Métodos Numéricos

UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA (UNAD)

INSTRUCTOR:

Miguel Andrés Heredia

ESTUDIANTE:

Jherson Leandro Leal Gordillo

1.098.665.923

CURSO:

Métodos Numéricos

CEAD Bucaramanga, Julio de 2012

Usar la regla de Simpson de 1/3 para aproximar la siguiente integral: (Recuerde que debe hallar f(-2), f(1) y f(4))

∫_a^b▒〖f(x)dx=h/3〗[f(a)+4f(m)+f(b)

Donde h= (b-a)/2 y m=(a+b)/2

∫_(-2)^4▒(1-x-〖4x〗^3+x^5 )dx

h= (4-(-2))/2=3

m=(-2+4)/2=1

3/3 [f(-2)+4f(1)+f(4)]

f(-2)=(1-(2)-4(-2)^3+(〖-2)〗^5=3

f(1)=(1-(1)-4(1)^3+(〖1)〗^5=-3

f(4)=(1-(4)-4(4)^3+(4)^5=765

∫_(-2)^4▒(1-x-〖4x〗^3+x^5 )dx=1(3+4(-3)+765=756

Aplicar el método de Runge-Kutta de orden cuatro para obtener la aproximación y(0,8) a la solución del siguiente problema de valor inicial, con h=0,2.

y^´=y-x^2+1

〖y(0)〗^´=〖0.5〗^2

x_0=0

y_0=0.25

h=0.2

f(x,y)=y-x^2+1

Sabiendo los valores iniciales de x_0,y_0 se procede a hallar los k, obteniendo:

Primera Iteración.

k_1=hf(x_0,y_0)

k_1=0.2(0.25+0^2+1)

k_1=0.2(1.25)

k_1=0.25

k_2=hf(x_0+h/2,y_0+k_1/2)

k_2=0.2[(0.5+0.25/2)-(0+0.2/2)^2+1]

k_2=0.2[(0.5+0.125)-(0.1)^2+1]

k_2=0.2[0.625-0.01+1]

k_2=0.2[1.615]

k_2=0.323

k_3=hf(x_0+h/2,y_0+k_2/2)

k_3=0.2[(0.5+0.323/2)-(0+0.2/2)^2+1]

k_3=0.2[(0.5+0.1615)-(0.1)^2+1]

k_3=0.2[0.6615-0.01+1]

k_3=0.2[1.6515]

k_3=0.3303

k_4=hf(x_0+h ,y_0+k_3)

k_4=0.2[(0.5+0.3303)-(0+0.2)^2+1]

k_4=0.2[(0.8303)-(0.2)^2+1]

k_4=0.2[0.8303-0.04+1]

k_4=0.2[1.7903]

k_4=0.35806

Segunda Iteración.

x_1=x_0+h

x_1=0+0.2

x_1=0.2

y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)

y_1=0.25+1/6(0.25+(2)(0.323)+(2)(0.3303)+0.35806)

y_1=0.25+1/6(0.25+0646+0.6606+0.35806)

y_1=0.25+1/6(1.91466)

y_1=0.25+0.31911

y_1=0.56911

k_1=hf(x_1,y_1)

k_1=0.2(0.56911-(0.2)^2+1)

k_1=0.2(0.56911-0.04+1)

k_1=0.2(1.52911)

k_1=0.305822

k_2=hf(x_1+h/2,y_1+k_1/2)

k_2=0.2[(0.56911+0.305822/2)-(0.2+0.2/2)^2+1]

k_2=0.2[(0.56911+0.152911)-(0.3)^2+1]

k_2=0.2[0.722021-0.09+1]

k_2=0.2[1.632021]

k_2=0.3264042

k_3=hf(x_1+h/2,y_1+k_2/2)

k_3=0.2[(0.56911+0.3264042/2)-(0.2+0.2/2)^2+1]

k_3=0.2[(0.56911+0.1632021)-(0.3)^2+1]

k_3=0.2[0.7323121-0.09+1]

k_3=0.2[1.6423121]

k_3=0.32846242

k_4=hf(x_1+h ,y_1+k_3)

k_4=0.2[(0.56911+0.32846242)-(0.2+0.2)^2+1]

k_4=0.2[(0.89757242)-(0.4)^2+1]

k_4=0.2[(0.89757242)-0.16+1]

k_4=0.2[1.73757242]

k_4=0.347514484

Tercera Iteración.

x_2=x_1+h

x_2=0.2+0.2

x_2=0.4

y_2=y_1+1/6(k_1+2k_2+2k_3+k_4)

y_2=0.56911+1/6(0.305822+(2)(0.3264042)+(2)(0.32846242)+0.347514484)

y_2=0.56911+1/6(0.305822+0.6528084+0.65692484+0.347514484)

y_2=0.56911+1/6(1.963069724)

y_2=0.56911+0.327178287

y_2=0.896288287

k_1=hf(x_2,y_2)

k_1=0.2(0.896288287-(0.4)^2+1)

k_1=0.2(0.896288287-0.16+1)

k_1=0.2(1.736288287)

k_1=0.347257657

k_2=hf(x_2+h/2,y_2+k_1/2)

...