AP Ecuaciones Diferenciales Ordinarias

8/16/2019 AP Ecuaciones Diferenciales Ordinarias

http://slidepdf.com/reader/full/ap-ecuaciones-diferenciales-ordinarias 1/21







hi = xx+1 − xi x

y = f (x, y)

y(n) = f (x,y,y

, y

,....,y(n+1))

y = f (x, y)

y = y(x)

y = f (x, y(x))

y = y(x)

y = f (x, y(x)) (xi, yi)

y = y(x)

(xi, yi)

(x, y)

x

dy

dx = x

dy =

x dx ⇒ y =

x2

2 + C

P = (x0, y0)

x0

P

y = x



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

L > 0

|f (x, y2) − f (x, y1)| ≤ L|y2 − y1|

D

y = f (x, y)

f (x, y)

D

xy

D

(x0, y0) ∈ D y(x)

y = f (x, y(x))

y(x0) = y0

h → 0

h → 0

h → 0



n

n

t = 0

y = y (x)

y

= f (x, y(x))

y = y(x)

m + 1

x0

x0

y(x) = y(x0)+y(x0)(x−x0)+y(x0)

2! (x−x0)2+

y(x0)

3! (x−x0)3+ ...+

ym(x0)

m! (x−x0)m+

ym+1(ξ )

(m + 1)!(x−x0)m+1

ξ ∈ (x0, x)

y = f (x, y(x)) y = f x + f · f y y = f xx + 2f · f xy + f 2 · f xy + f x · f y + f · f 2y

f x f x f y f y

h

xn = x0 + n · h n = 0, 1, 2, 3...

m = 2

yn+1 = yn + hf (xn, yn) + h2

2! (f x + f · f y)2 + ... +

h3

3! f (2)(ξ,y(ξ))

ξ ∈ (xn, xn+1) h = xn+1 − xn

y = −y + x + 1 [0;1] y0 = 1

x = 1

h = 0, 1



y = −y + x + 1

y = f x + f · f y = 1 + (−y + x + 1)(−1) = y + x

yn+1 = yn + h · (−yn + xn + 1) + h2

2! (yn − xn)2 + E T

y(x)

n = 0 ⇒ x0 = 0 ∧ y0 = 1

y1 = y0 + h · (−y0 + x0 + 1) + h2

2! (y0 − x0)2 = 1, 00500

n = 1 ⇒ x1 = 0, 1 ∧ y1 = 1, 00500

y2 = y1 + h · (−y1 + x1 + 1) + h2

2! (y1 − x1)2 = 1, 01859

n = 9 ⇒ x9 = 0, 9 ∧ y9 = 1, 30101

y10 = y9 + h · (−y9 + x9 + 1) + h2

2! (y9 − x9)2 = 1, 36171

y(1) ≈ 1, 36171

y = f (x, y(x))

y(x0) = y0

h

xn = x0 + n · h

n = 0, 1, 2, 3,... xn ≤ x ≤ xn+1 yn

x = xn

yn+1 = yn +

xn+1

xn

f (t, y(t)) dt

y(t)

y(t)

f

f

x

y

yn+1 = yn + h · f (xn, yn)

yn+1 = yn + h · y(xn, yn) + h2

2! y(ξ )

h2

h

O(h)



y = −y + x + 1

[0;1]

y0 = 1

x = 1 h = 0, 1

f (x, y) = −y + x + 1

yn+1 = yn + h (−yn + xn + 1)

n

yn

n = 0 ⇒ x0 = 0 ∧ y0 = 1

y1 = y0 + h (−y0 + x0 + 1) = 1,0000

n = 1 ⇒ x1 = 0, 1 ∧ y1 = 1

y2 = y1 + h (−y1 + x1 + 1) = 1,0100



n = 9 ⇒ x9 = 0, 9 ∧ y9 = 1, 28742

y10 = y9 + h (−y9 + x9 + 1) = 1,34868

y(1) ≈ 1, 34868

h

h

f (x, y)

xn, yn

mn = f (xn, yn) + f (xn+1, yn+1)

2

y = yn + h

2 ·

f (xn, yn) + f (xn+1, yn+1)

2 · (x − xn)

xn+1 = xn + h

yn+1 = yn + h

2 · [f (xn, yn) + f (xn+1, yn+1)]

xn+1

yn+1 = yn + h

2 [f (xn, yn) + f (xn+1, yn + h · f (xn, yn))]

h3 h2

O(h2)



y = −y + x + 1

[0;1]

y0 = 1

x = 1 h = 0, 1

f (x, y) = −y + x + 1

yn+1 = yn + h

2 [(−yn + xn + 1)

f (xn,yn)

+ (−yn + h (−yn + xn + 1) yn+1 Euler simple

+xn+1 + 1)

n yn

n = 0 ⇒ x0 = 0 , x1 = 0,1 ∧ y0 = 1

y1

= y0

+ h

2 [(−y

0 + x

0 + 1) + (−y

0 + h (−y

0 + x

0 + 1) + x

1 + 1) = 1, 00500

n = 1 ⇒ x1 = 0, 1 , x2 = 0,2 ∧ y1 = 1, 00500

y2 = y1 + h

2 [(−y1 + x1 + 1) + (−y1 + h (−y1 + x1 + 1) + x2 + 1) = 1, 01902

n = 9 ⇒ x0 = 0, 9 , x10 = 1 ∧ y9 = 1, 30723

y10 = y9 + h

2

[(−y9 + x9 + 1) + (−y9 + h (−y9 + x9 + 1) + x10 + 1) = 1, 36854

y(1) ≈ 1, 34854

xn+1

yn+1 = yn + h f (xn, yn)

yk+1n+1 = yn +

h

2 [f (xn, yn) + f (xn+1, ykn+1)]

y = −y + x + 1

[0;1]

y0 = 1

x = 0, 1

10−5



f (x, y) = −y + x + 1

ykn+1 = yn + h (−yn + xn + 1) n = 0, 1, 2, 3,...

k = 0 , n = 0 x0 ∧ y0y01 = y0 + h (−y0 + x0 + 1) = 1, 000000

yk+11 = y0 +

h

2 [f (x0, y0) + f (x1, yk1 )] k = 1, 2, 3,...

E kn = |ykn+1 − yk−1n+1|

k = 1 y11 = y0 + h2

[f (x0, y0) + f (x1, y01)] = 1, 005000 ⇒ E 10 = |y11 − y01 | = 0, 00500

k = 2 y

2

1 = y0 +

h

2 [f (x0, y0) + f (x1, y

1

1)] = 1, 004750 ⇒ E

2

0 = |y

2

1 − y

1

1 | = 0, 00025

k = 3 y31 = y0 + h2

[f (x0, y0) + f (x1, y21)] = 1, 004762 ⇒ E 30 = |y31 − y21 | = 0, 000012

k = 4 y41 = y0 + h2 [f (x0, y0) + f (x1, y31)] = 1, 004763 ⇒ E 40 = |y41 − y31 | < 10−5

n = 1, 2, 3, ..

yn+1 = yn + h φ(xn, yn, h)

φ(xn, yn, h)

[x0; x1]

f (x, y)

yn+1 = yn + w1 K 1 + w2 K 2

K 1 = h f (xn, yn) K 2 = h f (xn + α h , yn + β K 1)

w1, w2, α β

w1 + w2 = 1 w2 = 1

2 α w2 =

1

2 β

α = β = 1

yn+1 = yn + h (1

2K 1 +

1

2K 2)

K 1 = f (xn, yn)K 2 = f (xn + h, yn + h K 1)



α = β = 1/2

yn+1 = yn + h K 2

K 1 = f (xn, yn)

K 2 = f (xn + h2

, yn + h K 12

)

yn+1 = yn + h

6 · (K 1 + 2K 2 + 2K 3 + K 4)

K 1 = f (xn, yn)

K 2 = f (xn + h

2, yn +

h

2K 1)

K 3 = f (xn + h

2, yn +

h

2K 2)

K 4 = f (xn + h, yn + h K 3)

y = −y + x + 1 [0;1] y0 = 1

x = 1

h = 0, 1

K

y(xn)

n = 0 , x0 = 0 ∧ y0 = 1

K 1 = f (x0, y0) = −y0 + x0 + 1 = −1 + 0 + 1 = 0

K 2 = f (x0 + h

2, y0 +

1

2K 1h) = f (0, 05, 1) = −1 + 0, 05 + 1 = 0, 05000

K 3 = f (x0 + h

2, y0 +

1

2K 2h) = f (0, 05, 1, 0025) = −1, 0025 + 0, 05 + 1 = 0, 04750

K 4 = f (x0 + h, y0 + K 3h) = f (0, 1, 1, 00475) = −1, 00475 + 0, 1 + 1 = 0, 09525

y1 = y0 + h

6(K 1 + 2 K 2 + 2 K 3 + k4) = 1, 00484

n = 1 , x1 = 0, 1 ∧ y0 = 1, 00484

K 1 = f (x1, y1) = −y1 + x1 + 1 = −1, 00484 + 0, 1 + 1 = 0, 0951611

K 2 = f (x1 + h

2, y1 +

1

2K 1h) = f (0, 15, 1, 00960) = −1, 00960 + 0, 15 + 1 = 0, 14040

K 3 = f (x1 + h

2, y1 +

1

2K 2h) = f (0, 15, 1, 01186) = −1, 01186 + 0, 15 + 1 = 0, 13814

K 4 = f (x1 + h, y1 + K 3h) = f (0, 2, 1, 01865) = −1, 01865 + 0, 2 + 1 = 0, 018135

y2 = y1 + h

6(K 1 + 2 K 2 + 2 K 3 + k4) = 1, 01873

n = 9 , x9 = 0, 9 ∧ y0 = 1, 30657

y10 = y9 + h6

(K 1 + 2 K 2 + 2 K 3 + k4) = 1, 36788 ⇒ y(1) = 1, 36788



y = f (x, y(x)) y(x0) = y0 h

xn = x0 + n · h

n = 0, 1, 2, 3... y1 y2

y0

y1

y(x) xn

y(xn+h) = yn+1 = yn + h y n + h2

2! yn + ...

y(xn−h) = yn−1 = yn − h y n + h2

2! yn + ...

yn+1 = yn−1 + 2h yn n = 1, 2, 3...

E pT = h3

3 y(ξ )(3)

yn+1

yk+1n+1 = yn +

h

2 [f (xn, yn) + f (xn+1, ykn+1)]

E cT = − h3

12y(ξ )(3)

f (x, y)

N

(xi, yi)

y(x)

[xn−N , xn]

[xn− p, xn+1]

f (x, y)

N = 0 ⇒ yn+1 = yn + h f n

N = 1 ⇒ yn+1 = yn + h

2 (3 f n − f n−1)

N = 2 ⇒ yn+1 = yn + h

12 (23 f n − 16 f n−1 + 5 f n−2)

N = 3 ⇒ yn+1 = yn + h24

(55 f n − 59 f n−1 + 37 f n−2 − 9 f n−3)



y = −y + x + 1

[0;1]

y0 = 1

x = 1 N = 3 y1, y2, y3

h = 0, 1

n xn yn f (xn, yn)

N = 3

y4 = y3 + h

24 (55 f 3 − 59 f 2 + 37 f 1 − 9 f 0) = 1, 07032

y5 = y4 + h

24 (55 f 4 − 59 f 3 + 37 f 2 − 9 f 1) = 1, 10654

y10 = y9 + h

24 (55 f 9 − 59 f 8 + 37 f 7 − 9 f 6) = 1, 36789 ⇒ y(1) = 1, 36789

[xn−N , xn]

[xn− p, xn+1]

yn+1 yn+1

yn+1

f (x, y)

N = 0 ⇒ yn+1 = yn + h

2 · (f n + f n+1)

N = 1 ⇒ yn+1 = yn + h

3(

5

4 f n+1 + 2 f n −

1

4 f n−1)

N = 2 ⇒ yn+1 = yn + h

24(9 f n+1 + 19 f n − 5 f n−1 + f n−2)

N = 3 ⇒ yn+1 = yn + h

720(251 f n+1 + 646 f n − 264 f n−1 + 106 f n−2 − 19 f n−3)

y = −y + x + 1 [0;1] y0 = 1

x = 1

N = 1

y1, y2, y3

h = 0, 1



n xn yn f (xn, yn)

n = 3 y04 = y3 + h

12 (23 f 2 − 16 f 1 + 5 f 0) = 1, 070324

• y14 = y3 + h

3(

5

4 f 04 + 2 f 3 −

1

4 f 2) = 1, 070324

• y24 = y3 + h

3(

5

4 f 14 + 2 f 3 −

1

4 f 2) = 1, 070323

n = 9 y010 = y9 + h12

(23 f 8 − 16 f 7 + 5 f 6) = 1, 367879

• y110 = y9 + h

3(

5

4 f 010 + 2 f 9 −

1

4 f 8) = 1, 367897

• y210 = y9 + h

3(

5

4 f 110 + 2 f 9 −

1

4 f 8) = 1, 367897

⇒ y(1) = 1, 367897

y1 = f 1(x, y1, y2, y3,.....,yN )y2 = f 2(x, y1, y2, y3,.....,yN )y3 = f 3(x, y1, y2, y3,.....,yN )y4 = f 4(x, y1, y2, y3,.....,yN )...........................................yN = f N (x, y1, y2, y3,.....,yN )

y1(x = x0) = y1(0)y2(x = x0) = y2(0)y3(x = x0) = y3(0)y4(x = x0) = y4(0)....................yN (x = x0) = yN (0)

n n = 0, 1, 2, 3, 4,...

y1,n+1 = y1,n + h f 1(x, y1,n, y2,n, y3,n,.....,yN,n) n = 0, 1, 2, 3... y1(x = x0) = y1(0)y2,n+1 = y2,n + h f 2(x, y1,n, y2,n, y3,n,.....,yN,n) n = 0, 1, 2, 3... y2(x = x0) = y2(0)

...........................................yN,n+1 = yN,n + h f N (x, y1,n, y2,n, y3,n,.....,yN,n) n = 0, 1, 2, 3... yN (x = x0) = yN (0)



[ 0 ; 0, 6] h = 0, 2

y1 = f 1(x, y1, y2) = −y1 + 2 x + y2

y2 = f 2(x, y1, y2) = 0, 6 y1 + x y1(x0) = y1,0 = 1, 25

y2(x0) = y2,0 = 2, 34

n = 0 y1,1 = y1,0 + h (−y1,0 + 2 x0 + y2,0) = 1, 46800y2,1 = y2,0 + h (0, 6 y1,0 + x0) = 2, 49000

n = 1 y1,2 = y1,1 + h (−y1,1 + 2 x1 + y2,1) = 1, 75240y2,2 = y2,1 + h (0, 6 y1,1 + x1) = 2, 70616

n = 2 y1,3 = y1,2 + h (−y1,2 + 2 x2 + y2,2) = 2, 10315y2,3 = y2,2 + h (0, 6 y1,2 + x2) = 2, 99645

K 1

N

K 1,1 = f 1(x0, y1,0, y2,0,....,yN,0)K 1,2 = f 2(x0, y1,0, y2,0,....,yN,0)...........................................K 1,N = f N (x0, y1,0, y2,0,....,yN,0)

K 2

K 2,1 = f 1(x0 + h

2, y1,0 +

h

2K 1,1, y2,0 +

h

2K 1,2,....,yN,0 +

h

2K 1,N )

K 2,2 = f 2(x0 + h

2, y1,0 +

h

2K 1,1, y2,0 +

h

2K 1,2,....,yN,0 +

h

2K 1,N )

..................................................................................................

K 2,N = f N (x0 + h

2, y1,0 +

h

2K 1,1, y2,0 +

h

2K 1,2,....,yN,0 +

h

2K 1,N )

K 3

K 3,1 = f 1(x0 + h2

, y1,0 + h2

K 2,1, y2,0 + h2

K 2,2,....,yN,0 + h2

K 2,N )

K 3,2 = f 2(x0 + h

2, y1,0 +

h

2K 2,1, y2,0 +

h

2K 2,2,....,yN,0 +

h

2K 2,N )

..................................................................................................

K 3,N = f N (x0 + h

2, y1,0 +

h

2K 2,1, y2,0 +

h

2K 2,2,....,yN,0 +

h

2K 2,N )

K 3

K 4,1 = f 1(x0 + h, y1,0 + hK 3,1, y2,0 + hK 3,2,....,yN,0 + hK 3,N )K 4,2 = f 2(x0 + h, y1,0 + hK 3,1, y2,0 + hK 3,2,....,yN,0 + hK 3,N )..................................................................................................

K 4,N = f N (x0 + h, y1,0 + hK 3,1, y2,0 + hK 3,2,....,yN,0 + hK 3,N ) x = x1



y1,1 = y1,0 + h

6(K 1,1 + 2 K 2,1 + 2 K 3,1 + K 4,1)

y2,1 = y2,0 + h

6(K 1,2 + 2 K 2,2 + 2 K 3,2 + K 4,2)

...........................................

yN,1 = yN,6 +

h

6 (K 1,N + 2 K 2,N + 2 K 3,N + K 4,N )

y(m) = f (x, y

, y

, y(3),....,y(m−

1))

y(x = x0) = y0 , y(x = x0) = y0 , y(x = x0) = y 0 , y(3)(x = x0) = y(3)0 , ... , y(m)(x = x0) = y(m)

0

y1 = y(x) , y2 = y (x) , y3 = y (x) , y4 = y(3)(x) , ... , ym = y(m−1)(x)

y1 = f 1(x, y1, y2, y3,.....,ym)

y

2 = f 2(x, y1, y2, y3,.....,ym)...................................ym = f m(x, y1, y2, y3,.....,ym)

y1(x = x0) = y0

y2(x = x0) = y

0.............ym(x = x0) = y0

y + 2 y + 5 y = 3 x2 + 10

y(x0) = y0 = 4y(x0) = y 0 = −2

y1 = dy

dx = f 1(x, y1, y2) = y2

y2 = dy2

dx =

d2y2dx2

= y(x) = f 2(x, y1, y2) = 3 x2 + 10 − 5 y1 − 2 y2

y1(x0) = y0 = 4y2(x0) = y 0 = −2



m y(x) + c y(x) + k y(x) = p(x)y1(x = x0) = y0y2(x = x0) = y 0

yn = yn+1 − yn−1

2h yn =

yn+1 − 2yn + yn−1

h2

m yn+1 − 2yn + yn−1

h2 + c

yn+1 − yn−1

2h + k yn = pn

yn+1 =

pn − m

h2

− c

2hyn−1 − k −

2m

h2 yn

m

h2 +

c

2h

n − 1

y n − 1

h

T <

1

π

T = 2π k

m

n

n

x = x0

x = xn

i

y(x) + p(x)y(x) + q (x)y(x) = r(x)

α y(a) + β y(0) = µγ y(b) + δ y(b) = ν

p(x), q (x), r(x) [a, b] α,β,γ,δ,µ,ν



y(x)

y(x) + p(x)y(x) + q (x)y(x) = r(x)

y(x = a) = ya

y(x = b) = yb

[a; b] n h

yi = yi+1 − yi−1

2h

yi = yi+1 − 2yi + yi−1

h2

yi+1 − 2yi + yi−1

h2 + pi

yi+1 − yi−1

2h + q i yi = ri

i 1 −

h

2 pi

yi−1 + (h2 q i − 2) yi +

1 −

h

2 pi

yi+1 = h2 ri i = 1, 2, 3,...n − 1

i = 1 1 − h

2 p

1 y0

+ (h2 q 1

− 2) y1

+ 1 − h

2 p

1 y2

= h2 r1

i = 2

1 −

h

2 p2

y1 + (h2 q 2 − 2) y2 +

1 −

h

2 p2

y3 = h2 r2

i = 3

1 −

h

2 p3

y2 + (h2 q 3 − 2) y3 +

1 −

h

2 p3

y4 = h2 r3

...........................................................................................................

i = n − 1

1 −

h

2 pn−1

yn−2 + (h2 q n−1 − 2) yn−1 +

1 −

h

2 pn−1

yn = h2 rn−1

n − 1 n − 1

y + 4 y = −10 x2

y(x = 0) = 0y(x = 6) = 30

[0; 6]

h = 1

y(x)

yi+1 − 2yi + yi−1

h2 + 4 y

i = −10 x2

i

h = 1



yi−1 + 2 yi + yi+1 = −10 x2i

i = 1 y0 + 2 y1 + y2 = −10 x21

i = 2 y1 + 2 y2 + y3 = −10 x22

i = 3 y2 + 2 y3 + y4 = −10 x23

i = 4 y3 + 2 y4 + y5 = −10 x24

i = 5 y4 + 2 y5 + y6 = −10 x25

2 1 0 0 01 2 1 0 00 1 2 1 0

0 0 1 2 10 0 0 1 2

y1y2y3y4y5

=

−10−40−90

−160−280

⇒

y1y2y3y4y5

=

−2030

−80

40−160

x = 3 y(3) = −80

d

dx( p(x)y(x)) + (q (x) + λ r(x)) y(x) = 0

α y(a) + β y(a) = 0γ y(b) + δ y(b) = 0

p(x) > 0, q (x) > 0

p(x), p(x), q (x), r(x)

[a, b]

α , β , γ , δ

λ

h = 1

y + p y = 0

y(x = 0) = y0 = 0y(x = 4) = y4 = 0

y(x)

yi+1 − 2yi + yi−1

h2 − p yi = 0

h = 1

yi−1 + ( p − 2) yi + yi+1 = 0

i = 1 y0 + ( p − 2) y1 + y2 = 0

i = 2 y1

+ ( p − 2) y2

+ y3

= 0

i = 3 y2 + ( p − 2) y3 + y4 = 0



(2 − p) −1 0

−1 (2 − p) −10 −1 (2 − p)

y1

y2y3

=

0

00

(2 − p)3 − 2 (2 − p) = 0

p1 = 0,58578 p2 = 2,00000 p3 = 3,41421

p = p1 = 0,58578

1,41422 −1 0−1 1,41422 −10 −1 1,41422

y1y2y3

=

000

y1 = 1

φ1 =

y11

y21y31

=

1,0000

1,41421,0000

p = p2 = 2,00000

φ2 =

y12

y22y32

=

1,0000

0,0000−1,0000

p = p3 = 3,41421

φ3 =

y13

y23y33

=

1,0000

−1,41421,0000

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AP Ecuaciones Diferenciales Ordinarias