Antworten (2)

Steve Areola
Steve Areola am 5 Aug. 2023

0 Stimmen

clc
a=0;b=2;N=10; alph=0.5;
h=(b-a)/N; t(1)=a; w(1)=alph;
f = @(t,y) y-t^2+1;
for i=2:N+1
t(i)=a+(i+1)*h;
w(i)=w(i-1)+h*f(t(i-1),w(i-1));
end
disp("t w")
x=[t;w];
fprintf("%5.9f %5.9f\n",x)

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Gefragt:

am 13 Jan. 2021

Beantwortet:

am 5 Aug. 2023

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