社区居民需求的问题:我的c语言程序中函数问题

我的程序
#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*v, double*x,double *y,double (*function)(double,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;
v[0]=2.1367e+14;
for(i=0;i<n;i++)
{
x[i+1]=x[i]+h;
k1=function(x[i],v[0],y[i]);
k2=function(x[i]+h/2,v[0],y[i]+h*k1/2);
k3=function(x[i]+h/2,v[0],y[i]+h*k2/2);
k4=function(x[i]+h,v[0],y[i]+h*k3);
y[i+1]=y[i]+h*(k1+2*k2+2*k3+k4)/6;

}
return 1;
}
double function(double x,double ,double y)
{

return -k1/0.68*y-38*y+y*(-1.21e+7)*0.6-y*(7.31e+7);
}

int main()
{
double x[6],y[6],v[1];

printf("用四阶龙格-库塔方法\n");
RungeKutta(680,0,4,6,v,x,y,function);
for(int i=0;i<6;i++)
printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]);
return 1;
}
我想请问在return -k1/0.68*y-38*y+y*(-1.21e+7)*0.6-y*(7.31e+7); 语句中如何把y 代替为y(x,v[0]),需要再加些什么?,还有如何才能使用 K1 ? 在这里小弟先些过各位大虾 了!