Ejercicios Resueltos Ecuaciones Diferenciales

dy + 2y 2y = 0 dx  p(  p(x) = 2

´ 

e

2dx

e2x e2x ddxy + 2e 2e2x = 0

d 2x dx [e y ]

=0

´ 

d 2x dx [e y ]

=0

´ 

dx + c

e2x y = c y = ce−2x dy = 3y 3y dx dy dx

− 3y = 0  p(  p(x) = −3

´ 

e −3dx e−3x

e−3x ddxy

− 3e−3xy = 0

´  dy −3x ´  [ e y = 0 dx + c dx e−3x y = c y = ce3x

3

dy + 12y 12y = 4 dx dy dx

+ 4y 4y =

4 3

 p(  p(x) = 4 ´ 

e

4dx

e4x e4x ddxy + 4e 4e4x y = 43 e4x

´ 

d 4x dx [e y ]

=

´ 

e4x dx + c

e4x y = 14 e4x + c y=

1 4

+ ce−4x

y = 2y 2 y + x2 + 5



y

 − 2y = x2 + 5

´ 

e −2dx = e−2x e−2x y

´ 

5e−2x  − 2e−2xy = e−2xx2 + 5e

d 2x y] dx [e

−

=

´  −2x 2 ´  e x + 5 e−2x + c

(2x2 + 2x 2x + 1) + C  − 52 e−2x − 14 e−2x(2x y = − x2 − x2 − 14 + 52 + ce2x

e−2x y =

2

ydx

4(x + y 6 )dy )dy = 0 − 4(x

ydx = 4(x 4( x + y6 )dy )dy dx dy

=

4(x+y6 ) y

dx dy

=

4x y

+

4y6 y

dx dy

e−4

´ 

1 y dy

− 4yx = 4y5

log(y) e−4 log( elog(y )

4

−

1 dx y4 dy

− y1 4yx = y1 4y5 4

4

d 1 dy [ y 4 x] d 1 dy [ y4 x]

´ 

y −4 =

1 y4 x

= 4y

=4

´ 

ydy

= 2y2 + C 

x = 2y6 + cy4 xy + y = ex



y + x1 y =



´ 

e

1 x dx

= elog x = x

xy + xx y =



d xy] dx [xy]

´ 

ex x

d xy] dx [xy]

=

xex x

= ex

´ 

ex dx + c

xy = ex + c y = ex x−1 + cx−1 x dy dx

dy 2 +y = 2 dx y

+

y x

=

2 xy2

u = y 1−n

n=

−2

u = y1−(−2) = y 3 u1/3 = y

1 2/3 du 3u dx

−

=

dy dx

1 y4

1 2/3 du 3u dx

−

u1/3 x

+

=

2(u1/3 )2 x 1 2/3 3u

du dx

e3

´ 

1 x dx

+ 3 ux =

6 x

3

= e3log x = elog x = x3

x3 du 3x3 ux = x3 x6 dx + 3x d 3 dx [x u]

´ 

d 3 dx [x u]

= 6x2

=6

´ 

x2 + c

x3 u = 2x3 + c u = 2 + cx−3 u = y3 y 3 = 2 + cx−3 dy y 1/2 dx + y 3/2 = 1 dy dx

+

y 3/2 y 1/2

1 y 1 /2

=

u = y 1−n n =

−1/2

y (0) = 4

dy + y = y −1/2 ↔ dx

u = y 1−(−1/2) = y 3/2

u2/3 = y 2 1/3 du dx 3u

−

2 1/3 du dx 3u

−

=

dy dx

+ u2/3 = (u ( u2/3 )−1/2 2 1/3 3u

du dx

3

e2

´ 

dx

3

= e2x

+ 32 u =

3 2

3

3

3

3 3 2x u = e2x e 2 x du dx + e 2 2 3 d 2 x u] dx [e

´ 

3 d 2 x u] dx [e

=

3

= 32 e 2 x

´  3

2e

3

3 2x

dx + c

3

e 2 xu = e 2 x + c 3

u = 1 + ce− 2 x u = y3/2 3

y 3/2 = 1 + ce− 2 x y (0) = 4 43/2

3 = 1 + ce− 2 0

8 1=c c=7

−

3

y 3/2 = 1 + 7e 7e− 2 x y +



u = y 1−n

2 y= x

−2xy2

n =2

u = y 1−2 u = y −1 u−1 = y

−u−2 ddux = ddxy −u−2 ddux + x2 u−1 = −2x(u−1)2 −u2 du dx

2x − x2 u = 2x  p(  p(x) = − x2

e−2

´ 

1 x dx

x−2 ddux

= elog x

2

−

= x−2

− x−2 x2 u = x−22x

d 2 dx [x u]

−

= 2x−1

´ 

d 2 dx [x u]

−

=

´ 

2x−1 dx + c

x−2 u = 2 log log x + c u = 2x2 log x + cx2 u = y−1 1 2x2 log x+cx2

y=

y + xy = xy −1/2



n=

−1/2 u = y 1−n u = y 1−(−1/2) u = y 3/2 y = u2/3 dy dx

2 1/3 3u

= 23 u−1/3

+ xu2/3 = x(u2/3 )−1/2

−

2 1/3 3u du dx

+ 32 xu = 32 x

p(x) = 32 x

3

e2 3

e4x

´ 

2

du dx

´ 

xdx 3

2

+ e4x

2

3

= e4x

3

3 2 xu

2

= e4x

3 2x

d 34 x2 u dx e

= 32 xe 4 x dx + c

d 34 x2 u dx e

=

3

2

2

3

3 2

´  3

2

3

xe 4 x dx + c 2

e4x u = e4x + c 3

2

u = 1 + ce− 4 x u = y3/2

3

y 3/2 = 1 + ce− 4 x

2

1.(2x (2x M ( M (x, y) = 2x 2x ∂M  ∂y

=

1)dx + (3y (3y + 1)dy 1)dy = 0 − 1)dx

N (x, y) = 3y 3y + 1 − 1; N (

∂N  ∂x ∂M  ∂y

=0

∂N  ∂x

=0

f x (x, y) = M (x, y) f x (x, y) = 2x

−1 g (y )

´  ∂M  ∂x

=2

´ 

xdx

f ( f (x, y) = x2

∂f  ∂y

− ´  dx + g(y)

− x + g (y ) = g (y )

g  (y ) = 3y + 1

´   ´  ´  g (y) = 3 ydy + dy + c g(y ) = 32 y2 + y + c

x2

(seny

− x + 32 y2 + y = c

ysenx)dx + (cosx (cosx + xcosy − y )dy = 0 − ysenx) M ( M (x, y) = seny − ysenx N ( N (x, y) = cosx + xcosy − y ∂M  − senx ∂y = cosy ∂N  ∂x = −senx + cosy

∂M  ∂y

=

∂N  ∂x

f x (x, y) = seny

´ 

− ysenx ´ 

f x (x, y)dx = (seny

f ( f (x, y) = xseny

ysenx)dx − ysenx)

cosx ) + g (y )... − y(−cosx)

f y (x, y) = cosx + xcosy + g  (y ) = cosx + xcosy g  (y ) =

−y

−y

´   ´  g (y ) = − ydy + c

− 12 y2 + c

g (y ) =

f ( f (x, y) = xseny + ycosx

xseny + ycosx (3x (3x2 y + ey )dx =

− 12 y2

− 12 y2 = c

−(x3 + xey − 2y)dy

M ( M (x, y) = 3x 3 x2 y + ey N ( N (x, y) = x3 + xey

− 2y

M y (x, y) = 3x 3 x2 + ey N x (x, y) = 3x2 + ey M y (x, y) = N x (x, y) f x (x, y)

g (y ) f ( f (x, y) = (3x (3x2 y + ey )dx

´ 

f ( f (x, y) = x3 y + xey + g (y )

f y (x, y) = x3 + xey + g  (y) = x3 + xey g (y ) =



− 2y

− 2y

−2 ´  ydy + c g ( y ) = −y 2 + c

g (y ) =

x3 y + xey

− y2 = c

)dx + (3x (3x2 − 4xy)d xy )dyy = 0 − 2y2)dx M y (x, y) = 6x − 4y N x (x, y) = 6x − 4y (6xy (6xy

f x (x, y)

´ 

f ( f (x, y) = (6xy (6xy 3 x2 y f ( f (x, y) = 3x

− 2xy2 + g(y)

f y (x, y) = 3x2

3x2

− 2y2)dx

− 4xy + g(y)

− 4xy + g(y) = 3x2 − 4xy

g (y ) = 0



g (y ) = c

3x2 y

− 2xy2 = c

4x + 6)dx 6)dx + (2x (2x − 3x2 y 2 − 1)dy 1)dy = 0 − 2xy3 + 4x y(−1) = 0 M y = 2 − 6xy 2 = N X

(2y (2y

f x (x, y)

´ 

f ( f (x, y) = (2y (2y f ( f (x, y) = 2xy 2 xy

4x + 6)dx 6)dx − 2xy3 + 4x

− 3x2y3 + 2x 2x2 + 6x 6x + g (y )

f x (x, y) = 2x

− 3x2y2 + g(y)

N ( N (x, y) 2x

2 x − 3x2 y 2 − 1 g (y ) = −1 − 3x2y2 + g(y) = 2x

g (y ) =

2xy

−y + c

2x2 + 6x 6x − y = c − x2y3 + 2x y (−1) = 0 2(−1)2 + 6(−1) = c c = −4

2xy

− x2y3 + 2x 2x2 + 6x 6 x − y = −4

( xy sin x + 2y 2y cos x)dx )dx + 2x 2x cos xdy = 0;

−

µ(x, y) = xy M y (x, y) =

−x sin x + 2 cos cos x N x (x, y) = −2x sin x + 2 cos cos x N X  = M y xy( xy ( xy sin x + 2y 2y cos x)dx )dx + xy(2 xy (2x x cos x)dy )dy = 0

−

( x2 y2 sin x + 2xy 2xy 2 cos x)dx )dx + (2x (2x2 y cos x)dy )dy = 0

−

−2yx2 sin x + 4xy 4xy cos x N X (x, y) = 4xy cos x − 2x2 y sin x

M y (x, y) =

M Y  Y  = N X

f x (x, y) =

−x2y2 sin x + 2xy 2xy 2 cos x

f ( f (x, y) = ( x2 y 2 sin x + 2xy 2xy 2 cos x)dx

´ 

−

f ( f (x, y) = x2 y 2 cos x + g (y )

f y (x, y) = 2x2 y cos x + g (y )



N x 2x2 y cos x + g (y) = 2x 2 x2 y cos x



g (y ) = 0



g (y ) = c

f ( f (x, y) = x2 y 2 cos x + c

y = 2x2



´ 

y =2



´ 

x2 dx + c

y = 23 x3 + c1



´ 

y =



2 3

(x3 + c1 )dx + c2

´ 

y = ( 23 )( 14 )x4 + xc1 + c2 y = 16 x4 + c1 x + c2

y

´ 

 = sen( sen(kx) kx)

sen(kx) kx)dx + c1  = ´  sen( y = −kcos( kcos(kx) kx) + c1 ´  ´  ´  y = −k cos( cos(kx) kx)dx + c1 dx + c2 y = −k 2 sen( sen(kx) kx) + xc1 + c2 ´  ´  ´  ´  2 y = −k sen( sen(kx) kx)dx + c1 xdx + c2 dx + c3 y

y = k 3 cos( cos(kx) kx) + 12 c1 x2 + c2 x + c3 y

´ 

y

 = x1

 = ´  x1 dx + c1

y = log x + c1



´ 

y =



´ 

log xdx + c1

´ 

dx + c2

y = x log x

− 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 

´ 

´ 

2

´ 

y =

y y

= 12 x2 y = 12 = 16 x3

´ 

y



´ 



´ 

xdx +

´ 

sin xdx + c1

− cos x +´ c1 ´  x2 dx − cos xdx + c1 dx + c2  − sin x + c1x + c2 y  = x sin x, y(0) = 0 y (0) = 0 y (0) = 2 ´ 

 = ´  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 y = −3sin x + x cos x + 12 c1 x2 + c2 x + c3

xy + y = 0





dy dx

 p(  p(x) =

dp dx

d2 y dx2

=

xp + p = 0



1 x dx

− p1 d p ´  1 ´  1 d x = − x  p d p + c1 log x = − log p log p + log c1 =

log x = log( c p1 ) x=

c1  p

 p(  p(x) = x=

dy dx

c! dy/dx

x = c1 dx dy

´  1

x dx =

log x =

1 c1

´ 

1 c1 y

dy + c2

+ c2

y = c1 log x + c2 (x

− 1)y 1)y  − y

 p(  p(x) = (x

dy dx

dp dx

d2 y dx2

=

1) p − p = 0 − 1) p

(x

x 1 x 1 p

 − x−1 1 p = 0

− −

− 1)

 − x−1 1 p = 0

 p

dp dx

− x−1 1 p = 0

dp 1 dx = x 1 p 1 1  p dp = x 1 dx

− −

´  1

 p dp =

´ 

1 x 1 dx

−

+ c1

log( p log( p)) = log(x log(x

− 1) + log(c log(c1 ) log( p log( p)) = log[c log[c1 (x − 1)]  p = c1 (x − 1)  p =

dy dx

dy dx

− 1) dy = c1 (x − 1)dx 1)dx ´ 

= c1 (x

´ 

dy = c1 (x

y = c1 12 x2

y +y



− 1)dx 1)dx + c2

− x + c2

 − 2y = 0

m2 + m

−2= 0 (m + 2)(m 2)(m − 1) = 0 m1 = −2 m2 = 1

y = emx

y1 = e−2x y2 = ex y(x) = c1 e−2x + c2 ex y

(m

− 1)2 = 0

m1,2 = 1

y = emx y1 = ex y2 = y1

´  e´  p(y)dy

y2 = ex

´  e2x

y12

dx

e2x dx

y2 = ex x y(x) = c1 ex + c2 xex 4y

 − 8y + 5y 5y = 0

 − 2y + y = 0 m2 − 2m + 1 = 0

4m2

− 8m√ + 5 = 0

m1,2 = 8± m1,2 = 1

64 80 8 1 2i

−

± 1

1

y = c1 ex ei 2 x + c2 ex e−i 2 x 1

1

y = ex (c1 ei 2 x + c2 e−i 2 x ) y = ex (c1 cos 12 x + c2 sen 12 x) 3y

 − 2y − 8y = 0

3m2

− 2y − 8 = 0 (3m (3m + 4)(m 4)(m − 2)

m1 = 2 m2 =

− 43

y = emx

y1 = e2x y2 =

−e−

4 3x

4

y(x) = c1 e2x + c2 e 3 x yv

− 10 10yy + 9y 9y = 0

m5

10m m3 + 9m 9m = 0 − 10 m(m4 − 10 10m m2 + 9) = 0 m1 = 0 (m2 − 9)(m 9)(m2 − 1) m2,3 = ±3 m4,5 = ±1 y1 = e0 = 1 y2 = e3x y3 = e−3x y4 = ex y5 = e−x y(x) = c1 + c2 e3x + c3 e−3x + c4 ex + c5 ex y + 4y 4y + 3y 3y = 0 y (0) = 2 y (0) =





m2 + 4m 4m + 3 = 0



−3

√ 

m1,2 = −4±2 −36 m1,2 =

− 2 ± 3i

y(x) = e−2x (c1 cos3x cos3x + c2 sin3x sin3x) y (x) = e−2x ( 3c1 sin3x sin3x + 3c 3c2 cos3x cos3x)



cos3x + c2 sin3x sin3x) − 2e−2x(c1 cos3x y (0) = 2 y (0) = −3

−

y (0) = 2 2 = e0 (c1 cos cos 0 + c2 sin sin 0) 2 = c1 y  (0) =

−3

−3 = e0(−3c1 sin sin 0 + 3c 3c2 cos0) − 2e0 (c1 cos cos 0 + c2 sin sin 0) −3 = 3c2 − 2c1 −3 = 3c2 − 2(2) 3 c2 −3 + 4 = 3c

c2 =

1 3

(2cos3x + y(x) = e−2x (2cos3x d4 y dx4

m4

1 3

sin3x sin3x)

4

d y 18yy = 0 − 7 dx − 18 2

− 7m2 − 18 = 0

y + 3y 3y + 2y 2y = 6



yh = y + 3y 3y + 2y 2y = 0





m2 + 3m 3m + 2 = 0 (m

1)(m − 2) − 1)(m

m1 = 1 m2 = 2

yh = c1 ex + c2 e2x



A y p = A y  p  p = 0

 y p  p = 0 0 + 3(0) + 2A 2A = 6 A=3 y (x) = yh + y p y(x) = c1 ex + c2 e2x + 3 y + y = sin x



y +y = 0



m2 + 1 = 0

−1 m1,2 = ±√−1 m1,2 = α ± βi m1,2 = ±i m2 =

α=0

β=1

yh = c1 eαx cos βx + c2 eαx sin βx yh = c1 cos x + c2 sin x

sin x A sin x + B cos x y +y = 0



xn

y p = Ax sin x + Bx cos x y  p  p = A sin x + Ax cos x + B cos x

 − Bx sin x y p  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 2A cos x

− 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 = 0

−2B = 1 y p =

− 12 x cos x

A=0

B=

− 12

y(x) = yh + y y(x) = c1 cos x + c2 sin x y m2

− 12 x cos x

10yy  + 25y 25y = 30 30x x+3  − 10

10m m + 25 = 0 − 10

m1,2 = 5

yh = c1 e5x + c2 xe5x 30 30x x+3

Ax + B

y p = Ax + B y  p  p = A

 y p  p = 0

10(A) + 25(Ax 25(Ax + B ) = 30x 30x + 3 −10(A 25 25A A = 30

A=

6 5

25 25B B

− 10 10A A =3 25 25B B − 10( 65 ) = 3 25 25B B = 3 + 12 B= y p =

3 5 6 5x

+

3 5

y(x) = yh + y p y(x) = c1 e5x + c2 xe5x + 65 x + 1 4y

3 5

 + y + y = x2 − 2x

1 4y + y + y = 0 1 2 4m + m + 1 = 0





m1,2 =

−2

yh = c1 e−2x + c2 xe−2x f ( f (x) = x2

− 2x

y p = Ax2 + Bx + C  y p = 2Ax 2 Ax + B y p = 2A 1 (2A) + 2Ax 2 Ax + B + Ax2 + Bx + C  = 4 (2A 1 2 2Ax + Bx + C  = x2 2 A + B + Ax + 2Ax

x2

− 2x

− 2x

A=1 2A + B = 2 B =2 1 2A + 1 2A +

−2 = 0

B + C  = 0 C  = 0

− 12 A = − 12 y p = x2 − 12

C  =

y(x) = yh + y p

y(x) = c1 e−2x + c2 xe−2x + x2 y + 3y 3y =



y

−48 48x x2 e3x

− 12



m2 + 3 = 0

√−3 m = √3i 1,2 √ √ = c cos 3x + c sen 3x

m1,2 = yh

1

2

48x x2 e3x −48

y p = e3x (Ax2 + Bx + C )

3x 2 3x y  p (2Ax + B )  p = 3e (Ax + Bx + C ) + e (2Ax

 3x 2 3x y p (2Ax + B ) + 3e3x (2Ax (2Ax + B ) +  p = 9e (Ax + Bx + C ) + 3e (2Ax 3x 3x 2 3x 3x e (2A (2A) = 9e (Ax + Bx + C ) + 3e 3 e (4Ax (4Ax + 2B 2B ) + e (2A (2A)

9e3x (Ax2 + Bx + C ) + 3e3x (4Ax (4Ax + 2B ) + e3x (2A (2A) + 9e3x (Ax2 + Bx + C ) + 3e 3 e3x (2Ax (2Ax + B ) = 48 48x x2 e3x

−

9e3x Ax2 + 9e3x Bx + 9e3x C + C + 12 12ee3x Ax + 6e3x B + 2e3x A + 9e3x Ax2 + 9e3x Bx + 9e 9e3x C + C  + 6e 6e3x Ax + 3e 3e3x B = 48 48x x2 e3x 9A + 9A 9A = 18 18A A= A=

−48

−48

− 83

B =0

C  = 0 y

 − y = −3 y  m2 − m = 0 m(m − 1) = 0

m1 = 0 m2 = 1

−

yh = c1 e0x + c2 ex = c1 + c2 ex c1

−3

y p = Ax

y p = Ax y  p  p = A

 y p  p = 0 0

− A = −3

A=3

y p = 3x 3x

y(x) = yh + y p y(x) = c1 + c2 ex + 3x 3x y

− 6y = 3 − cosx

yh = y

m3

− 6m2 = 0 m2 (m − 6) = 0

− 6y = 0

m1,2 = 0 m3 = 6 yh = c1 + c2 x + c3 e6x 3 Bcosx + Csenx

− cosx

y p1 = A y p2 =

y p1

y p1 = Ax2 y p = Ax2 + Bcosx + Csenx y  p  p = 2Ax

 − Bsenx + Ccosx y p 2 A − Bcosx − Csenx  p = 2A y p  p = Bsenx − Ccosx Bsenx

12A A + 6Bcosx 6Bcosx + 6Csenx 6Csenx = 3 − Cosx − Ccosx − 12 12A A = 3 A = − 14 −12 6B − C  = 1 6C  + B = 0

B= C  = y p =

6 37 1 37 1 3 2x

y(x) =

6 1 37 cosx + 37 senx c1 + c2 x + c3 e6x 14 x2

+

−

+

6 37 cosx

+

1 37 senx

y + 2y 2y + y = senx + 3cos 3cos22x





yh = y + 2y 2y + y = 0





m2 + 2m 2m + 1 = 0 (m + 1)2 = 0 m1,2 =

−1

yh = c1 ex + c2 xex y p = Acosx + Bsenx + Ccos2 Ccos2x + Dsen2 Dsen2x y p =

Csen2x + 2Dcos 2Dcos22x −Asenx + Bcosx − 2Csen2

y = −Acosx − Bsenx − 4Ccos2 Ccos2x − 4Dsen2 Dsen2x  p

Ccos2x−4Dsen2 Dsen2x−2Asenx+2 Asenx+2Bcosx Bcosx−4Csen2 Csen2x+ −Acosx−Bsenx−4Ccos2 4Dcos2 Dcos2x + Acosx + Bsenx + Ccos2 Ccos2x + Dsen2 Dsen2x = senx + 3cos2 cos2x

−3Ccos2 Ccos2x − 3Dsen2 Dsen2x − 2Asenx + 2Bcosx − 4Csen2 Csen2x + 4Dcos2 Dcos2x = senx + 3Cos 3Cos22x

C  + 4D 4D = 3 −3C + −3D − 4C  = 0

C  =

D=

9 25 12 25

−2A = 1

A=

2B = 0 B = 0

− 12

y(x) = c1 ex + c2 xex

− 12 cosx + 259 cos2 cos2x + 12 sen2x 25 sen2

y + y = sec x



yh = y + y = 0



m2 + 1 = 0 m2 = m1,2 = m1,2 = α

√−1

± βi

−1 m1,2 =

±i

α=0β=1

yh = c1 cosx + c2 senx y1 = cosx

y2 = senx

y1 = cosx y2 = senx y1 =

W  =

y1 y1

W 1 =

y2 = y2

−senx y2 = cosx

cosx senx = [(cosx [(cosx)( )(cosx cosx)] )] senx senx cosx cosx cos2 x + sen2 x = 1

−

0 y2 0 senx = = [(0)(cosx [(0)(cosx)] )] [(senx [(senx)( )(secx secx)] )] =  f ( f (x) y2 secx secx cosx cosx senxsecx = senx cosx = tanx

−

−

W 2 =

u1 =

W 1 W 

− [(senx [(senx)( )(−senx)] senx)] =

y1 y1

0 = f ( f (x)

= −tanx = 1

−

−

cosx 0 = [(cosx [( cosx)( )(secx secx)) senx senx secx secx cosxsecx = cosx cosx = 1

−

−tanx u2 =

u1 =

W 2 W 

=

1 1

senx)] = − (0)(−senx)]

ln(cosx)] cosx)] = ln( ln(cosx) cosx) − ´  tanxdx = −[−ln( = 1 u2 =

´ 

dx = x

y p = u1 y1 + u2 y2 y p = ln( ln(cosx) cosx)cosx + xsenx y(x) = yh + y p y (x) = c1 cosx + c2 senxi + cosxln( cosxln(cosx) cosx ) + xsenx y + y = senx



yh = y + y = 0



m2 + 1 = 0 m2 = m1,2 =

±√−1

α=0

−1 m1,2 =

±i

β =1

yh = eαx (c1 cosβx + c2 senβx) senβx) yh = e0x (c1 cosβx + c2 senβx) senβx)

yh = c1 cosx + c2 senx y1 y2 y1 = cosx y1 =

−senx

y2 = senx y2 = cosx

W  =

cosx senx = cos2 x + sen2 x = 1 senx senx cosx cosx

−

W 1 =

0 senx = senx senx cosx cosx

−sen2x

cosx 0 = senxcosx senx senx senx senx

W 2 =

−

u1 u2 2

− sen1 x = −sen2x ´  u1 = − sen2 xdx = x2 − 14 sen2 sen2x u1 =

u2 = u2 =

´ 

senxcosx 1

= senxcosx

senxcosxdx = 12 sen2 x

− 14 sen2 sen2x)cosx + 12 sen2 x(senx) senx) y p = 12 xcosx − 14 cosxsen2 cosxsen2x + 12 sen3 x

y p = u1 y1 + u2 y2 = ( x2

y(x) = y p + yh y (x) = c1 cosx + c2 senx + 12 xcosx

cosxsen2x + 12 sen3 x − 14 cosxsen2

y  + y = cos2 x yh = y  + y = 0 yh = c1 cosx + c2 senx y1 y2 y1 = cosx y1 =

−senx

y2 = senx y2 = cosx

cosx senx = cos2 x + sen2 x = 1 senx senx cosx cosx

W  =

−

0 senx = 2 cos x cosx

W 1 =

cosx senx enx

W 2 =

−

−senxcos2x

0 = cos3 x cos2 x

u1 u2

−senxcos2x = −senxcos2x

u1 = u1 =

−

´ 

1

senxcos2 xdx =

u2 = u2 = y p = u1 y1 + u2 y2 =

´ 



− −



cos3 x cos3 x = 3 3

cos3 x = cos3 x 1

cos3 xdx = senx

 

3

− sen3 x

cos3 x (cosx) cosx) + senx 3

y p =

cos4 x + sen2 x 3



−



4

− sen3 x

cos4 x y(x) = c1 cosx + c2 senx + + sen2 x 3 y

 − y = cosh x y  − y = 0 m2 − 1 = 0 √ m2 = 1 m1,2 = ± 1 = ±1 yh = c1 ex + c2 e−x y1 y2 y1 = ex y1 = ex y2 = e−x y2 =

sen3 x 3

−e−x

−

sen4 x 3

(senx) senx)

W  =

ex ex

W 1 =

e−x = e−x

−

0 coshx

e−x = e−x

−

W 2 = u1

−e−x(ex) − ex(e−x) = −1 − 1 = −2

ex ex

coshx) = −e−x coshx −e−x(coshx)

0 = ex coshx coshx

u2 x u1 = −e −coshx = 12 e−x coshx 2 −

u1 =

1 2

´  −x e coshxdx = 18 e−2x (2e (2e2x x − 1) u2 =

ex coshx = 2

−

− 12 excoshx 2x

u2 = y p =

− 12 ´  excoshxdx = − 12 [ x2 + e4

ex [ 18 e 2x (2e (2e2x x

−

2x

− 1)] + (−e−x)(− x − e ) 4

y p = 18 e−x (2e (2e2x x

−x

− 1) + xe4

y (x) = c1 ex + c2 e−x + 18 e−x (2e (2e2x x y + 3y 3y  + 2y 2y =

+

1 1 + ex

(m + 2)(m 2)(m + 1) = 0

−2 m2 = −1

yh = c1 e−2x + c2 e−x y1 y2

y2 = e−x

−x

− 1) + xe4

m2 + 3m 3m + 2 = 0

y1 = e−2x y1 =

8

ex 8

yh = y  + 3y 3y  + 2y 2y = 0

m1 =

]

−2e−2x y2 = −e−x

+

ex 8

W  =

e−2x 2e−2x

−

e−x = (e ( e−2x )( e−x ) (e−x )( 2e−2x ) = e−x e−3x

−

−

W 1 = W 2 =

−

0 1 1+ex

e−x = e−x

−

e−2x 2e−2x

−

0 1 1+ex

−

−

−e−3x + 2e 2e−3x =

e−x 1 + ex

e−2x = 1 + ex

u1 ,u2

u1 =

−

e−x 1 + ex = e 3x −

u1 =

1 e−x = −2x = − x x 3 (e )(1 + e ) e (1 + ex )

− e−2x 1+ e−x

− ´  e−2x 1+ e−xdx = −ex + ln( ln(ex + 1) − 1

e−2x 1 e−2x ex = u2 = 1 + = e−3x (e−3x )(1 + ex ) e−x + 1 u2 =

´ 

1 dx = x + ln( ln(e−x + 1) − x e +1

y p = (e−2x )[ ex + ln( ln(ex + 1)

−

− 1] + [x[x + ln( ln(e−x + 1)](e 1)](e−x )

−e−x + e−2xln( ln(ex + 1) − e−2x + xe−x + e−x ln( ln(e−x + 1) y (x) = c1 e−2x + c2 e−x − e−x + e−2x ln( ln(ex + 1) − e−2x + xe−x + e−x ln( ln(e−x + 1) 3y − 6y  + 6y 6y = ex secx yh = 3y − 6y  + 6y 6y = 0 3m2 − 6m + 6 = 0 a = 3 b = −6 c = 6 y p =

y1 y2

√36 − 72 √−36 (−6) ± (−6)2 − 4(3)(6) − 6 m1,2 = = ± =1± = 1±i

 

2(3)

6

6

α=1 β=1 yh = ex (c1 cosx + c2 senx) senx) y1 y2

6

y1 = ex cosx y1 = ex cosx

− exsenx

y2 = ex senx y2 = ex senx + ex cosx

ex cosx ex senx = (ex cosx)( cosx)(eex senx + ex cosx) cosx) ex cosx ex senx ex senx + ex cosx (ex senx)( senx)(eex cosx ex senx) senx) = ex (cosxsenx + cos2 x cosxsenx + sen2 x)

W  =

−

−

−

−

W  = ex (cos2 x + sen2 x) = ex W 1 =

0 x e secx

ex senx = (ex senx)( senx)(eex secx) secx) = ex senx + ex cosx ex tanx

−

−

W 2 =

ex cosx ex cosx ex senx

−ex( senx )= cosx

cosx 0 = (ex cosx)( cosx)(eex secx) secx) = ex ( ) = ex e secx cosx x

−

u1 u2 x

= −tanx − e tanx ex ´  u1 = − tanxdx = −(−lncosx) lncosx) = lncosx u1 =

ex  u2 = x = 1 e

u2 =

´ 

dx = x

y p = lncosx( lncosx(ex cosx) cosx) + x(ex senx) senx) y (x) = ex (c1 cosx + c2 senx) senx) + ex cosxlncosx + xex senx

x2 y

 − 2y = 0

y = xm

y = xm y  = mxm−1 y  = (m

[(m x2 [(m

1)mxm−2 − 1)mx

1)mxm−2 ] − 2(x 2(xm ) = 0 − 1)mx

x2 [(m [(m

− 1)mx 1)mxm x−2 ] − 2(x 2(xm ) = 0 (m − 1)mx 1)mxm − 2xm = 0 xm [(m [(m − 1)m 1)m − 2] = 0 xm (m2 − m − 2) = 0 m2

−m−2 = 0 (m + 1)(m 1)(m − 2) = 0 m1 = −1 m2 = 2 y = c1 x−1 + c2 x2 x2 y  + y = 0 y = xm y = xm y  = mxm−1 y  = (m

x2 [(m [(m (m

− 1)mx 1)mxm−2

− 1)mx 1)mxm−2 ] + xm = 0

− 1)mx 1)mx2 xm x−2 + xm = 0 (m2 − m)xm + xm = 0 xm (m2 − m + 1) = 0 m2

α=

1 2

β=

1 2

√3

−m+1=0 √ m1,2 = 12 ± 12 3i 1

1

y = c1 x 2 + 2

√ 3i

1

1

+ c2 x 2 − 2

√ 3i

xiβ = (e ( elnx )iβ = eiβlnx

xiβ = cos( cos(βlnx) βlnx) + isen( isen(βlnx) βlnx) x−iβ = cos( cos(βlnx) βlnx)

− isen( isen(βlnx) βlnx)

xiβ + x−iβ = cos( cos(βlnx) βlnx) + isen( isen(βlnx) βlnx) + cos( cos(βlnx) βlnx) xiβ

− isen( isen(βlnx) βlnx) = 2cos( cos(βlnx) βlnx)

− x−iβ = cos( cos(βlnx) βlnx) + isen( isen(βlnx) βlnx) − cos( cos(βlnx) βlnx) + isen( isen(βlnx) βlnx) = 2isen 2 isen((βlnx) βlnx) y = C 1 xα+iβ + C 2 xα−iβ y1 = xα (xiβ + x−iβ ) = 2x 2 xα cos( cos(βlnx) βlnx) y2 = xα (xiβ

− x−iβ ) = 2xαisen( isen(βlnx) βlnx)

y1 = xα cos( cos(βlnx) βlnx)

= xα sen( sen(βlnx) βlnx )

y = xα [c1 cos( cos(βlnx) βlnx) + c2 sen( sen(βlnx)] βlnx)] 1

√

√

y = x 2 [c1 cos( cos( 12 3lnx) lnx) + c2 sen( sen( 12 3lnx)] lnx)] x2 y  + xy  + 4y 4y = 0

y = xm y  = mxm−1 y  = (m

x2 [(m [(m

− 1)mx 1)mxm−2

− 1)mx 1)mxm−2 ] + x(mxm−1 ) + 4(x 4(xm ) = 0 xm (m2 − m + m + 4) = 0 xm (m2 + 4) = 0 m2 =

−4

±√−4 m1,2 = ±2i

m1,2 =

α=0β=2 y = x0 (c1 cos2 cos2lnx + c2 sen2 sen2lnx) lnx) y = c1 cos2 cos2lnx + c2 sen2 sen2lnx x2 y

 − 3xy − 2y = 0 y = xm y  = mxm−1

− 1)mx 1)mxm−2

y  = (m

x2 [(m [(m

1)mxm−2 ] − 3x(mxm−1 ) − 2(x 2(xm ) = 0 − 1)mx xm [(m [(m2 − m) − 3m − 2] = 0 xm (m2 − 4m − 2) = 0 √ m1,2 = 2 ± 6

√ 6

y = c2 x2+

√ 6

+ c1 x2−

25 25x x2 y  + 25xy 25xy  + y = 0

y = xm y  = mxm−1 y  = (m

25 25x x2 [(m [(m

− 1)mx 1)mxm−2

− 1)mx 1)mxm−2 ] + 25x 25x(mxm−1 ) + xm = 0

xm [25m [25m2

25m m + 25m 25m + 1] = 0 − 25

25 25m m2 + 1 = 0

m1,2 =

 

± − 251 = ± 15 i

α = 0, 0, β =

1 5

1 1 y = x0 (c1 cos lnx + c2 sen lnx) lnx) 5 5 1 1 y = c1 cos lnx + c2 sen lnx 5 5 x2 y  + 5xy 5xy  + y = 0

y = xm y  = mxm−1 y  = (m

− 1)mx 1)mxm−2

[(m 1)mx 1)mxm−2 ] + 5x 5 x(mxm−1 ) + xm = 0 x2 [(m xm (m2 m + 5m 5m + 1) = 0 2 m + 4m 4m + 1 = 0 m1,2 = 2 √  3 √  y = c1 x2+ 3 + c2 x2− 3

− − ±√

xy 

− 4y = x4 y = xm

y  = mxm−1 y  = (m

x2 y 

1)mxm−2 − 1)mx

− 4xy = x5 yh = x2 y  − 4xy  = 0 x2 [(m [(m − 1)mx 1)mxm−2 ] − 4x(mxm−1 ) = 0

xm (m2 m 4m) = 0 m(m 5) = 0 m1 = 0 m2 = 5 yh = c1 x0 + c2 x5 yh = c1 + c2 x5

−

− −

P ( P (x)y + Q(x)y  = f ( f (x) y 

x

− 4 xy = x3

f ( f (x) = x3 y1 y2  y1 = 1 y1 = 0 y2 = x5 y2 = 5x 5 x4 1 x5 W  = = 5x4 0 = 5x4 0 5x4 0 x5 W 1 = 3 = 0 x8 = x8 x 5x4 1 0 = x3 W 2 = 0 x3 u1 u2 − x8  u1 = 5x4 ´ = 15 x4 1 5 u1 = 15 x4 dx = 25 x 3 x 1  u2 = 5x´  4 = 5x u2 = 15 x1 dx = 15 lnx 1 5 y p = 25 x (1) + 15 lnx( lnx(x5 ) 5 1 5 y p = 25 x + x5 lnx y (x) = yh + y p 5 1 5 y (x) = c + c2 x5 25 x + x5 lnx

− −

−

−

−

−

− −

−

x2 y 

− xy + y = 2x y = xm

y  = mxm−1 y  = (m yh = x2 y 

1)mxm−2 − 1)mx

− xy + y = 0 x2 [(m [(m − 1)mx 1)mxm−2 ] − x(mxm−1 ) + xm = 0 m2 − m − m + 1 = 0 m2 − 2m + 1 = 0

(m 1)2 m1,2 = 1 yh = c1 x + c2 xlnx

−

y 

− x1 y + x1 y = x2 2

f ( f (x) = x2 y1 = x y2 = xlnx

y1 = 1

y2 = lnx + 1

x lnx = (x)(lnx )(lnx + 1) (lnx)(1) lnx)(1) = xlnx + x 1 lnx + 1 lnx + x = xln xx + x = xln(1) xln(1) + x = x 0 lnx W 1 = 2 = (lnx ( lnx)( )( x2 ) = x2 lnx lnx + 1 x x 0 W 2 = = x2 x 0 = 2 1 x2 u1 u2 2  x lnx u1 = x = 2xlnx 2 ´  lnx+1 u1 = 2 lnx = x2 x u2 = x2 ´  u2 = 2 x1 = 2lnx y p = y1 u1 + y2 u2 = x( lnxx+1 ) + xlnx(2 xlnx(2lnx lnx)) = lnx + 1+ W  =

−

−

−

−

−

x2 y 

2y = x4 ex − 2xy + 2y y = xm y  = mxm−1

y  = (m x2 [(m [(m

1)mxm−2 − 1)mx

1)mxm−2 ] − 2x(mxm−1 ) + 2x 2 xm = x4 ex − 1)mx x2 y  − 2xy  + 2y 2y = 0 2 m x (m − m − 2m + 2) = 0 m2 − 3m + 2 = 0 (m − 2)(m 2)(m − 1) = 0 m1 = 2 m2 = 1 yh = c1 x2 + c2 x y 

− x2 y + x2 y = x2ex 2

y1 y2 f ( f (x) = x2 ex y1 = x2 y1 = 2x 2x  y2 = x y2 = 1

− lnx = xlnx −

x2 x = x2 2x2 = x2 2x 1 0 x W 1 = 2 x = 0 x3 ex = x3 ex x e 1 x2 0 W 2 = = x4 ex 2x x2 ex u1 u2 − x3 ex  u1 = −x2 = xex ´  u1 = xex dx = ex (x 1) 4 x u2 = x−xe2 = x2 ex ´  2 x u2 = x e dx = ex (x2 2x + 2) y p = u1 y1 + u2 y2 = [ex (x 1)]x 1)]x2 + [e [ex (x2 2 x x 2 y p = x e (x 1) + xe (x 2x + 2) y (x) = y p + yh y (x) = c1 x2 + c2 x + x2 ex (x 1) + xex (x2 W  =

− −

−

−

−

−

−

− − − −

−

y

− 2x + 2)]x 2)]x − 2x + 2)

xy = 0

 −   − −   − −  − −  − − ∞ c xn n=0 n

y=

y =





∞ (n−1)nc 1)ncn xn−2 n=2

∞ (n 1)nc n 1)ncn xn−2 x ( ∞ n=2 n=0 cn x ) = 0 ∞ (n 1)nc ∞ c xn+1 = 0 1)ncn xn−2 n=2 n=0 n ∞ n(n n=3

2(1)c 2(1)c2 x0 2c2

∞ n(n n=3

k=n 2 n = k+2 n = k

−

1)c 1)cn xn−2

∞ c xn+1 = 0 n=0 n

∞ c xn+1 = 0 n=0 n

1)c 1)cn xn−2

k = n+1

−1

∞ n(n − 1)c 1)cn xn−2 − n=3

∞ c xn+1 = 0 n=0 n k 2c 2 ∞ 2)(k + 1)c 1)ck+2 xk − ∞ k=1 (k + 2)(k k=1 ck−1 x = 0 2c2 ∞ [(k + 2)(k 2)(k + 1)c 1)ck+2 − ck−1 ]xk = 0 k=1 [(k

  

2c2

(k + 2)(k 2)(k + 1)c 1)ck+2 ck+2 =





− ck−1 = 0

ck−1 (k + 2)(k 2)(k + 1)

k 2c2 = 0 c2 = 0 c0 3(2)

k = 1 c3 = k = 2 c4 =

c1 = 4(3)

c2 = 5(4)

k = 3 c5 =

= 16 c0

1 20 c2

1 12 c1

=0

← c2 = 0

k = 4 c6 =

c3 = 6(5)

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 = k = 7 c9 =

c5 =0 8(7)

c6 = 9(8)

← c5 = 0

1 1 72 ( 180 )c0

k = 8 c10 =

c7 = 10(9)

k = 9 c11 =

c8 =0 11(10)

=

1 12960 c0

1 10(9)(504) c1

← c8 = 0

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 )

y + x2 y + xy = 0





n y= ∞ n=0 cn x ∞ y = n=1 cn nxn−1 y = ∞ 1)nc 1)ncn xn−2 n=2 (n

     − ∞ (n n=2

−

1)nc 1)ncn xn−2 + x2



∞ c nxn−1 + x [ n=1 n

 

cn xn ] = 0

∞ (n − 1)nc n+1 n+1 1)ncn xn−2 + ∞ + ∞ =0 n=2 n=1 cn nx n=0 cn x ∞ ∞ n−2 n+1 n+1 0 2c2 x + 6c3 x n=4 (n − 1)nc 1)ncn x + n=1 cn nx + c0 x1 ∞ =0 n=1 cn x





 −  

k=n 2

 



k = n+1

2c2 x0 + 6c3 x ∞ 1)(k + 2)c 2)ck+2 xk+2−2 + ∞ 1)x 1)xk−1+1 + k =2 (k + 2 1)(k k =2 ck−1 (k k−1+1 c0 x 1 ∞ =0 k=2 ck−1 x ∞ 2c2 + 6c 6c3 x + c0 x k=2 (k + 1)(k 1)(k + 2)c 2)ck+2 xk + ck−1 (k 1)x 1)xk + ck−1 xk = 0 ∞ 2c2 + 6c 6c3 x + c0 x k=2 [(k [(k + 1)(k 1)(k + 2)c 2)ck+2 + ck−1 (k 1) + ck−1 ]xk = 0 (k + 1)(k 1)(k + 2)c 2)ck+2 + ck−1 (k 1) + ck−1





−

−

−

−

−

2c 2 = 0 c 2 = 0 6c3 + c0 = 0 c3 = 16 c0

−

kck−1 (k + 1)(k 1)(k + 2) k = 2, 3, 4,... 2c1 1 c4 = 3(4) = 6 c1 3c2 c5 = 4(5) =0 c2 = 0 4c3 2 1 c6 = 5(6) = 15 ( 16 c0 ) = 45 c0 5c4 5 1 5 c7 = 6(7) = 42 ( 6 c1 ) = 136 c1 6c5 6 c8 = 7(8) = 56 (0) = 0 7c6 7 1 7 c9 = 8(9) = 72 ( 45 )c0 = 72(45) c0 8c7 4 5 5 c10 = 9(10) = 45 ( 136 c1 ) = 45(34) c1 9c8 9 c11 = 10(11) = 110 (0) = 0 10c9 5 7 7 c12 = 11(12) = 66 ( 72(45) c0 ) = 66(72)(9) c0 ck+2 =

[(k 1)+1] ck 1 (k +1)(k +2)

−

−

=

← −

−

−

−

−

−

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 ]

− y  − 2xy + y = 0

n y= ∞ n=0 cn x ∞ y = n=1 cn nxn−1 y = ∞ 1)nc 1)ncn xn−2 n=2 (n

    −  −  −−   −− 

∞ (n 1)nc ∞ c nxn−1 + ∞ c xn = 0 1)ncn xn−2 − 2x n n n=2 ∞ (n 1)nc ∞ n=1 ∞ c nx=0 n−2 n n − 1) nc x 2 c nx + = 0 n n=2 n=1 n n=0 n ∞ ∞ ∞ n−2 n 2c2 n=3 (n 1)nc 1)ncn x − 2 n=1 cnnx + c0 n=1 cnxn = 0

 

     −  

k=n 2 k=n ∞ c xk = 0 k +2−2 k 2c 2 ∞ ( k + 2 1)(k 1)( k + 2)c 2) c x 2 ∞ k+2 k k=1 k =1 ck kx + c0 ∞ c kxk + ∞ c xkk=1 k 2c2 + c0 ∞ ( k + 1)(k 1)( k + 2)c 2) c x + = 0 k +2 k =1 k =1 k k=1 k k k k 2c2 + c0 ∞ ( k + 1)(k 1)( k + 2)c 2) c x 2 c kx + c x = 0 k k k +2 k =1 ∞ k k k 2c2 + c0 k=1 (k + 1)(k 1)(k + 2)c 2)ck+2 x 2ck kx + ck x = 0

− −

2c2 + c0 = 0 c2 = 12 c0 (k + 1)(k 1)(k + 2)c 2)ck+2 xk 2ck kxk + ck xk = 0 [(k [(k + 1)(k 1)(k + 2)c 2)ck+2 2ck k + ck ]xk = 0 (k + 1)(k 1)(k + 2)c 2)ck+2 2ck k + ck (2k (2k + 1)c 1)ck ck+2 = (k + 1)(k 1)(k + 2) k = 1, 2, 3, 4,...

−

− −

−

3c1 2(3)

c3 = 5c2 3(4)

c4 =

c6 =

9c4 5(6)

c7 =

11c5 6(7)

c0

=

11 7 42 ( 40 c1 )

=

c9 =

15c7 8(9)

=

15 11 72 ( 6(40) c1 )

+

5 4 24 x

+

=

11 6(40) c1

18c9 10(11)

=

161 72(240) c1

17(13) − 9(10)(56)(16) c0

=

1 161 55 ( 8(240) )c1

11 7 240 x

1 6 16 x

+

161 161 + 72(240) x9 + 55(8)(240) x11 17(13) 13 8 10 56(16) x + 90(56)(16) x



(x2 + 2)y 2)y + 3xy 3xy



y =



−

−y = 0

∞ c xn n=0 n ∞ c nxn−1

 

y=

n=1 n

∞ (n − 1)nc 1)ncn xn−2 n=2 n−1 − ∞n=0 cnxn = 0 (x2 + 2) ∞ 1)ncn xn−2 + 3x 3x ∞ n=2 (n − 1)nc n=1 cn nx ∞ (n − 1)nc ∞ c xn = n 1)ncn xn−2 +2 ∞ 1)ncn xn−2 + ∞ n=2 n=2 (n − 1)nc n=1 3cn nx − n=0 n y =



x2

=

13 − 161 c0) = − 56(16) c0

17c8 9(10)

+

7 40 c1

− 245 c0) = − 161 c0

13 56 (

7 5 40 x

=

9 30 (

=

1 3 2x



7 1 20 ( 2 c1 )

=

13c6 7(8)

c11 = y = c1

− 12 c0) = − 245 c0

c8 =

c10 =



5 12 (

=

7c3 4(5)

c5 =

= 12 c1











0









∞ (n − 1)nc ∞ c xn = 0 n 1)ncn xn + ∞ 2(n − 1)nc 1)ncn xn−2 + ∞ n=2 n=2 2(n n=1 3cn nx − n=0 n ∞ (n − 1)nc n 1)ncn xn + ∞ 2(n − 1)nc 1)ncn xn−2 + 3c 3c1 x ∞ n=2 n=2 2(n n=2 3cn nx − c0 + ∞ c x c xn = 0







     − −  − −   −  − −  −   −  − 1



n=2 n

3c1 x + c0 + c1 x ∞ 1)nc 1)ncn xn + 2(2 1)2c 1)2c2 x2−2 + 2(3 n=2 (n ∞ c xn = 0 n 1)3c 1)3c3 x3−2 ∞ 2(n 1)nc 1)ncn xn−2 + ∞ n=4 2(n n=2 3cn nx n=2 n 3c1 x + c0 + c1 x + 4c 4c2 + 12c 12c3 x ∞ =2 (n ∞ 3c nnx n n n=2 k=n

1)nc 1)ncn xn + ∞ 2(n 2(n ∞ c xn = 0n=4 n=2 n

2

1)ncn xn−2 + − 1)nc

k=n

3c1 x + c0 + c1 x + 4c 4c2 + 12c 12c3 x 2)c 2)ck+2 xk+2−2 + 4c1 x + c0 + 4c2 + 12 12cc3 x

−

∞ (k 1)kc 1)kck xk + ∞ 2(k 2(k + 2 − 1)(k 1)(k + k=2 ∞ 3c kxk ∞ c kx=2 k = 0 k k=2 k=2 k

∞ [(k 1)kc 1)kck + 2(k 2(k +1)(k +1)(k + 2)c 2)ck+2 + 3ck k − ck ]xk = 0 k=2 [(k

4c1 + 12c 12c3 = 0 c3 = c31 c0 + 4c 4c2 = 0 c2 = c40 (k 1)kc 1)kck + 2(k 2(k + 1)(k 1)(k + 2)c 2)ck+2 + 3c 3ck k ck 3kck + ck (k 1)kc 1)kck [ 3k + 1 k 2 k ]ck [ 4 k + 1 k 2 ]c k ck+2 = = = 2(k 2(k + 1)(k 1)(k + 2) 2(k 2(k + 1)(k 1)(k + 2) 2(k 2(k + 1)(k 1)(k + 2) k = 2, 2 , 3, 4, 5,... c2 = c40 c3 = c31 −4−2)c2 = 11 ( 1 c ) = 11 c c4 = (−6+1 2(3)(4) 24 4 0 96 0

−

− −

−

− −

− −

−

−

− −

− − − −

−

− − − − − − −

9)c3 1 1 1 c5 = ( 12+1 = 20 2(4)(5) 40 c3 = 2 ( 3 c1 ) = 6 c1 16)c4 31(11) 31 11 c6 = ( 16+1 = 31 2(5)(6) 60 c4 = 60 ( 96 c0 ) = (60)(96) c0 25)c5 1 11 c7 = ( 20+1 = 8444 c5 = 11 2(6)(7) 21 ( 6 c1 ) = 126 c1 36)c6 11(31) 59 11(31) c8 = ( 24+1 = 112 ( 60(96) c0 ) = 112(60)(96) c0 2(7)(8) 31(11) 6 11(31) 1 3 4 8 y = c0 [ 14 x2 + 11 90 x + 60(96) x + 112(60)(96) x ] + c1 [ 3 x

− − − −

−

−

−

−

−

≤t