Calculadora del método de Runge-Kutta de cuarto orden

La calculadora encontrará la solución aproximada de la ecuación diferencial de primer orden usando el método clásico de Runge-Kutta de cuarto orden, con pasos mostrados.

$$$y^{\prime } = f{\left(t,y \right)}$$$:

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

Encuentre $$$y{\left(2 \right)}$$$ para $$$y^{\prime } = \sin{\left(t y \right)}$$$, cuando $$$y{\left(0 \right)} = 1$$$, $$$h = \frac{2}{5}$$$ usando el método clásico de Runge-Kutta de cuarto orden.

Solución

El método de Runge-Kutta establece que $$$y_{n+1} = y_{n} + \frac{h}{6} \left(k_{1} + 2 k_{2} + 2 k_{3} + k_{4}\right)$$$, donde $$$t_{n+1} = t_{n} + h$$$, $$$k_{1} = f{\left(t_{n},y_{n} \right)}$$$, $$$k_{2} = f{\left(t_{n} + \frac{h}{2},y_{n} + \frac{h k_{1}}{2} \right)}$$$, $$$k_{3} = f{\left(t_{n} + \frac{h}{2},y_{n} + \frac{h k_{2}}{2} \right)}$$$ y $$$k_{4} = f{\left(t_{n} + h,y_{n} + h k_{3} \right)}$$$.

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

Paso 1

$$$t_{1} = t_{0} + h = 0 + \frac{2}{5} = \frac{2}{5}$$$

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

$$$k_{2} = f{\left(t_{0} + \frac{h}{2},y_{0} + \frac{h k_{1}}{2} \right)} = f{\left(0 + \frac{\frac{2}{5}}{2},1 + \frac{\left(\frac{2}{5}\right)\cdot \left(0\right)}{2} \right)} = f{\left(\frac{1}{5},1 \right)} = 0.198669330795061$$$

$$$k_{3} = f{\left(t_{0} + \frac{h}{2},y_{0} + \frac{h k_{2}}{2} \right)} = f{\left(0 + \frac{\frac{2}{5}}{2},1 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.198669330795061\right)}{2} \right)} = f{\left(\frac{1}{5},1.03973386615901 \right)} = 0.206451342596583$$$

$$$k_{4} = f{\left(t_{0} + h,y_{0} + h k_{3} \right)} = f{\left(0 + \frac{2}{5},1 + \left(\frac{2}{5}\right)\cdot \left(0.206451342596583\right) \right)} = f{\left(\frac{2}{5},1.08258053703863 \right)} = 0.419625061196877$$$

$$$y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{2}{5} \right)} = y_{0} + \frac{h}{6} \left(k_{1} + 2 k_{2} + 2 k_{3} + k_{4}\right) = 1 + \frac{\frac{2}{5}}{6} \left(0 + 2 \cdot 0.198669330795061 + 2 \cdot 0.206451342596583 + 0.419625061196877\right) = 1.08199109386534$$$

Paso 2

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

$$$k_{1} = f{\left(t_{1},y_{1} \right)} = f{\left(\frac{2}{5},1.08199109386534 \right)} = 0.419411035089935$$$

$$$k_{2} = f{\left(t_{1} + \frac{h}{2},y_{1} + \frac{h k_{1}}{2} \right)} = f{\left(\frac{2}{5} + \frac{\frac{2}{5}}{2},1.08199109386534 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.419411035089935\right)}{2} \right)} = f{\left(\frac{3}{5},1.16587330088333 \right)} = 0.643853534490712$$$

$$$k_{3} = f{\left(t_{1} + \frac{h}{2},y_{1} + \frac{h k_{2}}{2} \right)} = f{\left(\frac{2}{5} + \frac{\frac{2}{5}}{2},1.08199109386534 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.643853534490712\right)}{2} \right)} = f{\left(\frac{3}{5},1.21076180076349 \right)} = 0.664225362212255$$$

$$$k_{4} = f{\left(t_{1} + h,y_{1} + h k_{3} \right)} = f{\left(\frac{2}{5} + \frac{2}{5},1.08199109386534 + \left(\frac{2}{5}\right)\cdot \left(0.664225362212255\right) \right)} = f{\left(\frac{4}{5},1.34768123875025 \right)} = 0.881081971595253$$$

$$$y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{4}{5} \right)} = y_{1} + \frac{h}{6} \left(k_{1} + 2 k_{2} + 2 k_{3} + k_{4}\right) = 1.08199109386534 + \frac{\frac{2}{5}}{6} \left(0.419411035089935 + 2 \cdot 0.643853534490712 + 2 \cdot 0.664225362212255 + 0.881081971595253\right) = 1.34310114720475$$$

Paso 3

$$$t_{3} = t_{2} + h = \frac{4}{5} + \frac{2}{5} = \frac{6}{5}$$$

$$$k_{1} = f{\left(t_{2},y_{2} \right)} = f{\left(\frac{4}{5},1.34310114720475 \right)} = 0.879343087787042$$$

$$$k_{2} = f{\left(t_{2} + \frac{h}{2},y_{2} + \frac{h k_{1}}{2} \right)} = f{\left(\frac{4}{5} + \frac{\frac{2}{5}}{2},1.34310114720475 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.879343087787042\right)}{2} \right)} = f{\left(1,1.51896976476216 \right)} = 0.998657304313516$$$

$$$k_{3} = f{\left(t_{2} + \frac{h}{2},y_{2} + \frac{h k_{2}}{2} \right)} = f{\left(\frac{4}{5} + \frac{\frac{2}{5}}{2},1.34310114720475 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.998657304313516\right)}{2} \right)} = f{\left(1,1.54283260806746 \right)} = 0.999609040694986$$$

$$$k_{4} = f{\left(t_{2} + h,y_{2} + h k_{3} \right)} = f{\left(\frac{4}{5} + \frac{2}{5},1.34310114720475 + \left(\frac{2}{5}\right)\cdot \left(0.999609040694986\right) \right)} = f{\left(\frac{6}{5},1.74294476348275 \right)} = 0.867452549636552$$$

$$$y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{6}{5} \right)} = y_{2} + \frac{h}{6} \left(k_{1} + 2 k_{2} + 2 k_{3} + k_{4}\right) = 1.34310114720475 + \frac{\frac{2}{5}}{6} \left(0.879343087787042 + 2 \cdot 0.998657304313516 + 2 \cdot 0.999609040694986 + 0.867452549636552\right) = 1.72598970236746$$$

Paso 4

$$$t_{4} = t_{3} + h = \frac{6}{5} + \frac{2}{5} = \frac{8}{5}$$$

$$$k_{1} = f{\left(t_{3},y_{3} \right)} = f{\left(\frac{6}{5},1.72598970236746 \right)} = 0.877394887797677$$$

$$$k_{2} = f{\left(t_{3} + \frac{h}{2},y_{3} + \frac{h k_{1}}{2} \right)} = f{\left(\frac{6}{5} + \frac{\frac{2}{5}}{2},1.72598970236746 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.877394887797677\right)}{2} \right)} = f{\left(\frac{7}{5},1.90146867992699 \right)} = 0.461368005308125$$$

$$$k_{3} = f{\left(t_{3} + \frac{h}{2},y_{3} + \frac{h k_{2}}{2} \right)} = f{\left(\frac{6}{5} + \frac{\frac{2}{5}}{2},1.72598970236746 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.461368005308125\right)}{2} \right)} = f{\left(\frac{7}{5},1.81826330342908 \right)} = 0.561356508370458$$$

$$$k_{4} = f{\left(t_{3} + h,y_{3} + h k_{3} \right)} = f{\left(\frac{6}{5} + \frac{2}{5},1.72598970236746 + \left(\frac{2}{5}\right)\cdot \left(0.561356508370458\right) \right)} = f{\left(\frac{8}{5},1.95053230571564 \right)} = 0.020739477392444$$$

$$$y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{8}{5} \right)} = y_{3} + \frac{h}{6} \left(k_{1} + 2 k_{2} + 2 k_{3} + k_{4}\right) = 1.72598970236746 + \frac{\frac{2}{5}}{6} \left(0.877394887797677 + 2 \cdot 0.461368005308125 + 2 \cdot 0.561356508370458 + 0.020739477392444\right) = 1.92222859520394$$$

Paso 5

$$$t_{5} = t_{4} + h = \frac{8}{5} + \frac{2}{5} = 2$$$

$$$k_{1} = f{\left(t_{4},y_{4} \right)} = f{\left(\frac{8}{5},1.92222859520394 \right)} = 0.06597893710495$$$

$$$k_{2} = f{\left(t_{4} + \frac{h}{2},y_{4} + \frac{h k_{1}}{2} \right)} = f{\left(\frac{8}{5} + \frac{\frac{2}{5}}{2},1.92222859520394 + \frac{\left(\frac{2}{5}\right)\cdot \left(0.06597893710495\right)}{2} \right)} = f{\left(\frac{9}{5},1.93542438262493 \right)} = -0.33553324651362$$$

$$$k_{3} = f{\left(t_{4} + \frac{h}{2},y_{4} + \frac{h k_{2}}{2} \right)} = f{\left(\frac{8}{5} + \frac{\frac{2}{5}}{2},1.92222859520394 + \frac{\left(\frac{2}{5}\right)\cdot \left(-0.33553324651362\right)}{2} \right)} = f{\left(\frac{9}{5},1.85512194590122 \right)} = -0.196342927593455$$$

$$$k_{4} = f{\left(t_{4} + h,y_{4} + h k_{3} \right)} = f{\left(\frac{8}{5} + \frac{2}{5},1.92222859520394 + \left(\frac{2}{5}\right)\cdot \left(-0.196342927593455\right) \right)} = f{\left(2,1.84369142416656 \right)} = -0.519093645672128$$$

$$$y_{5} = y{\left(t_{5} \right)} = y{\left(2 \right)} = y_{4} + \frac{h}{6} \left(k_{1} + 2 k_{2} + 2 k_{3} + k_{4}\right) = 1.92222859520394 + \frac{\frac{2}{5}}{6} \left(0.06597893710495 + 2 \left(-0.33553324651362\right) + 2 \left(-0.196342927593455\right) - 0.519093645672128\right) = 1.82110412475186$$$

Respuesta

$$$y{\left(2 \right)}\approx 1.82110412475186$$$A