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Lorenz 방정식의 상태-공간 표현 (4th Runge Kutta Methode)
방정식 초기값은 t = 0
dx/dt = -10x + 10y x = 5, y = 5, z = 5
dy/dt = 28x - y - xz
dz/dt = -2.6666667z + xy
Table 작성
t x y z k11 k12 k13
0.0 5.000000 5.000000 5.000000 0.000000 110.000000 11.666665
0.1 9.781470 17.079464 10.439
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f) = %lf u2(%1.1f) = %lf error = %lf\n", t, w2[i], t, u2(t), u2(t)-w2[i]);
}
}
double f1(double t, double w1, double w2){
return 3*w1 + 2*w2 - (2*t*t+1)*exp(2*t);
}
double f2(double t, double w1, double w2){
return 4*w1 + w2 + (t*t + 2*t - 4)*exp(2*t);
}
double u1(double t){
return exp(5*t)/3 - exp(
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++)
{
for(kount=1; kount<=100; kount++)
{
kstep=kstep+1;
time=h*kstep;
k1=h*z;
l1=-h*(bm*z+km*y);
k2=h*(z+l1/2);
l2=-h*(bm*(z+l1/2)+km*(y+k1/2));
k3=h*(z+l2/2);
l3=-h*(bm*(z+l2/2)+km*(y+k2/2));
k4=h*(z+l3);
l4=-h*(bm*(z+13)+km*(y+k3));
y =y+(k1+2*k2+2*k3+k4)/6;
z =z+(l1+2*l2+2*l3+l4)/6;
}
printf(
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