Метод рунге-кутта 3 порядка - C++ - Ответ 6186254

#include<stdio.h>
#include<conio.h>
#include<math.h>
float fun(float x, float y)
{
return exp(x)-2*y;
}
int main()
{
int i,j,N;
float x0,y0,x,h,y,k[200][200],Y[200],n,p,a;
//clrscr();
printf("\n Enter the initial value of X0:");
scanf("%f",&x0);
printf("\n Enter the initial value of Y0:");
scanf("%f",&y0);
printf("\n Enter the given value of X:");
scanf("%f",&x);
printf("\n Enter the value of step size h:");
scanf("%f",&h);
n=(x-x0)/h;
N=(int)n;
printf("\nNumber of iterations is:%d\n",N);
for(i=1;i<=N;i++)
{
k[i][1]=h*fun(x0,y0);
k[i][2]=h*fun(x0+0.5*h,y0+0.5*k[i][1]);
k[i][3]=h*fun(x0+h,y0+2*k[i][2]+k[i][1]);
y = y0+(k[i][1] + 4 * k[i][2] + k[i][3])/6;
 
}printf("Y(%f)=%f",a,Y[i]);
printf("\n");
printf("\nk1\t\tk2\t\tk3\t\tY\n");
printf("\n");
 
for(i=1;i<=N;i++)
{
for(j=1;j<=5;j++)
{
printf("%f\t",k[i][j]);
}
a=x0+h;
 
printf("\n");
 
}
printf("\nThe final value of Y:\t%f",y);
getch();;
}