C++ TUTORIAL 10

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 << "\nX\tY\t\tExact\t\tError\n";

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


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>


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;

cout<<"\nIterations\tx\ty\tK1\t\tK2"<<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 <<"\t\t"<< setprecision(1) << x[i]<<"\t";

cout << setprecision(5) << y[i] << "\t";

cout << setprecision(3)<< k[1][i] << "\t\t" << 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


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>



{

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;


cin >> x0 >> xn >> y0;

h=(xn-x0)/n ;

x=x0;

y=y0;

cout << "\nX\t\tY\n";

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 << "\t\t"<< y <<endl;

}

}

//Output

Enter no. of subintervals: 5


X Y

0.1 1.08635

0.2 1.17681

0.3 1.27082

0.4 1.36766

0.5 1.46647


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>



{

double func;

func= 2*x*y;

return func;

}

void main()

{

double x,x0,xn,y,y0, exact,error, h,k1,k2,k3,k4;


cin >> h;


cin >> x0 >> xn >> y0;

x=x0;

y=y0;

cout << "\nX\tY\tExact\tError\n";

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:



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


QUESTION 4

Do Problem 1 with RK4. Compare the results between RK2 and RK4.

#include<iostream>

#include<cmath>

#include<iomanip>



{

double func;

func= -x*y*y;

return func;

}

void main()

{

double x,x0,xn,y,y0,h,exact,error,k1,k2,k3,k4;


cin >> h;


cin >> x0 >> xn >> y0;

x=x0;

y=y0;

cout << "\nX\tY\tExact\t\tError\n";

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= 2/(x*x-2);

error= exact-y;


fixed << y << "\t";

cout <<setprecision(6) << exact << "\t" << fabs(error) << endl;

}

}

//Output:



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)


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>



{

double func;

func= y-x;

return func;

}

void main()

{

double x,x0,xn,y,y0,h,k1,k2,k3,k4;


cin >> h;


cin >> x0 >> xn >> y0;

x=x0;

y=y0;

cout << "\nX\tY\n";

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;


fixed << y << endl;

}

}

//Output:


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

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C++ TUTORIAL 10