Ejercicio con Método de Runge–Kutta de orden 4 (h = 0.1)
Ejercicio 19: Runge–Kutta de orden 4 (3 iteraciones, (h = 0.1))
Dada la EDO:
\[y' = y\,\tan(x), \quad y(0)=1\]El esquema clásico de RK4 es
\[\begin{aligned} k_1 &= f(x_n, y_n), \\ k_2 &= f\!\left(x_n + \tfrac{h}{2},\; y_n + \tfrac{h}{2}k_1\right),\\ k_3 &= f\!\left(x_n + \tfrac{h}{2},\; y_n + \tfrac{h}{2}k_2\right),\\ k_4 &= f\!\left(x_n + h,\; y_n + h k_3\right),\\[6pt] y_{n+1} &= y_n + \tfrac{h}{6}\bigl(k_1 + 2k_2 + 2k_3 + k_4\bigr), \end{aligned} \qquad f(x,y)=y\tan x.\]Paso 0 → 1 ((x = 0) a (0.1))
\[\begin{aligned} x_0 &= 0, &\quad y_0 &= 1,\\ k_1 &= 0.000000,\\ k_2 &= 0.050042,\\ k_3 &= 0.050167,\\ k_4 &= 0.100838,\\[6pt] y_1 &= y_0 + \tfrac{h}{6}\bigl(k_1 + 2k_2 + 2k_3 + k_4\bigr)=1.005021. \end{aligned}\]Paso 1 → 2 ((x = 0.1) a (0.2))
\[\begin{aligned} x_1 &= 0.1, &\quad y_1 &= 1.005021,\\ k_1 &= 0.100838,\\ k_2 &= 0.152656,\\ k_3 &= 0.153048,\\ k_4 &= 0.206830,\\[6pt] y_2 &= y_1 + \tfrac{h}{6}\bigl(k_1 + 2k_2 + 2k_3 + k_4\bigr)=1.020339. \end{aligned}\]Paso 2 → 3 ((x = 0.2) a (0.3))
\[\begin{aligned} x_2 &= 0.2, &\quad y_2 &= 1.020339,\\ k_1 &= 0.206833,\\ k_2 &= 0.263176,\\ k_3 &= 0.263895,\\ k_4 &= 0.323791,\\[6pt] y_3 &= y_2 + \tfrac{h}{6}\bigl(k_1 + 2k_2 + 2k_3 + k_4\bigr)=1.046752. \end{aligned}\]\[y(0.3) \approx 1.046752\]