7
1 −2s [f (t)] = e s
s s 1 4s 1 −2s 1 + e = e + e4s s s
−
s
−1
.
f (t) =
3t 0
0
≥
´ ´ = 01 e−st + 1∞ e−st (0)dt ´L1[f (t)] ´ 3tdt 1 −st − st e 3tdt = 3 te dt 0
0
u = t du = 1 ,
−st
−e s e−st 1 ´ 1 e−st e−st 1 1 e−st )| − 0 (− )dt = t(− )| − [ ] s 0 s s 0 s s
dv = e−st v = ,
−
= 3 t(
3 = t( e−st ) 10 s
−
3
[ e−s
−
´s ∞
|−
− =
1 [( e−s ) s
− 0] − 1s [(e−s) − 1]
− 1s e−s] = 3se−s(−1 − 1s) + s12
L
0dt = 0 3 [f (t)] = e−s ( 1 s
E
ncot
1
8
1 −st 1 [e ]0 s
− − 1s + 1s)
r
ar
f (t)
da
s
u
t
r
ans
f
or
m
ad
de
L
.
F (s) =
4
5
1 s2
apl
ac
e
F (s) ,done
f (t) = −1 F (s)
L {
}
L−1[ s1 ] L−−11[ s1n! ] = f (t) = tn L [s ] = t 2
n+1 2
U
s
enco
a
ntr
9
r
e
ar
l
t
e
o
F (s)
r
e
m
a
de
l
a
t
r
a
ns
f
o
r
m
a
da
de
l
a
de
r
i
v
a
da
f (t)
da
.
f (t) = t sin3 t f (t) = t sin3 t f (0) = 0 f (t) = sin 3 t + 3t cos3 t f (0) = 0 f (t) = 3co s 3 t + 3co s 3t 9t sin3 t f (0) = 6 [f (t)] = s2 F (s) f (0) f (0) [3cos3 t + 3co s 3t 9t sin3 t] = s2 [t sin3 t] 0 6 [cos 3t] 9 [t sin3 t] = s2 [t sin3 t] 6 [cos 3t] = s2 [t sin3 t] + 9 [t sin3 t] 6 [cos 3t] = (s2 + 9) [t sin3 t] s 6[ 2 ] = (s2 + 9) [t sin3 t] s +9 6s [t sin3 t] = 2 (s + 9)2 sean
,
L L L L L
,
− −
− L L
L L
− − L L
,
L
L 1
0
.
f (t) = t cosh t s
ean:
f (t) = t cosh t f (0) = 0 f (t) = cosh t + t sinh t f (0) = 1 f (t) = sinh t + sinh t + t cosh t [f (t)] = s2 F (s) f (0) f (0) [2 sinh t + t cosh t] = s2 [t cosh t] 0 1 2 [sinh t] + [t cosh t] = s2 [t cosh t] 1 2 [sinh t] = s2 [t cosh t] [t cosh t] 1 1 2 2 = (s2 1) [t cosh t] 1 s 1 2 + 1 = ( s2 1) [t cosh t] 2 (s 1) 2 2+s 1 = (s2 1) [t cosh t] (s2 1) s2 + 1 [t cosh t] = 2 (s 1)2 ,
L L L L − −
L −
− − L L L L −L − L − − L − − L ,
− − − −
−
4
6
− −0
de
una
f
unc
i
o
n
pa
r
a
1
1
.
f (t) = t2 cos3 t Sean:
f (t) = t2 cos3 t f (0) = 0 ,
f (t) = 2t cos3 t
− 3t2 sin3 t f (0) = 0 f (t) = 2co s 3 t − 6t sin3 t − (6t sin3 t+9t2 cos3 t) = 2co s 3 t − 12t sin3 t − 9t2 cos3 t
A
pl
i
c
ando
e
l
t
e
or
em
,
a
L[f (t)] = s2F (s) − f (0) − f (0) L[2cos3 t − 12t sin3 t − 9t2 cos3 t] = s2L[t2 cos3 t] − 0 − 0 2L[cos3 t] − 12L[t sin3 t] − 9L[t2 cos3 t] = s2 L[t2 cos3 t] 2L[cos3 t] − 12L[t sin3 t] = s2 L[t2 cos3 t] + 9 L[t2 cos3 t] s s2 + 9
2
t
r
= (s2 + 9)L[t2 cos3 t] ⇐ − 12 (s2 6s + 9)2
ans
f
or
m
ad
t sin3 t
de
e
6s (s2 + 9)2
s
an
te
ri
orm
pues
podem
os
ya
l
a
habi
obs
am
e
os
r
r
v
ar
e
s
que
uel
l
t
a
o
ent.
2s(s2 + 9) 36s = (s2 + 9) [t2 cos3 t] (s2 + 9)2 2s3 + 18s 36s = (s2 + 9) [t2 cos3 t] (s2 + 9)2 2s3 18s = [t2 cos3 t] (s2 + 9)2 (s2 + 9) 2s3 18s [t2 cos3 t] = 2 (s + 9)3
− −
L L
−
L
−
L .
H
al
l
f (t)
ar
m
e
di
ant
e
e
l
t
e
or
e
m
a
de
l
a
t
r
ans
F (s) 1
2
.
F (s) =
L−1 Sa
be
m
1 s(s o
s
e
4)
−
− 4) L−1 s −1 a qu
1 s(s
= eat
,
ent
on
4
7
ces
:
f
or
m
ad
de
l
a
i
nt
e
gr
al
da
L−1
1
=
´t
s(s 4) 1 1 = = (e4t 4 4 4 e4t
1
3
−
−
eaτ a
eaτ dτ =
0
at
|t0 = ea − a1
don
e
a =4
:
− 1)
.
1 s2 (s + 3)
F (s) =
L−1 s2(s1+ 3) = L−1 (s +1 a) = eat a=3 ´ L−1 s(s 1+ 3) = 0t e3τ dτ = 13 e3τ |t0 = 13 e3t − 13 S
i
,done
,
´ t 1 3t 1 = 0 e s2 (s + 3) 3 1 3t f (t) = (e 3t 1) 9
L−1
− 13
entocs
:
dτ =
1 3τ e 9
− 13 τ
t
1 3t e 9
= 0
− 13 t − 19
− −
1
4
.
3 s2 (s2
F (s) =
L−1 C
onci
− L − − 3 s2 (s2
L−1 L−1 1 t 3
R
ap
es
l
1
ol
s2
3
a2
= sinh at
a=3
,done
,
´t
:
−
−
ver
1 (sinh3 t 9
l
as
s
i
gui
|
−
−
t
1 sinh 3t 9
=
0
− 3t) ent
es
ec
uaci
ones
di
f
er
enci
al
es
,
ace
5
entocs
1 1 1 = 0 sinh3 τ dτ = cosh 3τ t0 = cosh3 t s(s2 9) 3 3 3 ´t 1 3 1 1 1 = 0 cosh 3τ dτ = sinh 3τ τ s2 (s2 9) 3 3 9 3
f (t) =
L
a
−1
endo
−
=
9)
9)
.
y + y = 0 , y(0) = 1
L{y + y} = L{0} 4
8
usand
o
l
a
t
r
an
s
f
or
m
ad
de
−
P
or
e
l
t
e
or
e
m
a
de
t
r
ans
f
or
m
ads
de
der
i
va
das
s2 Y (s)
− sy(0) − y(0) + Y (s) = 0 s2 Y (s) − s(1) − 0 + Y (s) = 0 Y (s)(s2 + 1) − 1 = 0 S
u
s
t
i
t
u
yen
do
pl
i
c
or
or
i
i
ci
al
2
s +1 1 = e−t s2 + 1
am
n
os
l
a
t
r
ans
f
or
m
ad
i
nve
r
s
a
Y (s)
a
L−1 p
val
1
Y (s) = A
el
l
i
n
eal
i
d
ad
y = e−x
1
6
.
y + 4y = 2 , y(0) = 0 , y (0) = 0
L [y + 4y] = L [2] 2 s2 Y (s) − sy(0) − y (0) + 4Y (s) = s S
u
s
t
i
t
u
yen
s2 Y (s)
do
el
val
or
i
n
i
ci
al
− 0 − 0 + 4Y (s) = 2s
2 Y (s)(s2 + 4) = 2 s Y (s) = s(s2 + 4) s
e
apl
i
ca
l
a
t
r
an
s
f
or
m
ad
a
i
nver
s
a
L−1 s(s22+ 4) = ω =2 L−1 s2 +ω ω2 = sin ωt ´ 2 L−1 s(s2 + 4) = 0t sin2 τ dτ = − 12 cos2 τ |t0 = − 12 cos2 t + 12 1 1 f (t) = − cos2 t 2 2 S
i
p
or
l
i
1
7
n
1 2
y=
eal
i
d
done
,
onces
ad
− 12 cos2 x
.
y + 16y = 4 ,
y(0) = 1 , y (0) = 0
A
ent
pl
i
c
am
os
t
r
ans
f
or
m
ad
de
L
apl
ac
e
a
L[y + 16y] = L[4] 4
9
l
a
e
c
uac
i
ón
4 s
s2 Y (s) + sy(0) + y (0) + 16Y (s) =
S
u
s
t
i
t
u
i
m
os
l
os
val
ore
s
i
ni
ci
al
es
,
4 Y (s)(s2 + 16) + s + 0 = s 4 4−s2 s s Y (s) = s =
−
(s2 + 16) A
pl
i
cam
os
t
r
ansf
y
d
m
ad
i
p
4
=
(s2 + 16)
or
es
ej
am
Y (s)
os
− s2
s(s2 + 16)
nver
s
a
2 4 s −1 4 s −1 = −1 = s(s2 + 16) s(s2 + 16) (s2 + 16) s −1 = cos ωt ω=4 s2 + ω 2 s −1 = cos 4t (s2 + 16) 4 −1 s(s2 + 16) ω −1 = sin ωt ω =4 s2 + ω 2 t ´t 4 1 1 −1 = 0 sin4 τ dτ = cos4 τ = cos4 t 2 s(s + 16) 4 4 0 1 1 cos4 t + 4 4 4 s2 1 1 3 1 1 − s(s2 + 16) = 4 cos4 t + 4 cos4 t = 4 cos4 t + 4
L L L E
s
−
nt
s
L L
p
or
l
i
eal
i
d
ad
,done
−
−
n
,done
L
a
−L
−
−
− 14
=
−
3 1 cos4 x + 4 4
y= 8
par
L
L
1
−
once
i
.
y A
pl
i
c
am
os
l
a
t
r
− 2y + 5y = 0 ,
ans
f
or
m
ad
de
L
y(0) = 2 , y (0) = 4;
apl
ac
e
a
l
a
e
c
uac
i
ón
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