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

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

## Tu aportación

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

### Paso 1

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

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

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

### Paso 2

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

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

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

### Paso 3

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

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

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

### Paso 4

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

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

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

### Paso 5

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

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

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

## Respuesta

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