把不同的方法合起来写，核心算法完全正确，请各位帮忙，我那里不对，还要怎么改
下面这几个方法都是单独验证的时候正确的，也能得到完全正确的解,步长step=0.001,区间a=0,b=0.01#include<stdio.h>
#include<math.h>
#define maxsize 100
double f1(double x,double y);
double f2(double x);
/*
** 欧拉折线法求解微分方程
**/
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");/*输入步长*/
scanf("%lf",&step);
printf("please input a= and b=\n");/*微分区间*/
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);
for(i=0;i<=n;i++)/*分区间*/
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]);/*算法核心,求解公式*/
}
for(i=0;i<=n;i++)
{
printf("x[%d]=%lf,y[%d]=%lf,f2=%lf",i,x[i],i,y[i],f2(x[i]));/*输出x[i],y[i],准确值*/
printf("\n");
}
}
/*
** 欧拉折线法改进求解微分方程
**/
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");/*输入步长*/
scanf("%lf",&step);
printf("please input a= and b=\n");/*微分区间*/
scanf("%lf %lf",&a,&b);
n=(int)((b-a)/step);/*分区间*/
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);/*算法核心,求解公式*/
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]));/*输出x[i],y[i],准确值*/
printf("\n");
}
}
/*
** 欧拉折线法改进求解微分方程
**/
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");/*输入步长*/
scanf("%lf",&step);
printf("please input a= and b=\n");/*微分区间*/
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);/*算法核心,求解公式*/
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");
}
}
/*
** 欧拉折线法改进求解微分方程
**/
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");/*输入步长*/
scanf("%lf",&step);
printf("please input a= and b=\n");/*微分区间*/
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);/*算法核心,求解公式*/
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]));/*输出x[i],y[i],准确值*/
printf("\n");
}
}
/*下面两个是子程序*/
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;
}
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 sel=0;
double x[maxsize],y[maxsize];
while(sel!=0)
{
menu();
printf("请输入命令号:\n");
scanf("%d",&sel);
switch(sel)
{
case 1:eulerian_method(x,y);
break;
case 2:modified_eulerian_method(x,y);
break;
case 3: runge_kutta(x,y);
break;
case 4:gill(x,y);
break;
case 0:exit(1);
break;
default:
printf("输入的命令号错误!请重新输入:\n");
break;
}
}
return 0;
}
