C++ TUTORIAL 9

TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
QUESTION 1
Write a C++ program using Taylor’s series method to solve the equation
=3x+y2 to approximate y when x=0.1, given that y=1 when x=0.
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main()
{
double x0=0,y0=1 , h=0.1,y1,y2,y3,y4, y;
y1=3*x0 + y0*y0;
y2=3+ 2*y0*y1;
y3=2*y1*y1 + 2*y0*y2;
y4=6*y1*y2 + 2*y0*y3;
y= y0+ (y1*h) + (y2*pow(h,2))/2 + (y3*pow(h,3))/6 + (y4*pow(h,4))/24;
cout << "The value of y when x=0.1 is " << setprecision(5) <<fixed << y << endl;
return 0;
}
//Output:
The value of y when x=0.1 is 1.12722
//Alternative way for question 1
#include <iostream>
#include <cmath>
#include <iomanip>
#define h 0.1
double f(double x,double y)
{
return 3*x + y*y;
}
double ff(double x,double y)
{

return 3 + 2*y*f(x,y);
}
double fff(double x, double y)
{
return 2*f(x,y)*f(x,y) + 2*y*ff(x,y);
}
double ffff(double x, double y)
{
return 6*f(x,y)*ff(x,y) + 2*y*fff(x,y);
}
void taylor(double x,double y[])
{
int i;
cout << "ittxtty " << endl;
for ( i=0;i<=4;i++)
{
y[i+1]=y[i]+h*f(x,y[i] ) + (h*h/2)*ff(x,y[i]) + (pow(h,3)/6)*fff(x,y[i])+
(pow(h,4)/24)*ffff(x,y[i]) ;
cout << i << "tt" << setprecision(1)<<fixed << x << "tt" << setprecision(5) <<
fixed << y[i] << endl;
x= x+ h;
}
}
int main()
{
double x0,y[100];
cout << "Enter x0 and y0 : ";
cin >> x0 >> y[0];
taylor(x0,y);
return 0;
}
//Output:
Enter x0 and y0 : 0 1
i x y
0 0.0 1.00000
1 0.1 1.12722
2 0.2 1.32071

QUESTION 2
Write a C++ program using Taylor’s series method to solve
+4y=x2 ,
y(0)=1 to approximate y(0.2).
#include <iostream>
#include <cmath>
#include <iomanip>
int main()
{
double x0=0,y0=1 , h=0.2,y1,y2,y3,y4, y;
y1= x0*x0 -4*y0;
y2= 2*x0- 4*y1;
y3= 2- 8*x0+16*y1; // y3= 2- 4*y2;
y4= -8+32*x0 -64*y1; // y4= -4*y3;
y= y0+ (y1*h) + (y2*pow(h,2))/2 + (y3*pow(h,3))/6 + (y4*pow(h,4))/24;
cout << "The value of y when x=0.2 is " << setprecision(5) <<fixed << y << endl;
return 0;
}
//Output:
The value of y when x=0.2 is 0.45387

//Alternative for question 2
#include <iostream>
#include <cmath>
#include <iomanip>
#define h 0.2
double f(double x,double y)
{
return x*x - 4*y;
}
double ff(double x,double y)
{
return 2*x - 4*f(x,y);
}
double fff(double x, double y)
{
return 2- 8*x + 16*f(x,y);
}
double ffff(double x, double y)
{
return -8 + 32*x- 64*f(x,y);
}
void taylor(double x,double y[])
{
int i;

cout << "ittxtty " << endl;
for ( i=0;i<=2;i++)
{
y[i+1]=y[i]+h*f(x,y[i] ) + (h*h/2)*ff(x,y[i]) + (pow(h,3)/6)*fff(x,y[i])+
(pow(h,4)/24)*ffff(x,y[i]) ;
cout << i << "tt" << setprecision(1)<<fixed << x << "tt" << setprecision(5) << fixed
<< y[i] << endl;
x= x+ h;
}
}
int main()
{
double x0,y[100];
cout << "Enter x0 and y0 : ";
cin >> x0 >> y[0];
taylor(x0,y);
return 0;
}
//Output:
Enter x0 and y0 : 0 1
i x y
0 0.0 1.00000
1 0.2 0.45387
2 0.4 0.21894

QUESTION 3
Write a C++ program to solve
= -xy2 , 푦(2)=1 in the interval 2< x <3
with h=0.1 using Euler’s method. Compare the results with exact
solution from 푦= 2/(x2 -2)
#include<iostream>
#include <cmath>
#include<iomanip>
#define F(x,y) -x*y*y
void main()
{
double y0,y,x,x0,xn,h, exact,error;
cout <<"Enter the value of range(x0 and xn): ";
cin >> x0 >> xn;
cout << "Enter the value of y0: ";
cin >> y0;
cout <<"Enter the h: ";
cin >> h;
cout << "nnx0= "<< x0 << "tt" << "y0= " << y0;
x=x0;
y=y0;
while(x<xn)
{

y= y + h * F(x,y);
x=x+h;
exact = 2/(x*x-2);
error= exact-y;
cout << "nx= " << setprecision(1) <<fixed << x << "t";
cout << setprecision(4) <<fixed << exact<< "t" <<setprecision(4) <<fixed
<< y << "t";
cout <<setprecision(4) << fixed << fabs(error) << endl;
}
}
//Output:
Enter the value of range(x0 and xn): 2 3
Enter the value of y0: 1
Enter the h: 0.1
x0= 2 y0= 1
x= 2.1 0.8299 0.8000 0.0299
x= 2.2 0.7042 0.6656 0.0386
x= 2.3 0.6079 0.5681 0.0398
x= 2.4 0.5319 0.4939 0.0380
x= 2.5 0.4706 0.4354 0.0352
x= 2.6 0.4202 0.3880 0.0322
x= 2.7 0.3781 0.3488 0.0292

x= 2.8 0.3425 0.3160 0.0265
x= 2.9 0.3120 0.2880 0.0240
x= 3.0 0.2857 0.2640 0.0217
//Alternative way for question no 3
#include<iostream>
#include<cmath>
#include<iomanip>
//Given dy/dx
float f(float(x),float(y))
{
return (-x*y*y);
}
int main()
{
double y[100],x[100], exact,error,percent_error;
int n,i;
float h;
//Entering the initial values of x & y
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<<"nIterationstxtytExact valuetErrort Percentage Error"<<endl;
//Calculating the value of x
for(i=1;i<=n;i++)
{
x[i]=x[i-1]+h;
}
//Calculating the value of y
for(i=1;i<=n;i++)
{
y[i]=y[i-1]+(h*f(x[i-1],y[i-1]));
}
//Printing result
for(i=0;i<=n;i++)
{
exact = 2/(x[i]*x[i]-2);
error= exact-y[i];
percent_error= (fabs(error)/exact)*100;
cout << i<<"tt"<< setprecision(2) <<x[i]<<"t";
cout << setprecision(4)<< y[i] << "t" << exact << "tt" ;
cout << setprecision(4) <<fixed << fabs(error) << "t ";

cout << setprecision(5) << percent_error << endl;
}
return 0;
}
//Output:
Enter the value of x0: 2
Enter The Value of y0: 1
Enter the number of Iterations: 10
Enter The Value of Step Size: 0.1
Iterations x y Exact value Error Percentage Error
0 2 1 1 0.0000 0.00000
1 2.10 0.8000 0.8299 0.0299 3.60000
2 2.20 0.6656 0.7042 0.0386 5.48480
3 2.30 0.5681 0.6079 0.0398 6.54182
4 2.40 0.4939 0.5319 0.0380 7.14753
5 2.50 0.4354 0.4706 0.0352 7.48768
6 2.60 0.3880 0.4202 0.0322 7.66332
7 2.70 0.3488 0.3781 0.0292 7.73341
8 2.80 0.3160 0.3425 0.0265 7.73413
9 2.90 0.2880 0.3120 0.0240 7.68862
10 3.00 0.2640 0.2857 0.0217 7.61210

QUESTION 4
Write a C++ program to find y(1) from
= x+y , y(0)=1 using Euler’s
method by using h=0.2. The exact solution is y=-1-x+2ex
#include<iostream>
#include<cmath>
#include<iomanip>
//Given dy/dx
{
return (x+y);
}
int main()
{
int n,i;
float h;
cin>>x[0];
cin>>y[0];

cin>>n;
cin>>h;
for(i=1;i<=n;i++)
{
x[i]=x[i-1]+h;
}
for(i=1;i<=n;i++)
{
y[i]=y[i-1]+(h*f(x[i-1],y[i-1]));
}
//Printing result
for(i=0;i<=n;i++)
{
exact = -1-x[i] +2*exp(x[i]);
error= exact-y[i];

}
return 0;
}
//Output:
0 0 1 1 0.0000 0.00000
1 0.2 1.2000 1.2428 0.0428 3.44427
2 0.4 1.4800 1.5836 0.1036 6.54497
3 0.6 1.8560 2.0442 0.1882 9.20821
4 0.8 2.3472 2.6511 0.3039 11.46256
5 1.0 2.9766 3.4366 0.4599 13.38324
QUESTION 5
= -2xy2, y(0)=1 in the interval
0≤ x ≤0.5 with h=0.1 using Euler’s method. The exact value is
y= 1/(x2+1)
#include<iostream>
#include<cmath>
#include<iomanip>

//Given dy/dx
{
return (-2*x*y*y);
}
int main()
{
int n,i;
float h;
cin>>x[0];
cin>>y[0];
cin>>n;
cin>>h;

for(i=1;i<=n;i++)
{
x[i]=x[i-1]+h;
}
for(i=1;i<=n;i++)
{
y[i]=y[i-1]+(h*f(x[i-1],y[i-1]));
}
//Printing result
for(i=0;i<=n;i++)
{
exact = (1/(x[i]*x[i]+1));
error= exact-y[i];
}
return 0;
}

//Output :
0 0 1 1 0.0000 0.00000
1 0.1 1.0000 0.9901 0.0099 1.00000
2 0.2 0.9800 0.9615 0.0185 1.92000
3 0.3 0.9416 0.9174 0.0242 2.63266
4 0.4 0.8884 0.8621 0.0263 3.05314
5 0.5 0.8253 0.8000 0.0253 3.15629
QUESTION 6
= x+y2 with y(1)=0 at x=1.3 using
Euler’s Method
#include<iostream>
#include<cmath>
#include<iomanip>
//Given dy/dx
{
return (x+ y*y);
}

int main()
{
double y[100],x[100];
int n,i;
float h;
cin>>x[0];
cin>>y[0];
cin>>n;
cin>>h;
cout<<"nIterationstxty"<<endl;
for(i=1;i<=n;i++)
{
x[i]=x[i-1]+h;
}
for(i=1;i<=n;i++)
{

y[i]=y[i-1]+(h*f(x[i-1],y[i-1]));
}
//Printing result
for(i=0;i<=n;i++)
{
cout << i<<"tt"<< setprecision(2) <<x[i]<< "t" << setprecision(5)<< y[i] <<
endl;
}
return 0;
}
//Output:
Iterations x y
0 1 0
1 1.1 0.1
2 1.2 0.211
3 1.3 0.33545

C++ TUTORIAL 9

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