Numerical solution of ordinary differential equations by using Runge-Kutta Method of Order Two and Runge-Kutta Method of Order Four How to write the C++ codes?

1. TUTORIAL 10 SJEM2231: STRUCTURED PROGRAMMING (C++)
QUESTION 1
Write a C++ program using second order RK method with h=0.1 to find
y(2.5) for
= -xy2, y(2)=1
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x,double y)
{
double func;
func= -x*y*y;
return func;
}
void main()
{
float x,x0,xn,y,y0, h,exact,error,k1,k2;
cout << "Enter the stepsize h: ";
cin >> h;
cout << "Enter x0,xn and y0: ";
cin >> x0 >> xn >> y0;
x=x0;
y=y0;
cout << "nXtYttExactttErrorn";

2. while(x < xn)
{
k1=h*f(x,y);
k2=h*f((x+h/2),(y + k1/2));
y= y+ k2;
x=x+h;
exact=2/(x*x -2);
error= exact-y;
cout << setprecision(1) <<fixed << x << "t"<< setprecision(6) << fixed << y << "t";
cout << setprecision(6) << exact << "t" << fabs(error) << endl;
}
}
//Output:
Enter the stepsize h: 0.1
Enter x0,xn and y0: 2 2.5 1
X Y Exact Error
2.1 0.833950 0.829876 0.004074
2.2 0.709463 0.704225 0.005238
2.3 0.613199 0.607903 0.005296
2.4 0.536859 0.531915 0.004944
2.5 0.475051 0.470588 0.004462
2.6 0.424136 0.420168 0.003967

3. TUTORIAL 10 SJEM2231: STRUCTURED PROGRAMMING (C++)
QUESTION 2
Write a C++ function program using RK2 method to solve
= sin(y) ,
with y(0)=1 from x=0 to 0.5 with step size h=0.1.
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double(x),double(y))
{
return sin(y);
}
int main()
{
long double y[100],x[100], k[100][100];
int n,i;
float h;
cout<<"Enter the value of x0: ";
cin>>x[0];
cout<<"Enter The Value of y0: ";
cin>>y[0];
cout<<"Enter the number of iterations: ";
cin>>n;
cout<<"Enter The Value of Step Size: ";
cin>>h;

4. cout<<"nIterationstxtytK1ttK2"<<endl;
for(i=1;i<=n;i++)
{
k[1][i]= h*f(x[i-1],y[i-1]);
k[2][i]= h*f(x[i-1]+h/2 ,y[i-1]+ (k[1][i])/2);
y[i]=y[i-1]+ k[2][i];
x[i]=x[i-1]+h;
}
for(i=0;i<=n;i++)
{
cout << i <<"tt"<< setprecision(1) << x[i]<<"t";
cout << setprecision(5) << y[i] << "t";
cout << setprecision(3)<< k[1][i] << "tt" << k[2][i] << endl;
}
return 0;
}
//Output:
Enter the value of x0: 0
Enter The Value of y0: 1
Enter the number of iterations: 5
Enter The Value of Step Size: 0.1
Iterations x y K1 K2
0 0 1 0.0841 0.0863
1 0.1 1.0863 0.0885 0.0905
2 0.2 1.1768 0.0923 0.094

5. TUTORIAL 10 SJEM2231: STRUCTURED PROGRAMMING (C++)
3 0.3 1.2708 0.0955 0.0968
4 0.4 1.3677 0.0979 0.0988
5 0.5 1.4665
//Alternative way for question 2
#include<iostream>
#include<cmath>
using namespace std;
double f(double x,double y)
{
double func;
func= sin(y);
return func;
}
void main()
{
float x,x0,xn,y,y0,n, h,k1,k2;
cout << "Enter no. of subintervals: ";
cin >> n;
cout << "Enter x0,xn and y0: ";
cin >> x0 >> xn >> y0;
h=(xn-x0)/n ;

6. x=x0;
y=y0;
cout << "nXttYn";
while(x<xn)
{
k1=h*f(x,y);
k2=h*f((x+h/2),(y + k1/2));
y= y+ k2;
x=x+h;
cout << x << "tt"<< y <<endl;
}
}
//Output
Enter no. of subintervals: 5
Enter x0,xn and y0: 0 0.5 1
X Y
0.1 1.08635
0.2 1.17681
0.3 1.27082
0.4 1.36766
0.5 1.46647

7. TUTORIAL 10 SJEM2231: STRUCTURED PROGRAMMING (C++)
QUESTION 3
Write a C++ program using fourth order RK method with h=0.1 to
approximate y(1.5) for
= 2xy, y(1)=1. Compute the errors using exact
solution, y=ex2-1
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x,double y)
{
double func;
func= 2*x*y;
return func;
}
void main()
{
double x,x0,xn,y,y0, exact,error, h,k1,k2,k3,k4;
cout << "Enter the stepsize h: ";
cin >> h;
cout << "Enter x0,xn and y0: ";
cin >> x0 >> xn >> y0;
x=x0;
y=y0;
cout << "nXtYtExacttErrorn";

8. while(x <xn)
{
k1=f(x,y);
k2=f((x+h/2),(y + k1*h/2));
k3=f((x+ h/2) ,(y+k2*h/2));
k4=f(x+h, y+h*k3);
y= y+ (h/6)*(k1+2*k2+2*k3+k4);
x=x+h;
exact=exp(x*x -1);
error= exact - y;
cout << setprecision(1) << fixed << x << "t"<< setprecision(4) <<
fixed << y <<"t" ;
cout << setprecision(4) <<fixed << exact << "t" << fabs(error) << endl;
}
}
//Output:
Enter the stepsize h: 0.1
Enter x0,xn and y0: 1 1.5 1
X Y Exact Error
1.1 1.2337 1.2337 0.0000
1.2 1.5527 1.5527 0.0000
1.3 1.9937 1.9937 0.0000
1.4 2.6116 2.6117 0.0001
1.5 3.4902 3.4903 0.0001

9. TUTORIAL 10 SJEM2231: STRUCTURED PROGRAMMING (C++)
QUESTION 4
Do Problem 1 with RK4. Compare the results between RK2 and RK4.
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x,double y)
{
double func;
func= -x*y*y;
return func;
}
void main()
{
double x,x0,xn,y,y0,h,exact,error,k1,k2,k3,k4;
cout << "Enter the stepsize h: ";
cin >> h;
cout << "Enter x0,xn and y0: ";
cin >> x0 >> xn >> y0;
x=x0;
y=y0;
cout << "nXtYtExactttErrorn";
while(x <xn)
{

10. k1=f(x,y);
k2=f((x+h/2),(y + k1*h/2));
k3=f((x+ h/2) ,(y+k2*h/2));
k4=f(x+h, y+h*k3);
y= y+ (h/6)*(k1+2*k2+2*k3+k4);
x=x+h;
exact= 2/(x*x-2);
error= exact-y;
cout << setprecision(1) << fixed << x << "t"<< setprecision(6) <<
fixed << y << "t";
cout <<setprecision(6) << exact << "t" << fabs(error) << endl;
}
}
//Output:
Enter the stepsize h: 0.1
Enter x0,xn and y0: 2 2.5 1
X Y Exact Error
2.1 0.829885 0.829876 0.000010
2.2 0.704237 0.704225 0.000011
2.3 0.607914 0.607903 0.000011
2.4 0.531924 0.531915 0.000010
2.5 0.470596 0.470588 0.000008
Thus, RK4 is better than RK2 since RK4 has smaller error difference than RK2 for y(2.5)

11. TUTORIAL 10 SJEM2231: STRUCTURED PROGRAMMING (C++)
QUESTION 5
Write a C++ function program using classical RK4 method with h=0.2 to
obtain y(1) for
= y-x, y(0)=2.
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x,double y)
{
double func;
func= y-x;
return func;
}
void main()
{
double x,x0,xn,y,y0,h,k1,k2,k3,k4;
cout << "Enter the stepsize h: ";
cin >> h;
cout << "Enter x0,xn and y0: ";
cin >> x0 >> xn >> y0;
x=x0;
y=y0;
cout << "nXtYn";
while(x <xn)
{
k1=f(x,y);

12. k2=f((x+h/2),(y + k1*h/2));
k3=f((x+ h/2) ,(y+k2*h/2));
k4=f(x+h, y+h*k3);
y= y+ (h/6)*(k1+2*k2+2*k3+k4);
x=x+h;
cout << setprecision(1) << fixed << x << "t"<< setprecision(6) <<
fixed << y << endl;
}
}
//Output:
Enter the stepsize h: 0.2
Enter x0,xn and y0: 0 1 2
X Y
0.2 2.421400
0.4 2.891818
0.6 3.422106
0.8 4.025521
1.0 4.718251