龙格-库塔方法

#include<stdlib.h>
#include<stdio.h>
/*n表示几等分，n+1表示他输出的个数*/
int RungeKutta(double y0,double a,double b,int n,double *x,double *y,int style,double (*function)(double,double))
{
 double h=(b-a)/n,k1,k2,k3,k4;
 int i;
// x=(double*)malloc((n+1)*sizeof(double));
// y=(double*)malloc((n+1)*sizeof(double));
 x[0]=a;
 y[0]=y0;
 switch(style)
 {
 case 2:
  for(i=0;i<n;i++)
  {
   x[i+1]=x[i]+h;
   k1=function(x[i],y[i]);
   k2=function(x[i]+h/2,y[i]+h*k1/2);
   y[i+1]=y[i]+h*k2;
  }
  break;
 case 3:
  for(i=0;i<n;i++)
  {
   x[i+1]=x[i]+h;
   k1=function(x[i],y[i]);
   k2=function(x[i]+h/2,y[i]+h*k1/2);
   k3=function(x[i]+h,y[i]-h*k1+2*h*k2);
   y[i+1]=y[i]+h*(k1+4*k2+k3)/6;
  }
  break;

 case 4:
  for(i=0;i<n;i++)
  {
   x[i+1]=x[i]+h;
      k1=function(x[i],y[i]);
   k2=function(x[i]+h/2,y[i]+h*k1/2);
   k3=function(x[i]+h/2,y[i]+h*k2/2);
   k4=function(x[i]+h,y[i]+h*k3);
   y[i+1]=y[i]+h*(k1+2*k2+2*k3+k4)/6;
  }
   break;
 default:
  return 0;
 }
 return 1;
}
double function(double x,double y)
{
 return y-2*x/y;
}
//例子求y'=y-2*x/y(0<x<1);y0=1;
/*
int  main()
{
 double x[6],y[6];
 printf("用二阶龙格-库塔方法\n");
 RungeKutta(1,0,1,5,x,y,2,function);
 for(int i=0;i<6;i++)
  printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]);
 printf("用三阶龙格-库塔方法\n");
 RungeKutta(1,0,1,5,x,y,3,function);
 for(i=0;i<6;i++)
  printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]);
 printf("用四阶龙格-库塔方法\n");
 RungeKutta(1,0,1,5,x,y,4,function);
 for(i=0;i<6;i++)
  printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]);
 return 1;
}

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