TALLER

1. Realice un mapa conceptual de los métodos iterativos empleados en la solución de ecuaciones diferenciales de valor inicial.

2. Plantee y solucione dos ejercicios sobre diferenciación numérica explicando paso a paso el procedimiento utilizado.

f ( x )=sin (x ) en x1=π4 y h=0.01

La fórmula para derivación numérica es: f ' (x1 )=f (x1+h )−f (x1 )

h

Así:

f ' (2 )=f ( π4 +0.01)−f ( π4 )

0.01=sin( π4 +0.01)−sin ( π4 )

0.01=0.714142−√2

20.01

=0.7035

f ( x )=x3 en x1=4 y h=0.001

f ' (4 )= f(4+0.001 )−f (4 )

0.001=4.001

3−43

0.001=64.048012−64

0.001=0.048012001

0.001=48.012001

3. Solucione el siguiente ejercicio utilizando la Regla del Trapecio, la Regla de Simpson 1/3 y la Regla de Simpson 3/8. Compare los resultados y haga un pequeño análisis.

Métodos iteracivos para E.D.O de V.I

Método de Euler Método de Taylor Método de Euler modificado

Método de Runge.Kutta de

cuarto orden

Método predictor-corrector

∫0

1x

2 x+1dx

Solución:

- Regla del trapecio:

∆ x=b−an

=1−04

=14=0,25

Veamos

∫a

b

f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]

Reemplazando

∫0

1x

2 x+1dx=1−0

2(4 )[ f (0 )+2 f (0,25 )+2 f (0,5 )+2 f (0,75 )+ f (1 ) ]

¿ 18 [0+2 0,25

2 (0,25 )+1+2 0,52 (0,5 )+1

+2 0,752 (0,75 )+1

+ 12 (1 )+1 ]

¿ 18 [0+13 + 1

2+ 35+13 ]

¿ 18 [ 5330 ]= 53

240=0,221

- Regla Simpson 1/3

Veamos

∫a

b

f ( x )dx=b−a3n [f (x0 )+3 f ( x1 )+3 f (x2 )+2 f ( x3 )+…+2 f (xn−2 )+4 f (xn−1 )+f (xn ) ]

Reemplazando

∫0

1x

2 x+1dx=1−0

3(4 ) [ f (x0 )+4 f (x1 )+2 f (x2 )+4 f (x3 )+ f (xn) ]

¿ 112

[ f (0 )+4 f (0,25 )+2 f (0,5)+4 f (0,75 )+ f (1 ) ]

¿ 112 [0+4 16+2 14 +4 3

10+ 13¿]

¿ 112 [ 2710 ]= 9

40=0,225

- Regla Simpson 3/8 Veamos

∫0

1x

2 x+1dx=

3(1−0)8(4) [f (x0 )+3 f (x1 )+3 f (x2 )+2 f (x3 )+ f (xn) ]

¿ 332

[ f (0 )+3 f (0,25 )+3 f (0,5)+2 f (0,75 )+f (1 ) ]

¿ 332 [0+3 16 +3 1

4+2 310

+ 13¿]

¿ 332 [ 13160 ]=131640=0,205

El valor real de esa integral es 0,225; luego podría decirse que los tres métodos son aproximados al valor real pero en este caso el mejor método el que más se aproxima es la Regla Simpson 1/3.

4. Solucione el siguiente ejercicio utilizando la regla del trapecio, la regla de Simpson 1/3 y la regla de Simpson 3/8. Compare los resultados y haga un pequeño análisis (dividiendo en 4 intervalos).

∫0

43√x exdx

REGLA DEL TRAPECIO:

∫a

b

f ( x )dx ≈ b−a2n

[ f ( x0 )+2 f (x1 )+2 f ( x2 )+…+2 f (xn−1 )+f (xn )]

x0=0 , x1=1 , x2=2 , x3=3 , x4=4

∫0

43√x exdx ≈ 4−0

2 (4 )[ 3√0e0+2 ( 3√1e1 )+2 ( 3√2e2)+2 ( 3√3e3 )+ 3√4 e4 ]=1

2[0+5.436563+18.619254+57.936713+86.669160 ]=84.330845

REGLA DE SIMPSON 1/3 Caso general

∫a

b

f ( x )dx ≈ h3 [ f (x0 )+4 f ( x1 )+2 f (x2 )+4 f (x3 )+2 f (x4 )+…+ f (xn) ]

h=b−an

h=4−04

=1

∫0

43√x exdx=1

3[ 3√0e0+4 ( 3√1e1)+2 ( 3√2e2 )+4 ( 3√3e3 )+ 3√4e4 ]=1

3[0+10.873127+18.619254+115.873428+86.669160 ]=232.034969

3=77.34498966

REGLA DE SIMPSON 3/8

∫a

b

f ( x )dx=3h8

¿ con h=b−a3

=43

x0=0 , x1=43, x2=

83, x3=4

∫0

43√x exdx=12

24 [ 3√0e0+3( 3√ 43 e4 /3)+3( 3√ 83 e83 )+ ( 3√4 e4 )]=12 [0+12.526415+59.8727837+86.6691608 ]=159.0683595

2=79.53417975

El valor real aproximado de esta integral es: 76.9621

Por lo tanto El método que más se aproxima es el de la regla de Simpson 1/3.

5. Solucione el siguiente ejercicio utilizando la integración de Romberg. Usando segmentos de longitud 1, 1/3 y 1/6.

∫0

1

ex2

dx

Solución: Para n=1Usamos la fórmula del trapecio, esto es;

∫a

b

f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]

I 1=∫0

1

ex2

dx=1−02(1) [ f (x0 )+ f (x1 ) ]

¿ 12

[ f (0 )+ f (1 ) ]

¿ 12

[e (0 )2+e(1)2 ]

¿ 12

[1+e ]

¿1,86

Para n=3

∆ x=b−an

=1−03

=13

Usamos la fórmula del trapecio, esto es;

∫a

b

f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]

I 2=∫0

1

ex2

dx=1−02(3) [ f (x0 )+2 f (x1 )+2 f (x2 )+f (x3 ) ]

¿ 16

[ f (0 )+2 f (1/3 )+2 f (2/3 )+ f (1 ) ]

¿ 16

[ e(0)2+2e(1/3)2+2e(2/3)2+e (1)2 ]

¿ 16

[1+2e 19+2e 49+e]¿1,512

Para n=6

∆ x=b−an

=1−06

=16

Usamos la fórmula del trapecio, esto es;

∫a

b

f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]

I 3=∫0

1

ex2

dx=1−02(6) [f (x0 )+2 f (x1 )+2 f (x2 )+2 f (x3 )+2 f (x4 )+2 f ( x5 )+ f (x6)]

¿ 16

[ f (0 )+2 f (1/6 )+2 f (1/3 )+2 f (1/2 )+2 f (2 /3 )+2 f (5 /6 )+ f (1 ) ]

¿ 16

[ e(0)2+2e(1/6)2+2e(1/3)2+2e(1/2)2+2e(2/3 )2+2e(5 /6)2+e (1)2 ]

¿ 16

[1+2e 136+2e 19+2e14+2e 49+2e 2536+e]¿1,475

NIVELI

NIVELII

NIVELIII

I 1 I 4=43I 2−

13I 1

I 5=43I3−

13I2

1615I 5−

115I 4

I 2

I 3

NIVELI

NIVELII

NIVELIII

1,86I 4=

43

(1,512 )−13

(1,86 )=1,396

I 5=43

(1,475 )−13

(1,512 )=1,4626

1615

(1,4626 )− 115

(1,396 )=1,467041,512

1,475

6. Solucione el siguiente ejercicio utilizando la integración de Romberg. Usando segmentos de longitud 1, ½, 1/4, 1/8

∫1

2

ex ln ( x )dx

Formulas del trapecio: Cuando n=1 h2[ f (a )+ f (b )]

Cuando n=2,3 ,…,∞ h2[ f (a )+2∑

j=1

n−1

f (a+ jh )+ f (b )] y

h=limite superior−limite inferior

n

Segmentos: 1, ½, ¼, 1/8.

Cuando h=1 n=1

I 1=12

[e1ln (1 )+e2 ln (2 ) ]=12

[0+5.121703 ]=2.560851

Cuando h=1/2 n=2

I 2=14 [e1 ln (1 )+2(e1+

12 ln(1+12 ))+e2 ln (2 )]= 14 [0+3.634337+5.121703 ]=2.189010

Cuando h=1/4 n=4

I 3=18¿

Cuando h=1/8 n=8

I 4=116

¿

Nivel 1 Nivel 2 Nivel 3 Nivel 4I 1=2.560851 I 5=2.065063 I 8=2.0625862 I=2.062585184I 2=2.189010 I 6=2.062741 I 9=2.0625852I 3=2.0943085 I 7=2.062595I 4=2.070524

43I 2−

13I1=2.065063

43I 3−

13I2=2.062741

43I 4−

13I 3=2.062595

1615I 6−

115I 5=2.0625862

1615I 7−

115I 6=2.0625852

6463I 9−

163I 8=2.062585184

7. Usar el Método de Euler mejorado para aproximar y (2,3) tomando h=0.1 dada la ecuación diferencial

y '=2 x+ y−3y (2 )=1

x1=2+0.1=2.1

u1=1+0.1 (2 (2 )+1−3 )=1.2

y1=1+0.1( 12 ) [(2 (2 )+1−3 )+(2 (2.1 )+(1.2 )−3 ) ]=1.22

x2=2.1+0.1=2.2

u2=1.22+0.1 (2 (2.1 )+1.22−3 )=1.462

y2=1.22+0.1( 12 )[(2 (2.1 )+1.22−3 )+(2 (2.2 )+(1.462 )−3 ) ]=1.4841

x3=2.2+0.1=2.3

u3=1.4841+0.1 (2 (2.2 )+1.4841−3 )=1.77251

y3=1.4841+0.1 (12 )[(2 (2.2 )+1.4841−3 )+(2 (2.3 )+(1.77251 )−3 )]=1.7969305

x4=2.3+0.1=2.4

u4=1.797+0.1 (2 (2.3 )+1.797−3 )=2.1367

y4=1.797+0.1( 12 ) [(2 (2.3 )+1.797−3 )+(2 (2.4 )+(2.1367 )−3 )]=2.1637

x5=2.4+0.1=2.5

u5=2.1637+0.1 (2 (2.4 )+2.1637−3 )=2.56

y5=2.1637+0.1( 12 )[ (2 (2.4 )+2.1637−3 )+ (2 (2.5 )+ (2.56 )−3 )]=2.59

x6=2.5+0.1=2.6

u6=2.59+0.1 (2 (2.5 )+2.59−3 )=3.5

y6=2.59+0.1( 12 )[ (2 (2.5 )+2.59−3 )+(2 (2.6 )+(3.5 )−3 )]=3.1

x7=2.6+0.1=2.7

u7=3.1+0.1 (2 (2.6 )+3.1−3 )=3.63

y7=3.1+0.1(12 )[ (2 (2.6 )+3.1−3 )+ (2 (2.7 )+(3.63 )−3 )]=3.66

x8=2.7+0.1=2.8

u8=3.66+0.1 (2 (2.7 )+3.66−3 )=4.26

y8=3.66+0.1( 12 )[ (2 (2.7 )+3.66−3 )+(2 (2.8 )+(4.26 )−3 )]=4.3

x9=2.8+0.1=2.9

u9=4.3+0.1 (2 (2.8 )+4.3−3 )=5

y9=4.3+0.1( 12 ) [(2 (2.8 )+4.3−3 )+(2 (2.9 )+(5 )−3 )]=5.035

x10=2.9+0.1=3

u10=5.035+0.1 (2 (2.9 )+5.035−3 )=5.8185

y10=5.035+0.1(12 )[(2 (2.9 )+5.035−3 )+(2 (3 )+ (5.8185 )−3 )]=5.8676

8. Utilizar el método de Runge-Kutta para aproximar y (0.5 )

Dada la siguiente ecuación diferencial:

y '=2 xy

y (0 )=1

El método de Runge-Kutta de segundo orden se presenta con las siguientes expresiones:

k 1=f (xn , y n) ; k2=f (xn+h , yn+hk1 )

yn+1= yn+h2(k1+k2)

h=0.25 , x0=0 , y0=1k 1=f (x0 , y0 )=f (0,1 )=0

Para esta iteración x1=0.25k 2=f (x0+h , y0+hk1 )=f (0+0.25 ,1+0.25 (0 ) )= f (0.25 ,1 )=0.5

y1= y (0.25 )= y0+h2

(k 1+k2 )=1+ 0.252

(0+0.5 )=1716

=1.0625

La segunda iteración es:

k 1=f (x1 , y1 )=f (0.25 ,1.0625 )=1732

=0.53125

k 2=f (x1+h , y1+hk1 )=f (0.25+0.25 ,1.0625+0.25 (0.53125 ))=f (0.5 ,1.1953125 )=1.1953125

y2= y (0.5 )= y1+h2

(k1+k 2 )=1.0625+ 0.252

(0.53125+1.1953125 )=1.278320313

9. Hallar la solución aproximada que proporciona el Método de Adams-Bashorth de segundo, tercer y cuarto orden para la ecuación: h=0.1, x en [1,2]

y '=4−2 xy2

y (1 )=1- Segundo orden

Por el método de Euler

x0=1

y0=1

f 0=( 4−2 (1 )12 )=2

x1=1.1

y1=1+0.1( 4−2 (1 )12 )=1.2

f 1=( 4−2 (1.1 )1.22 )=1.25

Por método de Adam’s Bashorth

x3=0.3

y3=1.2+0.12

(3 (1.25 )−2 )=1.3

x4=0.4

y4=1.3+0.12

(3 (1.25 )−2 )=1.3875

x5=0.5

y5=1.3875+0.12

(3 (1.25 )−2 )=1.475

x6=0.6

y6=1.475+0.12

(3 (1.25 )−2 )=1.5625

x7=0.7

y7=1.5625+0.12

(3 (1.25 )−2 )=1.65

x8=0.8

y8=1.65+0.12

(3 (1.25 )−2 )=1.7375

x9=0.9

y9=1.7375+0.12

(3 (1.25 )−2 )=1.825

x10=1

y10=1.825+0.12

(3 (1.25 )−2 )=1.9

- Tercer orden

x2=1.2

y2=1.2+0.1( 4−2 (1.1 )(1.2)2 )=1.325

f 2=( 4−2 (1.2 )1.3252 )=0.911

Por método de Adam’s Bashorth

x3=1.3

y3=1.325+0.112

[23 (0.911)−16 (1.25 )+5 (2 ) ]=1.416275

x4=1.4

y4=1.416275+0.112

[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.50755

x5=1.5

y5=1.50755+0.112

[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.598825

x6=1.6

y6=1.5 98825+0.112

[23 (0.911)−16 (1.25 )+5 (2 ) ]=1.6901

x7=1.7

y7=1.6901+0.112

[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.781375

x8=1.8

y8=1.781375+0.112

[23 (0.911)−16 (1.25 )+5 (2 ) ]=1.87265

x9=1.9

y9=1.87265+0.112

[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.963925

x10=2

y10=1.963925+0.112

[23 (0.911 )−16 (1.25 )+5 (2 ) ]=2.0552

- Cuarto orden

x3=1.3

y3=1.325+0.1( 4−2 (1.2 )(1.325)2 )=1.42

f 3=( 4−2 (1.3 )1.422 )=0.695

Por método de Adam’s Bashorth

x4=0.4

y4=1.42+0.124

[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.473

x5=0.5

y5=1.473+0.124

[55 (0.695 )−59 (0.911)+37 (1.25 )−9 (2 ) ]=1.526

x6=0.6

y6=1.526+0.124

[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.58

x7=0.7

y7=1.58+0.124

[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.633

x8=0.8

y8=1.633+0.124

[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.686

x9=0.9

y9=1.686+0.124

[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.74

x10=1

y10=1.74+0.124

[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.793