Lösung
Die modifizierte Eulersche Methode besagt, dass yn+1=yn+hf(tn+2h,yn+2hf(tn,yn)), wobei tn+1=tn+h.
Wir haben, dass h=51, t0=0, y0=1 und f(t,y)=2t−y.
Schritt 1
t1=t0+h=0+51=51
f(t0,y0)=f(0,1)=−1
y1=y(t1)=y(51)=y0+hf(t0+2h,y0+2hf(t0,y0))=1+5f(0+251,1+251(−1))=0.86
Schritt 2
t2=t1+h=51+51=52
f(t1,y1)=f(51,0.86)=−0.46
y2=y(t2)=y(52)=y1+hf(t1+2h,y1+2hf(t1,y1))=0.86+5f(51+251,0.86+251(−0.46))=0.8172
Schritt 3
t3=t2+h=52+51=53
f(t2,y2)=f(52,0.8172)=−0.0172
y3=y(t3)=y(53)=y2+hf(t2+2h,y2+2hf(t2,y2))=0.8172+5f(52+251,0.8172+251(−0.0172))=0.854104
Schritt 4
t4=t3+h=53+51=54
f(t3,y3)=f(53,0.854104)=0.345896
y4=y(t4)=y(54)=y3+hf(t3+2h,y3+2hf(t3,y3))=0.854104+5f(53+251,0.854104+251⋅0.345896)=0.95636528
Schritt 5
t5=t4+h=54+51=1
f(t4,y4)=f(54,0.95636528)=0.64363472
y5=y(t5)=y(1)=y4+hf(t4+2h,y4+2hf(t4,y4))=0.95636528+5f(54+251,0.95636528+251⋅0.64363472)=1.1122195296