## Aplicar el método de Euler modificado paso a paso

La calculadora encontrará la solución aproximada de la ecuación diferencial de primer orden utilizando el método de Euler modificado, con los pasos que se muestran.

O $y^{\prime } = f{\left(x,y \right)}$.
O $x_{0}$.
$y_0=y(t_0)$ o $y_0=y(x_0)$.
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 $y^{\prime } = 2 t - y$, cuando $y{\left(0 \right)} = 1$, $h = \frac{1}{5}$ usando el método de Euler modificado.

### Solución

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

Tenemos que $h = \frac{1}{5}$, $t_{0} = 0$, $y_{0} = 1$ y $f{\left(t,y \right)} = 2 t - y$.

### Paso 1

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

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

$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{5} \right)} = y_{0} + h f{\left(t_{0} + \frac{h}{2},y_{0} + \frac{h}{2} f{\left(t_{0},y_{0} \right)} \right)} = 1 + \frac{f{\left(0 + \frac{\frac{1}{5}}{2},1 + \frac{\frac{1}{5}}{2} \left(-1\right) \right)}}{5} = 0.86$

### Paso 2

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

$f{\left(t_{1},y_{1} \right)} = f{\left(\frac{1}{5},0.86 \right)} = -0.46$

$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{2}{5} \right)} = y_{1} + h f{\left(t_{1} + \frac{h}{2},y_{1} + \frac{h}{2} f{\left(t_{1},y_{1} \right)} \right)} = 0.86 + \frac{f{\left(\frac{1}{5} + \frac{\frac{1}{5}}{2},0.86 + \frac{\frac{1}{5}}{2} \left(-0.46\right) \right)}}{5} = 0.8172$

### Paso 3

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

$f{\left(t_{2},y_{2} \right)} = f{\left(\frac{2}{5},0.8172 \right)} = -0.0172$

$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{5} \right)} = y_{2} + h f{\left(t_{2} + \frac{h}{2},y_{2} + \frac{h}{2} f{\left(t_{2},y_{2} \right)} \right)} = 0.8172 + \frac{f{\left(\frac{2}{5} + \frac{\frac{1}{5}}{2},0.8172 + \frac{\frac{1}{5}}{2} \left(-0.0172\right) \right)}}{5} = 0.854104$

### Paso 4

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

$f{\left(t_{3},y_{3} \right)} = f{\left(\frac{3}{5},0.854104 \right)} = 0.345896$

$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{4}{5} \right)} = y_{3} + h f{\left(t_{3} + \frac{h}{2},y_{3} + \frac{h}{2} f{\left(t_{3},y_{3} \right)} \right)} = 0.854104 + \frac{f{\left(\frac{3}{5} + \frac{\frac{1}{5}}{2},0.854104 + \frac{\frac{1}{5}}{2} \cdot 0.345896 \right)}}{5} = 0.95636528$

### Paso 5

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

$f{\left(t_{4},y_{4} \right)} = f{\left(\frac{4}{5},0.95636528 \right)} = 0.64363472$

$y_{5} = y{\left(t_{5} \right)} = y{\left(1 \right)} = y_{4} + h f{\left(t_{4} + \frac{h}{2},y_{4} + \frac{h}{2} f{\left(t_{4},y_{4} \right)} \right)} = 0.95636528 + \frac{f{\left(\frac{4}{5} + \frac{\frac{1}{5}}{2},0.95636528 + \frac{\frac{1}{5}}{2} \cdot 0.64363472 \right)}}{5} = 1.1122195296$

### Respuesta

$y{\left(1 \right)}\approx 1.1122195296$A