Verbesserter Rechner für die Euler-Methode (Heun's)

Anwendung der Heun'schen Methode Schritt für Schritt

Der Rechner findet die ungefähre Lösung der Differentialgleichung erster Ordnung mit Hilfe des verbesserten Euler-Verfahrens (Heun-Verfahren), wobei die Schritte angezeigt werden.

Verwandte Rechner: Rechner für die Eulersche Methode, Rechner für die modifizierte Eulersche Methode

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

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Ihr Beitrag

Finde y(1)y{\left(1 \right)} für y(t)=3t+yy^{\prime }\left(t\right) = 3 t + y, wenn y(0)=7y{\left(0 \right)} = 7, h=15h = \frac{1}{5} mit Hilfe der verbesserten Eulerschen Methode.

Lösung

Die verbesserte Eulersche Methode besagt, dass yn+1=yn+h2(f(tn,yn)+f(tn+1,y~n+1))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), wobei y~n+1=yn+hf(tn,yn)\tilde{y}_{n+1} = y_{n} + h\cdot f{\left(t_{n},y_{n} \right)} und tn+1=tn+ht_{n+1} = t_{n} + h.

Wir haben, dass h=15h = \frac{1}{5}, t0=0t_{0} = 0, y0=7y_{0} = 7 und f(t,y)=3t+yf{\left(t,y \right)} = 3 t + y.

Schritt 1

t1=t0+h=0+15=15t_{1} = t_{0} + h = 0 + \frac{1}{5} = \frac{1}{5}

y~1=y~(t1)=y~(15)=y0+hf(t0,y0)=7+hf(0,7)=7+157=8.4\tilde{y}_{1} = \tilde{y}{\left(t_{1} \right)} = \tilde{y}{\left(\frac{1}{5} \right)} = 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

y1=y(t1)=y(15)=y0+h2(f(t0,y0)+f(t1,y~1))=7+h2(f(0,7)+f(15,8.4))=7+152(7+9)=8.6y_{1} = y{\left(t_{1} \right)} = y{\left(\frac{1}{5} \right)} = 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

Schritt 2

t2=t1+h=15+15=25t_{2} = t_{1} + h = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}

y~2=y~(t2)=y~(25)=y1+hf(t1,y1)=8.6+hf(15,8.6)=8.6+159.2=10.44\tilde{y}_{2} = \tilde{y}{\left(t_{2} \right)} = \tilde{y}{\left(\frac{2}{5} \right)} = 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 9.2 = 10.44

y2=y(t2)=y(25)=y1+h2(f(t1,y1)+f(t2,y~2))=8.6+h2(f(15,8.6)+f(25,10.44))=8.6+152(9.2+11.64)=10.684y_{2} = y{\left(t_{2} \right)} = y{\left(\frac{2}{5} \right)} = 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(9.2 + 11.64\right) = 10.684

Schritt 3

t3=t2+h=25+15=35t_{3} = t_{2} + h = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}

y~3=y~(t3)=y~(35)=y2+hf(t2,y2)=10.684+hf(25,10.684)=10.684+1511.884=13.0608\tilde{y}_{3} = \tilde{y}{\left(t_{3} \right)} = \tilde{y}{\left(\frac{3}{5} \right)} = 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

y3=y(t3)=y(35)=y2+h2(f(t2,y2)+f(t3,y~3))=10.684+h2(f(25,10.684)+f(35,13.0608))=10.684+152(11.884+14.8608)=13.35848y_{3} = y{\left(t_{3} \right)} = y{\left(\frac{3}{5} \right)} = 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

Schritt 4

t4=t3+h=35+15=45t_{4} = t_{3} + h = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}

y~4=y~(t4)=y~(45)=y3+hf(t3,y3)=13.35848+hf(35,13.35848)=13.35848+1515.15848=16.390176\tilde{y}_{4} = \tilde{y}{\left(t_{4} \right)} = \tilde{y}{\left(\frac{4}{5} \right)} = 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

y4=y(t4)=y(45)=y3+h2(f(t3,y3)+f(t4,y~4))=13.35848+h2(f(35,13.35848)+f(45,16.390176))=13.35848+152(15.15848+18.790176)=16.7533456y_{4} = y{\left(t_{4} \right)} = y{\left(\frac{4}{5} \right)} = 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

Schritt 5

t5=t4+h=45+15=1t_{5} = t_{4} + h = \frac{4}{5} + \frac{1}{5} = 1

y~5=y~(t5)=y~(1)=y4+hf(t4,y4)=16.7533456+hf(45,16.7533456)=16.7533456+1519.1533456=20.58401472\tilde{y}_{5} = \tilde{y}{\left(t_{5} \right)} = \tilde{y}{\left(1 \right)} = 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

y5=y(t5)=y(1)=y4+h2(f(t4,y4)+f(t5,y~5))=16.7533456+h2(f(45,16.7533456)+f(1,20.58401472))=16.7533456+152(19.1533456+23.58401472)=21.027081632y_{5} = y{\left(t_{5} \right)} = y{\left(1 \right)} = 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

Antwort

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