### Run the code online here:

https://onlinegdb.com/N-6KusNbR

### Code:

//Modified Euler's Method for differential equations #include<iostream> #include<math.h> #include<iomanip> 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+h,y_i); //calculate slope or dy/dx at x0+h (new x),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; }

### Run the code online here:

https://onlinegdb.com/N-6KusNbR

### Sample output

### Exact (ideal) solution

**Explaination of the code**:

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