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Calculadora del método Runge-Kutta de cuarto orden

Aplica el método Runge-Kutta de cuarto orden paso a paso

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

O y′(x)=f(x,y)y^{\prime }\left(x\right) = f{\left(x,y \right)}.
O x0x_{0}.
y0=y(t0)y_0=y(t_0) o y0=y(x0)y_0=y(x_0).
O x1x_{1}.

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Su opinión

Encuentre y(2)y{\left(2 \right)} para y′(t)=sin⁡(ty)y^{\prime }\left(t\right) = \sin{\left(t y \right)}, cuando y(0)=1y{\left(0 \right)} = 1, h=25h = \frac{2}{5} utilizando el método clásico Runge-Kutta de cuarto orden.

Solución

El método Runge-Kutta establece que yn+1=yn+h6(k1+2k2+2k3+k4)y_{n+1} = y_{n} + \frac{h}{6} \left(k_{1} + 2 k_{2} + 2 k_{3} + k_{4}\right), donde tn+1=tn+ht_{n+1} = t_{n} + h, k1=f(tn,yn)k_{1} = f{\left(t_{n},y_{n} \right)}, k2=f(tn+h2,yn+hk12)k_{2} = f{\left(t_{n} + \frac{h}{2},y_{n} + \frac{h k_{1}}{2} \right)}, k3=f(tn+h2,yn+hk22)k_{3} = f{\left(t_{n} + \frac{h}{2},y_{n} + \frac{h k_{2}}{2} \right)}, y k4=f(tn+h,yn+hk3)k_{4} = f{\left(t_{n} + h,y_{n} + h k_{3} \right)}.

Tenemos que h=25h = \frac{2}{5}, t0=0t_{0} = 0, y0=1y_{0} = 1, y f(t,y)=sin⁡(ty)f{\left(t,y \right)} = \sin{\left(t y \right)}.

Paso 1

t1=t0+h=0+25=25t_{1} = t_{0} + h = 0 + \frac{2}{5} = \frac{2}{5}

k1=f(t0,y0)=f(0,1)=0k_{1} = f{\left(t_{0},y_{0} \right)} = f{\left(0,1 \right)} = 0

k2=f(t0+h2,y0+hk12)=f(0+252,1+(25)⋅(0)2)=f(15,1)=0.198669330795061k_{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

k3=f(t0+h2,y0+hk22)=f(0+252,1+(25)⋅(0.198669330795061)2)=f(15,1.03973386615901)=0.206451342596583k_{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

k4=f(t0+h,y0+hk3)=f(0+25,1+(25)⋅(0.206451342596583))=f(25,1.08258053703863)=0.419625061196877k_{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

y1=y(t1)=y(25)=y0+h6(k1+2k2+2k3+k4)=1+256(0+2⋅0.198669330795061+2⋅0.206451342596583+0.419625061196877)=1.08199109386534y_{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

t2=t1+h=25+25=45t_{2} = t_{1} + h = \frac{2}{5} + \frac{2}{5} = \frac{4}{5}

k1=f(t1,y1)=f(25,1.08199109386534)=0.419411035089935k_{1} = f{\left(t_{1},y_{1} \right)} = f{\left(\frac{2}{5},1.08199109386534 \right)} = 0.419411035089935

k2=f(t1+h2,y1+hk12)=f(25+252,1.08199109386534+(25)⋅(0.419411035089935)2)=f(35,1.16587330088333)=0.643853534490712k_{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

k3=f(t1+h2,y1+hk22)=f(25+252,1.08199109386534+(25)⋅(0.643853534490712)2)=f(35,1.21076180076349)=0.664225362212255k_{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

k4=f(t1+h,y1+hk3)=f(25+25,1.08199109386534+(25)⋅(0.664225362212255))=f(45,1.34768123875025)=0.881081971595253k_{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

y2=y(t2)=y(45)=y1+h6(k1+2k2+2k3+k4)=1.08199109386534+256(0.419411035089935+2⋅0.643853534490712+2⋅0.664225362212255+0.881081971595253)=1.34310114720475y_{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

t3=t2+h=45+25=65t_{3} = t_{2} + h = \frac{4}{5} + \frac{2}{5} = \frac{6}{5}

k1=f(t2,y2)=f(45,1.34310114720475)=0.879343087787042k_{1} = f{\left(t_{2},y_{2} \right)} = f{\left(\frac{4}{5},1.34310114720475 \right)} = 0.879343087787042

k2=f(t2+h2,y2+hk12)=f(45+252,1.34310114720475+(25)⋅(0.879343087787042)2)=f(1,1.51896976476216)=0.998657304313516k_{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

k3=f(t2+h2,y2+hk22)=f(45+252,1.34310114720475+(25)⋅(0.998657304313516)2)=f(1,1.54283260806746)=0.999609040694986k_{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

k4=f(t2+h,y2+hk3)=f(45+25,1.34310114720475+(25)⋅(0.999609040694986))=f(65,1.74294476348275)=0.867452549636552k_{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

y3=y(t3)=y(65)=y2+h6(k1+2k2+2k3+k4)=1.34310114720475+256(0.879343087787042+2⋅0.998657304313516+2⋅0.999609040694986+0.867452549636552)=1.72598970236746y_{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

t4=t3+h=65+25=85t_{4} = t_{3} + h = \frac{6}{5} + \frac{2}{5} = \frac{8}{5}

k1=f(t3,y3)=f(65,1.72598970236746)=0.877394887797677k_{1} = f{\left(t_{3},y_{3} \right)} = f{\left(\frac{6}{5},1.72598970236746 \right)} = 0.877394887797677

k2=f(t3+h2,y3+hk12)=f(65+252,1.72598970236746+(25)⋅(0.877394887797677)2)=f(75,1.90146867992699)=0.461368005308125k_{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

k3=f(t3+h2,y3+hk22)=f(65+252,1.72598970236746+(25)⋅(0.461368005308125)2)=f(75,1.81826330342908)=0.561356508370458k_{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

k4=f(t3+h,y3+hk3)=f(65+25,1.72598970236746+(25)⋅(0.561356508370458))=f(85,1.95053230571564)=0.020739477392444k_{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

y4=y(t4)=y(85)=y3+h6(k1+2k2+2k3+k4)=1.72598970236746+256(0.877394887797677+2⋅0.461368005308125+2⋅0.561356508370458+0.020739477392444)=1.92222859520394y_{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

t5=t4+h=85+25=2t_{5} = t_{4} + h = \frac{8}{5} + \frac{2}{5} = 2

k1=f(t4,y4)=f(85,1.92222859520394)=0.06597893710495k_{1} = f{\left(t_{4},y_{4} \right)} = f{\left(\frac{8}{5},1.92222859520394 \right)} = 0.06597893710495

k2=f(t4+h2,y4+hk12)=f(85+252,1.92222859520394+(25)⋅(0.06597893710495)2)=f(95,1.93542438262493)=−0.33553324651362k_{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

k3=f(t4+h2,y4+hk22)=f(85+252,1.92222859520394+(25)⋅(−0.33553324651362)2)=f(95,1.85512194590122)=−0.196342927593455k_{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

k4=f(t4+h,y4+hk3)=f(85+25,1.92222859520394+(25)⋅(−0.196342927593455))=f(2,1.84369142416656)=−0.519093645672128k_{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

y5=y(t5)=y(2)=y4+h6(k1+2k2+2k3+k4)=1.92222859520394+256(0.06597893710495+2(−0.33553324651362)+2(−0.196342927593455)−0.519093645672128)=1.82110412475186y_{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(2)≈1.82110412475186y{\left(2 \right)}\approx 1.82110412475186A