euler

1
/*Eulers Method*/ #include<stdio.h> #include<conio.h> #include<math.h> #define f(x0,y0) ((2*y0/x0)+(x0*x0*x0)) main() { int i=1,n; float x0,y0,h=0.2; printf("enter x0,y0, no. of iterations\n"); scanf("%f %f %d",&x0,&y0,&n); for(i=1;i<n+1;i++) { printf(" the values at %d iteration is ",i); y0=y0+(h*f(x0,y0)); x0=x0+h; printf("y=%f,x=%f \n",y0,x0); } } Output: enter x0,y0, no. of iterations 1 0.5 2 the values at 1 iteration is y=0.900000,x=1.200000 the values at 2 iteration is y=1.545600,x=1.400000

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eulers program

TRANSCRIPT

Page 1: Euler

/*Eulers Method*/

#include<stdio.h>#include<conio.h>#include<math.h>#define f(x0,y0) ((2*y0/x0)+(x0*x0*x0))main(){

int i=1,n;float x0,y0,h=0.2;printf("enter x0,y0, no. of iterations\n");scanf("%f %f %d",&x0,&y0,&n);for(i=1;i<n+1;i++){

printf(" the values at %d iteration is ",i);y0=y0+(h*f(x0,y0));x0=x0+h;printf("y=%f,x=%f \n",y0,x0);

}}

Output:

enter x0,y0, no. of iterations10.52 the values at 1 iteration is y=0.900000,x=1.200000 the values at 2 iteration is y=1.545600,x=1.400000