The document presents several C++ programming tasks utilizing Taylor's series and Euler's method to solve differential equations. Each task includes code examples, input requirements, and outputs to approximate solutions of equations within specified intervals. The content covers methods to calculate values of y given initial conditions and different step sizes.

1. 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>
using namespace std;
#define h 0.1
double f(double x,double y)
{
return 3*x + y*y;
}
double ff(double x,double y)
{

2. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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

3. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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>
using namespace std;
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

4. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
//Alternative for question 2
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
#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;

5. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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

6. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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>
using namespace std;
#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)
{

7. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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

8. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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>
using namespace std;
//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];

9. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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 ";

10. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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

11. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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>
using namespace std;
//Given dy/dx
float f(float(x),float(y))
{
return (x+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: ";

12. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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 = -1-x[i] +2*exp(x[i]);
error= exact-y[i];
percent_error= (fabs(error)/exact)*100;
cout << i<<"tt"<< setprecision(1) <<x[i]<<"t";
cout << setprecision(4)<< y[i] << "t" << exact << "tt" ;

13. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
cout << setprecision(4) <<fixed << fabs(error) << "t ";
cout << setprecision(5) << percent_error << 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: 0.2
Iterations x y Exact value Error Percentage Error
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
Write a C++ program to solve
= -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>

14. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
using namespace std;
//Given dy/dx
float f(float(x),float(y))
{
return (-2*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;

15. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
//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 = (1/(x[i]*x[i]+1));
error= exact-y[i];
percent_error= (fabs(error)/exact)*100;
cout << i<<"tt"<< setprecision(1) <<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;
}

16. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
//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 Exact value Error Percentage Error
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
Write a C++ program to solve
= x+y2 with y(1)=0 at x=1.3 using
Euler’s Method
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
//Given dy/dx
float f(float(x),float(y))
{
return (x+ y*y);
}

17. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
int main()
{
double y[100],x[100];
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<<"nIterationstxty"<<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++)
{

18. TUTORIAL 9; SJEM2231: STRUCTURED PROGRAMMING (C++)
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:
Enter the value of x0: 1
Enter The Value of y0: 0
Enter the number of Iterations: 3
Enter The Value of Step Size: 0.1
Iterations x y
0 1 0
1 1.1 0.1
2 1.2 0.211
3 1.3 0.33545