Runga Kutta Mthod 4

6 Ansichten (letzte 30 Tage)
Ali
Ali am 13 Nov. 2020
Bearbeitet: Alan Stevens am 14 Nov. 2020
I want to solve second order equation through Ruga kutta method .Please check what is error in this code
%input paramters
m=100;
k=25000;
e=100;
F=15000;
w=(k/m)^0.5;
t(1)=0;
x(1)=0;
v(1)=1;
h=0.1;
tfinal=100;
N=ceil((tfinal-t(1))/h);
f1=@(t,x,v) v;
f2=@(t,x,v) -(k/m)*v-(e/m)*x+(F/m)*cos(w*t);
for j=1:N;
t(j+1)=t(j)+h;
K1x=f1(t(j),x(j),v(j));
K1v=f2(t(j),x(j),v(j));
K2x=f1(t(j)+h/2,x(j)+h/2*K1x,v(j)+h/2*K1v);
K2v=f2(t(j)+h/2,x(j)+h/2*K1x,v(j)+h/2*K1v);
K3x=f1(t(j)+h/2,x(j)+h/2*K2x,v(j)+h/2*K2v);
K3v=f2(t(j)+h/2,x(j)+h/2*K2x,v(j)+h/2*K2v);
K4x=f1(t(j)+h,x(j)+h*K3x,v(j)+h*K3v);
K4v=f2(t(j)+h,x(j)+h*K3x,v(j)+h*K3v);
x(j+1)=x(j)+h/6*(K1x+2*K2x+2*K3x+K4x);
v(j+1)=v(j)+h/6*(K1v+2*K2v+2*K3v+K4v);
end
disp(x(N));
figure;
plot(t,x,'lineWidth',1);

Akzeptierte Antwort

Alan Stevens
Alan Stevens am 14 Nov. 2020
Bearbeitet: Alan Stevens am 14 Nov. 2020
As before, your value for h is too large. Set h = 0.01 and it works!

Weitere Antworten (0)

Kategorien

Mehr zu Fractals finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by