Metodo De Runge-Kutta De 4To Orden

METODO DE RUNGE-KUTTA DE 4TO ORDEN
y i +1(x i +1) = y i +{ h /6 *[K1 +(2 * K2) +(2 * K3) + K4] }
* K1 =f [x i, y i]
* K2 =f [x i +(h /2), y i +(h *K1 /2)]
* K3 =f [x i +(h /2), y i +(h *K2 /2)]
* K4 =f [x i +h, y i +(h *K3)]
EJEMPLO:
Determine y (0.5) utilizando el método de Runge-Kutta de cuarto orden, en el intervalo de interés [0, 0.5], en 5 intervalos.
PVI { y’ =4e0.8x – 0.5y ;y(0) =2 ; y(0.5) =? }
h =0.5 – 0 / 5 h =0.1
por lo tanto x0 =0, x1 =0.1, x2 =0.3, x4 =0.4, x5 =0.5
ITERACIÓN I i =0 ; x0 =0 ; y0 =2
K1 =f [0, 2] =4e(0.8*0) – (0.5 * 2)
K1 =3

K2 =f [0 +0.1/2, 2 +(0.1 *3) /2] =f [0.05, 2.15] =4e(0.8*0.05) – (0.5 * 2.15)
K2 =3.088243

K3 =f [0 +0.1/2, 2 +(0.1 *3.088243) /2] =f [0.05, 2.154412]
K3 =4e(0.8*0.05) – (0.5 * 2.154412)
K3=3.086037

K4 =f [0 +0.1, 2 +(0.1 *3.086037)] =f [0.1, 2.308603]
K4 =4e(0.8*0.1) – (0.5 * 2.308603)
K4 =3.178846

y1(0.1) =2 +{0.1 /6 [3 +(2 *3.088243) +(2 *3.086037) +3.178846]}
y1(0.1) =2.308790

ITERACIÓN II i =1 ; x1 =0.1 ; y1 =2.308790
K1 =f [0.1, 2.308790] =4e(0.8*0.1) – (0.5 * 2.308790)
K1 =3.178753

K2 =f [0.1 +0.1/2, 2.308790 +(0.1 *3.178753) /2] =f [0.15,2.467727]
K2 =4e(0.8*0.15) – (0.5 * 2.467727)
K2 =3.276123

K3 =f [0.1 +0.1/2, 2.308790 +(0.1 *3.276123) /2] =f [0.15, 2.472596]
K3 =4e(0.8*0.15) – (0.5 * 2.472596)
K3 =3.273689

K4 =f [0.1 +0.1, 2.308790 +(0.1 *3.273689)] =f [0.2, 2.636158]
K4 =4e(0.8*0.2) – (0.5 * 2.636158)
K4 =3.375964

y2(0.2) =2.308790 +{0.1 /6 [3.178753 +(2 *3.276123) +(2 *3.273689) +3.375964]}
y2(0.2) =2.636362ITERACIÓN III i =2 ; x2 =0.2 ; y2 =2.636362
K1 =f [0.2, 2.636362] =4e(0.8*0.2) – (0.5 * 2.636362)
K1 =3.375862

K2 =f [0.2 +0.1/2, 2.6366362 +(0.1 *3.375862) /2] =f [0.25, 2.805155]
K2 =4e(0.8*0.25) – (0.5 * 2.805155)
K2 =3.483033

K3 =f [0.2 +0.1/2, 2.636362 +(0.1 *3.483033) /2] =f [0.25, 2.810513]
K3 =4e(0.8*0.25) – (0.5 * 2.810513)
K3 =3.480354

K4 =f [0.2 +0.1,2.636362 +(0.1 *3.480354)] =f [0.3, 2.984397]
K4 =4e(0.8*0.3) – (0.5 * 2.984397)
K4 =3.592798

y3(0.3) =2.636362 +{0.1 /6 [3.375862 +(2 *3.483033) +(2 *3.480354) +3.592798]}
y2(0.3) =2.984619

ITERACIÓN IV i =3 ; x3 =0.3 ; y3 =2.984619
K1 =f [0.3, 2.984619] =4e(0.8*0.3) – (0.5 * 2.984619)
K1 =3.592687

K2 =f [0.3 +0.1/2, 2.984619 +(0.1 *3.592687) /2] =f [0.35, 3.164253]
K2=4e(0.8*0.35) – (0.5 * 3.164253)
K2 =3.710392

K3 =f [0.3 +0.1/2, 2.984619 +(0.1 *3.710392) /2] =f [0.35, 3.170138]
K3 =4e(0.8*0.35) – (0.5 * 3.170138)
K3 =3.707450

K4 =f [0.3 +0.1, 2.984619 +(0.1 *3.707450)] =f [0.4, 3.355364]
K4 =4e(0.8*0.4) – (0.5 * 3.355364)
K4 =3.830829

y4(0.4) =2.984619 +{0.1 /6 [3.592687 +(2 *3.710392) +(2 *3.707450) +3.830829]}
y2(0.4) =3.355606ITERACIÓN V i =4 ; x4 =0.4 ; y4 =3.355606
K1 =f [0.4, 3.355606] =4e(0.8*0.4) – (0.5 * 3.355606)
K1 =3.830708

K2 =f [0.4 +0.1/2, 3.355606 +(0.1 *3.830708) /2] =f [0.45, 3.547141]
K2 =4e(0.8*0.45) – (0.5 * 3.547141)
K2 =3.959747

K3 =f [0.4 +0.1/2, 3.355606 +(0.1 *3.959747) /2] =f [0.45, 3.553593]
K3 =4e(0.8*0.45) – (0.5 * 3.553593)
K3 =3.956521

K4 =f [0.4 +0.1, 3.355606 +(0.1*3.956521)] =f [0.5, 3.751258]
K4 =4e(0.8*0.5) – (0.5 * 3.751258)
K4 =4.091669

y5(0.5) =3.355606 +{0.1 /6 [3.830708 +(2 *3.959747) +(2 *3.956521) +4.091669]}
La solución requerida es y5(0.5) =3.751521
 

Resolver aplicando el método de Runge-Kutta.

SOLUCIÓN
De la condición inicial del problema se tiene que X = 0, y Y = 1; además, h = 0.1., obtener y(0.5)=? Sustituyendoestos valores en se obtiene:

Llevando estos valores a (16) y el resultante a (12) se obtiene que para X = 0.1 la solución del problema es

Los valores de las ki para este punto obtenido de la solución, son:

luego

Después
K_1=0.5(1+0.2) (1.1236)^2=0.7575
K_2=0.5(1+(0.2+0.1/2)) (1.1236+((0.2)(0.7575))/2)^2=0.8431
K_3=0.5(1+(0.2+0.1/2)) (1.1236+((0.2)(0.6798))/2)^2=0.8494...