//Modified Euler's Method for differential equations
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
double df(double x, double y)
{
double a=x+y; //function for defining dy/dx
return a;
}
int main()
{
double x0,y0,x,y_i,dy1,dy2,dy_avg,y_n,h; //for initial values, width, etc.
cout<<"\nEnter the initial values of x and y respectively:\n";
cin>>x0>>y0; //Initial values
cout<<"\nEnter the value of x for which you would like to find y:\n";
cin>>x;
cout<<"\nEnter the interval width,h:\n";
cin>>h; //input width
cout<<"x"<<setw(16)<<"y"<<setw(16)<<"hy'"<<setw(16)<<"y_new"<<setw(16)<<"hy_new'"<<setw(16)<<"hy'avg"<<setw(16)<<"y_n+1"<<endl;
cout<<"--------------------------------------------------------------------------------------------------\n";
while(fabs(x-x0)>0.0000001 //I couldn't just write "while(x0<x)" as they both are floating point nos. It is dangerous to compare two floating point nos. as they are not the same in binary as they are in decimal. For instance, a computer cannot exactly represent 0.1 or 0.7 in binary just like decimal can't represent 1/3 exactly without recurring digits.
{
dy1=h*df(x0,y0); //calculate slope or dy/dx at x0,y0
y_i=y0+dy1; //calculate new y, which is y0+h*dy/dx
dy2=h*df(x0,y_i); //calculate slope or dy/dx at x0,new y
dy_avg=(dy1+dy2)/2.0; //calculate the average of the slopes at y0 and new y
y_n=y0+dy_avg; //calculate new y, which is y0+h*average(dy/dx)
cout<<x0<<setw(16)<<y0<<setw(16)<<dy1<<setw(16)<<y_i<<setw(16)<<dy2<<setw(16)<<dy_avg<<setw(16)<<y_n<<endl;
x0=x0+h; //calculate new x.
y0=y_n; //pass this new y as y0 in the next iteration.
}
cout<<x0<<setw(16)<<y0<<endl;
cout<<"The approximate value of y at x=0 is "<<y0<<endl; //print the solution.
return 0;
}

I'm a physicist specializing in theoretical, computational and experimental condensed matter physics. I like to develop Physics related apps and softwares from time to time. Can code in most of the popular languages. Like to share my knowledge in Physics and applications using this Blog and a YouTube channel.