尤拉公式解初值问题.cpp

来自「数值分析中的尤拉公式解初值问题问题」· C++ 代码 · 共 71 行

// 尤拉公式解初值问题.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include "iostream.h"


double f(double x, double y)	//函数f
{
	return -y;
}

int main(int argc, char* argv[])
{
	cout << "  **************************************" << endl;
	cout << "  **                                  **" << endl;
	cout << "  **         尤拉公式解初值问题       **" << endl;
	cout << "  **                                  **" << endl;
	cout << "  **************************************" << endl << endl;

	int n;	//等分数
	cout << "以步长h=0.1将区间(0,0.2)等分" << endl;
	
	double a=0,b=0.2;
	double h=0.1;
	n=(b-a)/h;
	double *X;	//存放x序列
	X=new double[n];
	double *Y;	//存放y序列
	Y=new double[n];
	Y[0]=1;	//初值
	int k;

	for (k=0;k<n;k++)
	{
		X[k]=a+k*h;
		Y[k+1]=Y[k]+h*f(X[k], Y[k]);
	}

	cout << "X=[ ";		//显示x序列
	for (k=0;k<n;k++)
		cout << X[k] << ", ";
	cout << b << " ]" <<endl;
	cout << "Y=[ ";		//显示y序列
	for (k=0;k<n;k++)
		cout << Y[k] << ", ";
	cout << Y[n] << " ]" <<endl;

//-------------------------------------------------------
	
	cout <<endl << "----------------------------------------" << endl;
	cout << "改进的欧拉公式:" << endl << endl;

	for (k=0;k<n;k++)
	{
		double yp,yc;
		yp=Y[k]+h*f(X[k], Y[k]);
		yc=Y[k]+h*f(X[k+1], yp);
		Y[k+1]=(yp+yc)/2;
	}

	cout << "Y=[ ";		//显示y序列
	for (k=0;k<n;k++)
		cout << Y[k] << ", ";
	cout << Y[n] << " ]" <<endl;

	return 0;

	delete [n] X;
	delete [n] Y;
}