Ejercicio con método de Taylor de orden 2
Ejemplo 3.3: Método de Taylor de orden 2 (2 iteraciones, h = 0.1)
Dada la EDO:
\[y' = x^2 - y, \quad y(0) = 1\]Aplicando la serie de Taylor:
\[y_{n+1} = y_n + h y'_n + \frac{h^2}{2!} y''_n\]Se tiene:
\[f(x, y) = x^2 - y\]Derivadas parciales:
\[f_x = 2x, \quad f_y = -1\]Entonces:
\[y'' = f_x + f_y \cdot f = 2x - (x^2 - y) = 2x - x^2 + y\]Paso 0 → 1 (x = 0 a x = 0.1)
\[x_0 = 0\] \[y_0 = 1\] \[f_0 = 0^2 - 1 = -1\] \[y''_0 = 2(0) - 0^2 + 1 = 1\] \[y_1 = y_0 + h f_0 + \frac{h^2}{2} y''_0\] \[y_1 = 1 + 0.1(-1) + \frac{0.01}{2}(1) = 0.905\]Paso 1 → 2 (x = 0.1 a x = 0.2)
\[x_1 = 0.1\] \[y_1 = 0.905\] \[f_1 = 0.1^2 - 0.905 = -0.895\] \[y''_1 = 2(0.1) - 0.1^2 + 0.905 = 1.095\] \[y_2 = y_1 + h f_1 + \frac{h^2}{2} y''_1\] \[y_2 = 0.905 + 0.1(-0.895) + \frac{0.01}{2}(1.095) \approx 0.820975\]\[y(0.2) \approx 0.82098\]