Calculadora mejorada del método de Euler (Heun)

La calculadora encontrará la solución aproximada de la ecuación diferencial de primer orden usando el método mejorado de Euler (Heun), con pasos mostrados.

Calculadoras relacionadas: Calculadora del método de Euler, Calculadora del método de Euler modificado

$$$y^{\prime } = f{\left(t,y \right)}$$$:

O $$$y^{\prime } = f{\left(x,y \right)}$$$.

dividir usando:

$$$h$$$:

$$$t_{0}$$$:

O $$$x_{0}$$$.

$$$y_{0}$$$:

$$$y_0=y(t_0)$$$ o $$$y_0=y(x_0)$$$.

$$$t_{1}$$$:

O $$$x_{1}$$$.

Si la calculadora no calculó algo o ha identificado un error, o tiene una sugerencia/comentario, escríbalo en los comentarios a continuación.

Tu aportación

Encuentra $$$y{\left(1 \right)}$$$ para $$$y^{\prime } = 3 t + y$$$, cuando $$$y{\left(0 \right)} = 7$$$, $$$h = \frac{1}{5}$$$ usando el método de Euler mejorado.

Solución

El método de Euler mejorado establece que $$$y_{n+1} = y_{n} + \frac{h}{2} \left(f{\left(t_{n},y_{n} \right)} + f{\left(t_{n+1},\tilde{y}_{n+1} \right)}\right)$$$, donde $$$\tilde{y}_{n+1} = y_{n} + h\cdot f{\left(t_{n},y_{n} \right)}$$$ y $$$t_{n+1} = t_{n} + h$$$.

Tenemos que $$$h = \frac{1}{5}$$$, $$$t_{0} = 0$$$, $$$y_{0} = 7$$$ y $$$f{\left(t,y \right)} = 3 t + y$$$.

Paso 1

$$$t_{1} = t_{0} + h = 0 + \frac{1}{5} = \frac{1}{5}$$$

$$$\tilde{y}_{1} = \tilde{y}{\left(t_{1} \right)} = \tilde{y}{\left(\frac{1}{5} \right)} = y_{0} + h\cdot f{\left(t_{0},y_{0} \right)} = 7 + h\cdot f{\left(0,7 \right)} = 7 + \frac{1}{5} \cdot 7 = 8.4$$$

$$$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{5} \right)} = y_{0} + \frac{h}{2} \left(f{\left(t_{0},y_{0} \right)} + f{\left(t_{1},\tilde{y}_{1} \right)}\right) = 7 + \frac{h}{2} \left(f{\left(0,7 \right)} + f{\left(\frac{1}{5},8.4 \right)}\right) = 7 + \frac{\frac{1}{5}}{2} \left(7 + 9\right) = 8.6$$$

Paso 2

$$$t_{2} = t_{1} + h = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$$$

$$$\tilde{y}_{2} = \tilde{y}{\left(t_{2} \right)} = \tilde{y}{\left(\frac{2}{5} \right)} = y_{1} + h\cdot f{\left(t_{1},y_{1} \right)} = 8.6 + h\cdot f{\left(\frac{1}{5},8.6 \right)} = 8.6 + \frac{1}{5} \cdot 9.199999999999999 = 10.44$$$

$$$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{2}{5} \right)} = y_{1} + \frac{h}{2} \left(f{\left(t_{1},y_{1} \right)} + f{\left(t_{2},\tilde{y}_{2} \right)}\right) = 8.6 + \frac{h}{2} \left(f{\left(\frac{1}{5},8.6 \right)} + f{\left(\frac{2}{5},10.44 \right)}\right) = 8.6 + \frac{\frac{1}{5}}{2} \left(9.199999999999999 + 11.639999999999999\right) = 10.684$$$

Paso 3

$$$t_{3} = t_{2} + h = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$$$

$$$\tilde{y}_{3} = \tilde{y}{\left(t_{3} \right)} = \tilde{y}{\left(\frac{3}{5} \right)} = y_{2} + h\cdot f{\left(t_{2},y_{2} \right)} = 10.684 + h\cdot f{\left(\frac{2}{5},10.684 \right)} = 10.684 + \frac{1}{5} \cdot 11.884 = 13.0608$$$

$$$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{5} \right)} = y_{2} + \frac{h}{2} \left(f{\left(t_{2},y_{2} \right)} + f{\left(t_{3},\tilde{y}_{3} \right)}\right) = 10.684 + \frac{h}{2} \left(f{\left(\frac{2}{5},10.684 \right)} + f{\left(\frac{3}{5},13.0608 \right)}\right) = 10.684 + \frac{\frac{1}{5}}{2} \left(11.884 + 14.8608\right) = 13.35848$$$

Paso 4

$$$t_{4} = t_{3} + h = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$$$

$$$\tilde{y}_{4} = \tilde{y}{\left(t_{4} \right)} = \tilde{y}{\left(\frac{4}{5} \right)} = y_{3} + h\cdot f{\left(t_{3},y_{3} \right)} = 13.35848 + h\cdot f{\left(\frac{3}{5},13.35848 \right)} = 13.35848 + \frac{1}{5} \cdot 15.15848 = 16.390176$$$

$$$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{4}{5} \right)} = y_{3} + \frac{h}{2} \left(f{\left(t_{3},y_{3} \right)} + f{\left(t_{4},\tilde{y}_{4} \right)}\right) = 13.35848 + \frac{h}{2} \left(f{\left(\frac{3}{5},13.35848 \right)} + f{\left(\frac{4}{5},16.390176 \right)}\right) = 13.35848 + \frac{\frac{1}{5}}{2} \left(15.15848 + 18.790176\right) = 16.7533456$$$

Paso 5

$$$t_{5} = t_{4} + h = \frac{4}{5} + \frac{1}{5} = 1$$$

$$$\tilde{y}_{5} = \tilde{y}{\left(t_{5} \right)} = \tilde{y}{\left(1 \right)} = y_{4} + h\cdot f{\left(t_{4},y_{4} \right)} = 16.7533456 + h\cdot f{\left(\frac{4}{5},16.7533456 \right)} = 16.7533456 + \frac{1}{5} \cdot 19.1533456 = 20.58401472$$$

$$$y_{5} = y{\left(t_{5} \right)} = y{\left(1 \right)} = y_{4} + \frac{h}{2} \left(f{\left(t_{4},y_{4} \right)} + f{\left(t_{5},\tilde{y}_{5} \right)}\right) = 16.7533456 + \frac{h}{2} \left(f{\left(\frac{4}{5},16.7533456 \right)} + f{\left(1,20.58401472 \right)}\right) = 16.7533456 + \frac{\frac{1}{5}}{2} \left(19.1533456 + 23.58401472\right) = 21.027081632$$$

Respuesta

$$$y{\left(1 \right)}\approx 21.027081632$$$A