Mathc initiation/Fichiers h : x 19c1c
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c1c.c |
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/* ---------------------------------- */
/* save as c1c.c */
/* ---------------------------------- */
#include "x_hfile.h"
#include "fc.h"
/* --------------------------------- */
int main(void)
{
double n;
double a;
double b;
double y0;
n = 10.0;
a = 0.0;
b = 2.0;
y0 = -1.5;
clrscrn();
printf(" Euler's method to approximate the solution\n"
" of the differential equation.\n\n"
" y' = %s\n\n",Ypeq);
p_Euler_s_Method(a,
b,
n,
y0,
Yp);
printf(" y_n = %.10f\n\n",
Euler_s_Method(a,
b,
n,
y0,
Yp)
);
printf(" The exact value is y = -1.012136331\n\n");
stop();
clrscrn();
n = 10000000.0;
printf(" Euler's method to approximate the solution\n"
" of the differential equation\n\n"
" y' = %s,\n\n with n = %.0f \n\n",Ypeq, n);
printf(" y_n = %.10f\n\n",
Euler_s_Method(a,
b,
n,
y0,
Yp)
);
printf(" The exact value is y = -1.012136331\n\n");
stop();
return 0;
}
/* --------------------------------- */
/* --------------------------------- */
Calculons la solution numérique de l'équation
y' = -(y+1)*(y+3) pour 0 < x < 2
avec comme condition initial y0 = -1.5 quand x = 0
Exemple de sortie écran 1 :
Euler's method to approximate the solution
of the differential equation.
y' = -(y+1)*(y+3)
k | x_k | y_k
--------------------------
1 | 0.200 | -1.3500000000
2 | 0.400 | -1.2345000000
3 | 0.600 | -1.1516980500
4 | 0.800 | -1.0956212897
5 | 1.000 | -1.0592014600
6 | 1.200 | -1.0362218386
7 | 1.400 | -1.0219955075
8 | 1.600 | -1.0132940649
9 | 1.800 | -1.0080117854
10 | 2.000 | -1.0048199090
y_n = -1.0048199090
The exact value is y = -1.012136331
Press return to continue.
Exemple de sortie écran : 2
Euler's method to approximate the solution
of the differential equation
y' = -(y+1)*(y+3),
with n = 10000000
y_n = -1.0121363227
The exact value is y = -1.012136331
Press return to continue.