C語言 例題5-1利用向前Euler尤拉 向前近似法 解一階常微分方程ODE y' = -y+x+1 0<= x <= 1 , y(0)=1 , h=0.2

C語言 例題5-1利用向前Euler尤拉 向前近似法 解一階常微分方程ODE y' = -y+x+1 0<= x <= 1 , y(0)=1 , h=0.2

/***********************************************
****EULER METHOD FOR DIFFERENTIAL EQUATIONS*****
***********************************************/
#include<stdio.h>
#include<math.h>
/*Define the RHS of the first order differential equation here(Ex: dy/dx=f(x,y))  */
double f(double x, double y){
return -y+x+1;
}
int main()
{
int i;
double y,xi,yi,xf,h;
yi=1;
xi=0;
xf=1;
h=0.2;

printf("x\t\ty\t\ty'\t\thy'\t\ty+hy'\n");
printf("_________________________________________________\n");
//Begin Euler Routine
while(xi<=xf){
y=yi+h*f(xi,yi);
printf("%0.2lf\t%0.4lf\t%0.4lf\t%0.4lf\t%0.4lf\n",xi,yi,f(xi,yi),h*f(xi,yi),y);
yi=y;
xi=xi+h;
}
return 0;
}


輸出畫面
y y' hy' y+hy'
_________________________________________________
0.00 1.0000 0.0000 0.0000 1.0000
0.20 1.0000 0.2000 0.0400 1.0400
0.40 1.0400 0.3600 0.0720 1.1120
0.60 1.1120 0.4880 0.0976 1.2096
0.80 1.2096 0.5904 0.1181 1.3277
1.00 1.3277 0.6723 0.1345 1.4621