以y'=x+y,0<x<1,y(0)=1为例,取步长h=0.1,已知精确值为y=-x-1+2e^x,用来进行精度比较。
#include<stdio.h>
using namespace std;
double cor[10000];
double f(double x,double y)//改写函数
{
return x+y;
}
double correctf(double x)//精确解函数
{
return -x-1+2*exp(x);
}
void Euler(double h,double l,double r,double *a,double *b,double tol)//欧拉法
{
double sum=0;
for(int i=1; i<=tol; i++)
{
b[i]=b[i-1]+h*f(a[i-1],b[i-1]);
sum+=fabs(b[i]-cor[i])/cor[i];
}
for(int i=1; i<=tol; i++)
printf("当x=%lf时,近似解为:%lf,准确解为:%lf\n",a[i],b[i],cor[i]);
printf("精度为:%lf\n\n",sum/tol);
}
void improvedEuler(double h,double l,double r,double *a,double *b,double tol)//改进的欧拉法
{
double b1,sum=0;
for(int i=1; i<=tol; i++)
{
b1=b[i-1]+h*f(a[i-1],b[i-1]);
b[i]=b[i-1]+h/2*(f(a[i-1],b[i-1])+f(a[i],b1));
}
for(int i=1; i<=tol; i++)