欧拉公式求微分
#include<stdio.h>
#include<math.h>
#define maxsize 100
int main(void)
{
double a,b,step,x[maxsize],y[maxsize];
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
y[i+1]=y[i]+step*f1(x[i+1],y[i]);
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
return 0;
}
double f1(double x,double y)
{
double z;
z=622*sin(314*x)-20*y;
return z;
}
double f2(double x)
{
double z;
z=48827*(exp(-20*x)+10*sin(314*x)/157-cos(314*x))/24749;
return z;
}
改进欧拉公式求微分#include<stdio.h>
#include<math.h>
#define maxsize 100
int main(void)
{
double a,b,step,x[maxsize],y[maxsize],k1,k2;
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
k1=step*f1(x[i],y[i]);
k2=step*f1(x[i]+step,y[i]+k1);
y[i+1]=y[i]+(k1+k2)/2;
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
return 0;
}
double f1(double x,double y)
{
double z;
z=622*sin(314*x)-20*y;
return z;
}
double f2(double x)
{
double z;
z=48827*(exp(-20*x)+10*sin(314*x)/157-cos(314*x))/24749;
return z;
}
龙格-库塔公式求微分
#include<stdio.h>
#include<math.h>
#define maxsize 100
int main(void)
{
double a,b,step,x[maxsize],y[maxsize],k1,k2,k3,k4;
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
k1=step*f1(x[i],y[i]);
k2=step*f1(x[i]+step/2,y[i]+k1/2);
k3=step*f1(x[i]+step/2,y[i]+k2/2);
k4=step*f1(x[i]+step,y[i]+k3);
y[i+1]=y[i]+(k1+2*k2+2*k3+k4)/6;
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
return 0;
}
double f1(double x,double y)
{
double z;
z=622*sin(314*x)-20*y;
return z;
}
double f2(double x)
{
double z;
z=48827*(exp(-20*x)+10*sin(314*x)/157-cos(314*x))/24749;
return z;
}
基尔公式求微分
#include<stdio.h>
#include<math.h>
#define maxsize 100
int main(void)
{
double a,b,step,x[maxsize],y[maxsize],k1,k2,k3,k4;
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
k1=f1(x[i],y[i]);
k2=f1(x[i]+step/2,y[i]+k1*step/2);
k3=f1(x[i]+step/2,y[i]+(sqrt(2)-1)*step*k1/2+(1-sqrt(2)/2)*step*k2);
k4=f1(x[i]+step,y[i]-sqrt(2)*step*k2/2+(1+sqrt(2)/2)*step*k3);
y[i+1]=y[i]+(k1+(2-sqrt(2))*k2+(2+sqrt(2))*k3+k4)*step/6;
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
return 0;
}
double f1(double x,double y)
{
double z;
z=622*sin(314*x)-20*y;
return z;
}
double f2(double x)
{
double z;
z=48827*(exp(-20*x)+10*sin(314*x)/157-cos(314*x))/24749;
return z;
}
我想把这几种方法和在一起，就是这个意思，以上单独求微分的方法都完全正确，我现在需要做的，就是想合起来
#include<stdio.h>
#include<math.h>
#define maxsize 100
void eulerian_method(double x[maxsize],double y[maxsize])
{
double a,b,step;
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
y[i+1]=y[i]+step*f1(x[i+1],y[i]);
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
}
void modified_eulerian_method(double x[maxsize],double y[maxsize])
{
double a,b,step,k1,k2;
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
k1=step*f1(x[i],y[i]);
k2=step*f1(x[i]+step,y[i]+k1);
y[i+1]=y[i]+(k1+k2)/2;
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
}
void runge_kutta(double x[maxsize],double y[maxsize])
{
double a,b,step,k1,k2,k3,k4;
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
k1=step*f1(x[i],y[i]);
k2=step*f1(x[i]+step/2,y[i]+k1/2);
k3=step*f1(x[i]+step/2,y[i]+k2/2);
k4=step*f1(x[i]+step,y[i]+k3);
y[i+1]=y[i]+(k1+2*k2+2*k3+k4)/6;
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
}
void gill(double x[maxsize],double y[maxsize])
{
double a,b,step,k1,k2,k3,k4;
double f1(double x,double y);
double f2(double x);
int i,n;
printf("please input step=\n");
scanf("%lf",&step);
printf("please input a= and b=\n");
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)
x[i]=a+i*step;
y[0]=0;
for(i=0;i<=n-1;i++)
{
k1=f1(x[i],y[i]);
k2=f1(x[i]+step/2,y[i]+k1*step/2);
k3=f1(x[i]+step/2,y[i]+(sqrt(2)-1)*step*k1/2+(1-sqrt(2)/2)*step*k2);
k4=f1(x[i]+step,y[i]-sqrt(2)*step*k2/2+(1+sqrt(2)/2)*step*k3);
y[i+1]=y[i]+(k1+(2-sqrt(2))*k2+(2+sqrt(2))*k3+k4)*step/6;
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));
printf("\n");
}
}
void menu()
{
printf("********************************\n");
printf("请选择命令号!*******************\n");
printf("1欧拉折线法求解微分!
*\n");
printf("2欧拉改进法求解微分!
*\n");
printf("3龙格-库塔公式求解微分!
*\n");
printf("4-基尔公式求解微分!
*\n");
printf("0退出!
*\n");
printf("********************************\n");
}
int main(void)
{
int i,sel;
double (double x[i],double y[i]);
while(1)
{
menu();
printf("请输入命令号:\n");
scanf("%d",&sel);
switch(sel)
{
case 1:eulerian_method(x[i],y[i]);
case 2:modified_eulerian-method(x[i],y[i]);
case 3: runge_kutta(x[i],y[i]);
case 4:gill(x[i],y[i]);
case 0:exit(1);break;
default:
printf("输入的命令号错误!请重新输入:\n");
break;
}
}
return 0;
}
double f1(double x,double y)
{
double z;
z=622*sin(314*x)-20*y;
return z;
}
double f2(double x)
{
double z;
z=48827*(exp(-20*x)+10*sin(314*x)/157-cos(314*x))/24749;
return z;
}
