Encuentre y(1/2) para d/(dt)(y) = e^(-t^2), cuando y(0) = 1, h = 1/10 usando el método mejorado de Euler

La calculadora encontrará $$$y{\left(\frac{1}{2} \right)}$$$ para $$$y^{\prime } = e^{- t^{2}}$$$, cuando $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{1}{10}$$$ 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 $$$y{\left(\frac{1}{2} \right)}$$$ para $$$y^{\prime } = e^{- t^{2}}$$$, cuando $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{1}{10}$$$ 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 $$$h = \frac{1}{10}$$$, $$$t_{0} = 0$$$, $$$y_{0} = 1$$$ y $$$f{\left(t,y \right)} = e^{- t^{2}}$$$.

Paso 1

$$$t_{1} = t_{0} + h = 0 + \frac{1}{10} = \frac{1}{10}$$$

$$$\tilde{y}_{1} = \tilde{y}{\left(t_{1} \right)} = \tilde{y}{\left(\frac{1}{10} \right)} = y_{0} + h\cdot f{\left(t_{0},y_{0} \right)} = 1 + h\cdot f{\left(0,1 \right)} = 1 + \frac{1}{10} \cdot 1 = 1.1$$$

$$$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{10} \right)} = y_{0} + \frac{h}{2} \left(f{\left(t_{0},y_{0} \right)} + f{\left(t_{1},\tilde{y}_{1} \right)}\right) = 1 + \frac{h}{2} \left(f{\left(0,1 \right)} + f{\left(\frac{1}{10},1.1 \right)}\right) = 1 + \frac{\frac{1}{10}}{2} \left(1 + 0.990049833749168\right) = 1.09950249168746$$$

Paso 2

$$$t_{2} = t_{1} + h = \frac{1}{10} + \frac{1}{10} = \frac{1}{5}$$$

$$$\tilde{y}_{2} = \tilde{y}{\left(t_{2} \right)} = \tilde{y}{\left(\frac{1}{5} \right)} = y_{1} + h\cdot f{\left(t_{1},y_{1} \right)} = 1.09950249168746 + h\cdot f{\left(\frac{1}{10},1.09950249168746 \right)} = 1.09950249168746 + \frac{1}{10} \cdot 0.990049833749168 = 1.19850747506238$$$

$$$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{1}{5} \right)} = y_{1} + \frac{h}{2} \left(f{\left(t_{1},y_{1} \right)} + f{\left(t_{2},\tilde{y}_{2} \right)}\right) = 1.09950249168746 + \frac{h}{2} \left(f{\left(\frac{1}{10},1.09950249168746 \right)} + f{\left(\frac{1}{5},1.19850747506238 \right)}\right) = 1.09950249168746 + \frac{\frac{1}{10}}{2} \left(0.990049833749168 + 0.960789439152323\right) = 1.19704445533253$$$

Paso 3

$$$t_{3} = t_{2} + h = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$$$

$$$\tilde{y}_{3} = \tilde{y}{\left(t_{3} \right)} = \tilde{y}{\left(\frac{3}{10} \right)} = y_{2} + h\cdot f{\left(t_{2},y_{2} \right)} = 1.19704445533253 + h\cdot f{\left(\frac{1}{5},1.19704445533253 \right)} = 1.19704445533253 + \frac{1}{10} \cdot 0.960789439152323 = 1.29312339924777$$$

$$$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{10} \right)} = y_{2} + \frac{h}{2} \left(f{\left(t_{2},y_{2} \right)} + f{\left(t_{3},\tilde{y}_{3} \right)}\right) = 1.19704445533253 + \frac{h}{2} \left(f{\left(\frac{1}{5},1.19704445533253 \right)} + f{\left(\frac{3}{10},1.29312339924777 \right)}\right) = 1.19704445533253 + \frac{\frac{1}{10}}{2} \left(0.960789439152323 + 0.913931185271228\right) = 1.29078048655371$$$

Paso 4

$$$t_{4} = t_{3} + h = \frac{3}{10} + \frac{1}{10} = \frac{2}{5}$$$

$$$\tilde{y}_{4} = \tilde{y}{\left(t_{4} \right)} = \tilde{y}{\left(\frac{2}{5} \right)} = y_{3} + h\cdot f{\left(t_{3},y_{3} \right)} = 1.29078048655371 + h\cdot f{\left(\frac{3}{10},1.29078048655371 \right)} = 1.29078048655371 + \frac{1}{10} \cdot 0.913931185271228 = 1.38217360508083$$$

$$$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{2}{5} \right)} = y_{3} + \frac{h}{2} \left(f{\left(t_{3},y_{3} \right)} + f{\left(t_{4},\tilde{y}_{4} \right)}\right) = 1.29078048655371 + \frac{h}{2} \left(f{\left(\frac{3}{10},1.29078048655371 \right)} + f{\left(\frac{2}{5},1.38217360508083 \right)}\right) = 1.29078048655371 + \frac{\frac{1}{10}}{2} \left(0.913931185271228 + 0.852143788966211\right) = 1.37908423526558$$$

Paso 5

$$$t_{5} = t_{4} + h = \frac{2}{5} + \frac{1}{10} = \frac{1}{2}$$$

$$$\tilde{y}_{5} = \tilde{y}{\left(t_{5} \right)} = \tilde{y}{\left(\frac{1}{2} \right)} = y_{4} + h\cdot f{\left(t_{4},y_{4} \right)} = 1.37908423526558 + h\cdot f{\left(\frac{2}{5},1.37908423526558 \right)} = 1.37908423526558 + \frac{1}{10} \cdot 0.852143788966211 = 1.4642986141622$$$

$$$y_{5} = y{\left(t_{5} \right)} = y{\left(\frac{1}{2} \right)} = y_{4} + \frac{h}{2} \left(f{\left(t_{4},y_{4} \right)} + f{\left(t_{5},\tilde{y}_{5} \right)}\right) = 1.37908423526558 + \frac{h}{2} \left(f{\left(\frac{2}{5},1.37908423526558 \right)} + f{\left(\frac{1}{2},1.4642986141622 \right)}\right) = 1.37908423526558 + \frac{\frac{1}{10}}{2} \left(0.852143788966211 + 0.778800783071405\right) = 1.46063146386746$$$

Respuesta

$$$y{\left(\frac{1}{2} \right)}\approx 1.46063146386746$$$A

Encuentre $$$y{\left(\frac{1}{2} \right)}$$$ para $$$y^{\prime } = e^{- t^{2}}$$$, cuando $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{1}{10}$$$ usando el método mejorado de Euler