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Calculadora del método de Euler
La calculadora encontrará la solución aproximada de la ecuación diferencial de primer orden utilizando el método de Euler, con los pasos que se muestran.
Tu aportación
Encuentre $$$y{\left(1 \right)}$$$ para $$$y^{\prime } = t y$$$, cuando $$$y{\left(0 \right)} = 3$$$, $$$h = \frac{1}{5}$$$ usando el método de Euler.
Solución
El método de Euler establece que $$$y_{n+1} = y_{n} + h\cdot f{\left(t_{n},y_{n} \right)}$$$, donde $$$t_{n+1} = t_{n} + h$$$.
Tenemos eso $$$h = \frac{1}{5}$$$, $$$t_{0} = 0$$$, $$$y_{0} = 3$$$ $$$f{\left(t,y \right)} = t y$$$.
Paso 1
$$$t_{1} = t_{0} + h = 0 + \frac{1}{5} = \frac{1}{5}$$$
$$$y{\left(\frac{1}{5} \right)} = y{\left(t_{1} \right)} = y_{1} = y_{0} + h\cdot f{\left(t_{0},y_{0} \right)} = 3 + h\cdot f{\left(0,3 \right)} = 3 + \frac{1}{5} \cdot 0 = 3$$$
Paso 2
$$$t_{2} = t_{1} + h = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$$$
$$$y{\left(\frac{2}{5} \right)} = y{\left(t_{2} \right)} = y_{2} = y_{1} + h\cdot f{\left(t_{1},y_{1} \right)} = 3 + h\cdot f{\left(\frac{1}{5},3 \right)} = 3 + \frac{1}{5} \cdot 0.6 = 3.12$$$
Paso 3
$$$t_{3} = t_{2} + h = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$$$
$$$y{\left(\frac{3}{5} \right)} = y{\left(t_{3} \right)} = y_{3} = y_{2} + h\cdot f{\left(t_{2},y_{2} \right)} = 3.12 + h\cdot f{\left(\frac{2}{5},3.12 \right)} = 3.12 + \frac{1}{5} \cdot 1.248 = 3.3696$$$
Paso 4
$$$t_{4} = t_{3} + h = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$$$
$$$y{\left(\frac{4}{5} \right)} = y{\left(t_{4} \right)} = y_{4} = y_{3} + h\cdot f{\left(t_{3},y_{3} \right)} = 3.3696 + h\cdot f{\left(\frac{3}{5},3.3696 \right)} = 3.3696 + \frac{1}{5} \cdot 2.02176 = 3.773952$$$
Paso 5
$$$t_{5} = t_{4} + h = \frac{4}{5} + \frac{1}{5} = 1$$$
$$$y{\left(1 \right)} = y{\left(t_{5} \right)} = y_{5} = y_{4} + h\cdot f{\left(t_{4},y_{4} \right)} = 3.773952 + h\cdot f{\left(\frac{4}{5},3.773952 \right)} = 3.773952 + \frac{1}{5} \cdot 3.0191616 = 4.37778432$$$