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Kalkulator zmodyfikowanej metody Eulera

Zastosowanie zmodyfikowanej metody Eulera krok po kroku

Kalkulator znajdzie przybliżone rozwiązanie równania różniczkowego pierwszego rzędu przy użyciu zmodyfikowanej metody Eulera, z pokazanymi krokami.

Powiązane kalkulatory: Kalkulator metody Eulera, Ulepszony kalkulator metody Eulera (Heuna)

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

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Twój wkład

Znaleźć y(1)y{\left(1 \right)} dla y′(t)=2t−yy^{\prime }\left(t\right) = 2 t - y, gdy y(0)=1y{\left(0 \right)} = 1, h=15h = \frac{1}{5} używając zmodyfikowanej metody Eulera.

Rozwiązanie

Zmodyfikowana metoda Eulera mówi, że yn+1=yn+hf(tn+h2,yn+h2f(tn,yn))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)}, gdzie tn+1=tn+ht_{n+1} = t_{n} + h.

Mamy, że h=15h = \frac{1}{5}, t0=0t_{0} = 0, y0=1y_{0} = 1 i f(t,y)=2t−yf{\left(t,y \right)} = 2 t - y.

Krok 1

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

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

y1=y(t1)=y(15)=y0+hf(t0+h2,y0+h2f(t0,y0))=1+f(0+152,1+152(−1))5=0.86y_{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

Krok 2

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

f(t1,y1)=f(15,0.86)=−0.46f{\left(t_{1},y_{1} \right)} = f{\left(\frac{1}{5},0.86 \right)} = -0.46

y2=y(t2)=y(25)=y1+hf(t1+h2,y1+h2f(t1,y1))=0.86+f(15+152,0.86+152(−0.46))5=0.8172y_{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

Krok 3

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

f(t2,y2)=f(25,0.8172)=−0.0172f{\left(t_{2},y_{2} \right)} = f{\left(\frac{2}{5},0.8172 \right)} = -0.0172

y3=y(t3)=y(35)=y2+hf(t2+h2,y2+h2f(t2,y2))=0.8172+f(25+152,0.8172+152(−0.0172))5=0.854104y_{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

Krok 4

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

f(t3,y3)=f(35,0.854104)=0.345896f{\left(t_{3},y_{3} \right)} = f{\left(\frac{3}{5},0.854104 \right)} = 0.345896

y4=y(t4)=y(45)=y3+hf(t3+h2,y3+h2f(t3,y3))=0.854104+f(35+152,0.854104+152⋅0.345896)5=0.95636528y_{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

Krok 5

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

f(t4,y4)=f(45,0.95636528)=0.64363472f{\left(t_{4},y_{4} \right)} = f{\left(\frac{4}{5},0.95636528 \right)} = 0.64363472

y5=y(t5)=y(1)=y4+hf(t4+h2,y4+h2f(t4,y4))=0.95636528+f(45+152,0.95636528+152⋅0.64363472)5=1.1122195296y_{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

Odpowiedź

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