Encontre $$$y{\left(\frac{1}{2} \right)}$$$ para $$$y^{\prime } = e^{- t^{2}}$$$, quando $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{1}{10}$$$ usando o método de Euler modificado

A calculadora encontrará $$$y{\left(\frac{1}{2} \right)}$$$ para $$$y^{\prime } = e^{- t^{2}}$$$, quando $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{1}{10}$$$ usando o método de Euler modificado, com as etapas mostradas.

Calculadoras relacionadas: Calculadora do Método de Euler, Calculadora do método de Euler (Heun) aprimorada

Ou $$$y^{\prime } = f{\left(x,y \right)}$$$.
Ou $$$x_{0}$$$.
$$$y_0=y(t_0)$$$ ou $$$y_0=y(x_0)$$$.
Ou $$$x_{1}$$$.

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Sua entrada

Encontre $$$y{\left(\frac{1}{2} \right)}$$$ para $$$y^{\prime } = e^{- t^{2}}$$$, quando $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{1}{10}$$$ usando o método de Euler modificado.

Solução

O método de Euler modificado afirma que $$$y_{n+1} = y_{n} + h f{\left(t_{n} + \frac{h}{2},y_{n} + \frac{h}{2} f{\left(t_{n},y_{n} \right)} \right)}$$$, onde $$$t_{n+1} = t_{n} + h$$$.

Temos que $$$h = \frac{1}{10}$$$, $$$t_{0} = 0$$$, $$$y_{0} = 1$$$ e $$$f{\left(t,y \right)} = e^{- t^{2}}$$$.

Passo 1

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

$$$f{\left(t_{0},y_{0} \right)} = f{\left(0,1 \right)} = 1$$$

$$$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{10} \right)} = y_{0} + h f{\left(t_{0} + \frac{h}{2},y_{0} + \frac{h}{2} f{\left(t_{0},y_{0} \right)} \right)} = 1 + \frac{f{\left(0 + \frac{\frac{1}{10}}{2},1 + \frac{\frac{1}{10}}{2} \cdot 1 \right)}}{10} = 1.09975031223975$$$

Passo 2

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

$$$f{\left(t_{1},y_{1} \right)} = f{\left(\frac{1}{10},1.09975031223975 \right)} = 0.990049833749168$$$

$$$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{1}{5} \right)} = y_{1} + h f{\left(t_{1} + \frac{h}{2},y_{1} + \frac{h}{2} f{\left(t_{1},y_{1} \right)} \right)} = 1.09975031223975 + \frac{f{\left(\frac{1}{10} + \frac{\frac{1}{10}}{2},1.09975031223975 + \frac{\frac{1}{10}}{2} \cdot 0.990049833749168 \right)}}{10} = 1.19752543595908$$$

Passo 3

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

$$$f{\left(t_{2},y_{2} \right)} = f{\left(\frac{1}{5},1.19752543595908 \right)} = 0.960789439152323$$$

$$$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{10} \right)} = y_{2} + h f{\left(t_{2} + \frac{h}{2},y_{2} + \frac{h}{2} f{\left(t_{2},y_{2} \right)} \right)} = 1.19752543595908 + \frac{f{\left(\frac{1}{5} + \frac{\frac{1}{10}}{2},1.19752543595908 + \frac{\frac{1}{10}}{2} \cdot 0.960789439152323 \right)}}{10} = 1.29146674224043$$$

Passo 4

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

$$$f{\left(t_{3},y_{3} \right)} = f{\left(\frac{3}{10},1.29146674224043 \right)} = 0.913931185271228$$$

$$$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{2}{5} \right)} = y_{3} + h f{\left(t_{3} + \frac{h}{2},y_{3} + \frac{h}{2} f{\left(t_{3},y_{3} \right)} \right)} = 1.29146674224043 + \frac{f{\left(\frac{3}{10} + \frac{\frac{1}{10}}{2},1.29146674224043 + \frac{\frac{1}{10}}{2} \cdot 0.913931185271228 \right)}}{10} = 1.37993733273478$$$

Passo 5

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

$$$f{\left(t_{4},y_{4} \right)} = f{\left(\frac{2}{5},1.37993733273478 \right)} = 0.852143788966211$$$

$$$y_{5} = y{\left(t_{5} \right)} = y{\left(\frac{1}{2} \right)} = y_{4} + h f{\left(t_{4} + \frac{h}{2},y_{4} + \frac{h}{2} f{\left(t_{4},y_{4} \right)} \right)} = 1.37993733273478 + \frac{f{\left(\frac{2}{5} + \frac{\frac{1}{10}}{2},1.37993733273478 + \frac{\frac{1}{10}}{2} \cdot 0.852143788966211 \right)}}{10} = 1.46160598099459$$$

Responder

$$$y{\left(\frac{1}{2} \right)}\approx 1.46160598099459$$$A