Lösung
Die verbesserte Eulersche Methode besagt, dass yn+1=yn+2h(f(tn,yn)+f(tn+1,y~n+1)), wobei y~n+1=yn+h⋅f(tn,yn) und tn+1=tn+h.
Wir haben, dass h=51, t0=0, y0=7 und f(t,y)=3t+y.
Schritt 1
t1=t0+h=0+51=51
y~1=y~(t1)=y~(51)=y0+h⋅f(t0,y0)=7+h⋅f(0,7)=7+51⋅7=8.4
y1=y(t1)=y(51)=y0+2h(f(t0,y0)+f(t1,y~1))=7+2h(f(0,7)+f(51,8.4))=7+251(7+9)=8.6
Schritt 2
t2=t1+h=51+51=52
y~2=y~(t2)=y~(52)=y1+h⋅f(t1,y1)=8.6+h⋅f(51,8.6)=8.6+51⋅9.2=10.44
y2=y(t2)=y(52)=y1+2h(f(t1,y1)+f(t2,y~2))=8.6+2h(f(51,8.6)+f(52,10.44))=8.6+251(9.2+11.64)=10.684
Schritt 3
t3=t2+h=52+51=53
y~3=y~(t3)=y~(53)=y2+h⋅f(t2,y2)=10.684+h⋅f(52,10.684)=10.684+51⋅11.884=13.0608
y3=y(t3)=y(53)=y2+2h(f(t2,y2)+f(t3,y~3))=10.684+2h(f(52,10.684)+f(53,13.0608))=10.684+251(11.884+14.8608)=13.35848
Schritt 4
t4=t3+h=53+51=54
y~4=y~(t4)=y~(54)=y3+h⋅f(t3,y3)=13.35848+h⋅f(53,13.35848)=13.35848+51⋅15.15848=16.390176
y4=y(t4)=y(54)=y3+2h(f(t3,y3)+f(t4,y~4))=13.35848+2h(f(53,13.35848)+f(54,16.390176))=13.35848+251(15.15848+18.790176)=16.7533456
Schritt 5
t5=t4+h=54+51=1
y~5=y~(t5)=y~(1)=y4+h⋅f(t4,y4)=16.7533456+h⋅f(54,16.7533456)=16.7533456+51⋅19.1533456=20.58401472
y5=y(t5)=y(1)=y4+2h(f(t4,y4)+f(t5,y~5))=16.7533456+2h(f(54,16.7533456)+f(1,20.58401472))=16.7533456+251(19.1533456+23.58401472)=21.027081632