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++) 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 RungeKutta(double h,double l,double r,double *a,double *b,double tol)//四阶龙格库塔法 { double k1,k2,k3,k4,sum=0; for(int i=1; i<=tol; i++) { k1=f(a[i-1],b[i-1]); k2=f(a[i-1]+h/2,b[i-1]+h/2*k1); k3=f(a[i-1]+h/2,b[i-1]+h/2*k2); k4=f(a[i-1]+h,b[i-1]+h*k3); b[i]=b[i-1]+h/6*(k1+2*k2+2*k3+k4); } 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); } int main() { double h,a[10000],b[10000],l,r; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(cor,0,sizeof(cor)); printf("请输入步长:"); scanf("%lf",&h); printf("请输入区间下限:"); scanf("%lf",&l); printf("请输入区间上限:"); scanf("%lf",&r); printf("请赋予初始值:"); scanf("%lf",&b[0]); double tol=(r-l)/h; for(int i=0; i<=tol; i++) a[i]=l+i*h; for(int i=1; i<=tol; i++) cor[i]=correctf(a[i]); printf("以下为欧拉法求解结果:\n"); Euler(h,l,r,a,b,tol); printf("以下为改进的欧拉法求解结果:\n"); improvedEuler(h,l,r,a,b,tol); printf("以下为四阶龙格库塔法求解结果:\n"); RungeKutta(h,l,r,a,b,tol); return 0; }
