C Program Of NA Using Runga-Kutta Mthod Of 4th Order

/* Runge-kutta method of 4th order to solve 10*dy/dx=(x*x)+(y*y)*/

# include <stdio.h>

# include <conio.h>

void main()

{

int c=0;

float x,y0,xp,h,i,y1,m1,m2,m3,m4;

float f(float,float);

clrscr();

printf("Solution by Runge kutta 2nd order Method\n");

printf("Enter initial Boundry condition x,y : ");

scanf("%f%f",&x,&y0);

printf("Enter Value of X at which Y is required : ");

scanf("%f",&xp);

printf("Enter Interval ,h : ");

scanf("%f",&h);

printf(" No. X\t m1 \t m2\t m3 \t m4 \t Y\n");

printf("-------------------------------------------------------------------\n");

for(i=x;i<=xp;i=i+h)

{

c++;

m1=h*f(i,y0);

m2=h*f(i+(h/2),y0+(m1/2));

m3=h*f(i+(h/2),y0+(m2/2));

m4=h*f(i+h,y0+m3);

y1=y0+((m1+2*m2+2*m3+m4)/6);

printf("%2d %5.6f %5.6f %5.6f %5.6f %5.6f %5.6f\n",c,i,m1,m2,m3,m4,y1);

y0=y1;

}

getch();

}

float f(float x,float y)

{

return ((x*x)+(y*y))/10;

}

OUTPUT

Solution by Runge kutta 2nd order Method

Enter initial Boundry condition x,y : 0 1

Enter Value of X at which Y is required : .09

Enter Interval ,h : .03

No. X m1 m2 m3 m4 Y

-------------------------------------------------------------------------------

1 0.000000 0.003000 0.003010 0.003010 0.003021 1.003010

2 0.030000 0.003021 0.003033 0.003033 0.003047 1.006043

3 0.060000 0.003047 0.003062 0.003062 0.003079 1.009106

4 0.090000 0.003079 0.003097 0.003097 0.003117 1.012204