Question

Hey

Can someone write me a c++ pogramm using 4th order runge kutta method? h=0.1

y' = 3y, y(0) = 1

Answer 1

Runge Kutta Method- With this method one can solve approximate value of 'y' of diffrential equation where one bounty conditions is also given.

Formula-

K1= h*f(xn, yn )

K2= h*f( xn + (h/2) , yn + (K1/2))

K3= h*f( xn + (h/2), yn + (K2/2) )

K4= h*f( xn + h, yn + K3)

K= (1/6)*(K1 + 2*K2 + 2*K3 + K4)

Yn+1 = yn + K

Here the value of 'n' are 1,2,3,4,5,.....,n

Also here 'h' is a step height and xn+1 = x0+h

Question- y' = 3*y, y(0)=1 , h=0.1

So initially when x0=0   y0 = 1

                             x1=x0+h=0+0.1=0.1    y1 = ?

                             x2=x1+h=0.1+0.1=0.2     y2 = ?

program-

#include<iostream>
using namespace std;
float equation(float,float);
int main()
{
    float x0= 0 , y =1 ,h = 0.1; // here the y value is y0
    int n=1;// no of iteration
    float k1,k2,k3,k4,k5;
    for(int i=1; i<=n ; i++)
    {
        k1=h* equation(x0,y);
        k2=h* equation(x0 + 0.5*h, y + 0.5*k1);
        k3=h* equation(x0 + 0.5*h, y + 0.5*k2);
        k4=h* equation(x0 + h, y + k3);

        k5= ((1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4));

        y=y + k5;

        x0=x0+h;
    }
   cout<<"The value of y is"<<y;

}

float equation(float x , float y)
{
    return 3*y;
}

{ 1 #include<iostream> 2 using namespace std; float equation(float, float); 4 int main() 5 - { 6 float xe= 0, y =1 ,h = 0.1; // here the y value is yo 7 int n=1;// no of iteration 8 float k1, k2, k3, k4, k5; 9 for(int i=1; i<=n; i++) 10 11 k1=h* equation(x0,y); 12 k2=h* equation(x + 0.5*h, y + 0.5*k1); 13 k3=h* equation(xo + 0.5*h, y + 0.5*k2); 14 k4=h* equation(x + h, y + k3); 15 16 k5= ((1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4)); 17 18 y=y + k5; 19 20 x0=x0+h; 21 } 22 cout<<"The value of y is"<<y; 23 24 25 } 26 27 float equation(float x, float y) 28 - { 29 return 3*) y; 30 } 31 input The value of y is1.34984 ... Program finished with exit code 0 Press ENTER to exit console.

When we run the loop 2 times the output is -

The value of y is1.82206 ... Program finished with exit code 0 Press ENTER to exit console. I

When we run the loop 3 times the output is -

The value of y is2.45949 ... Program finished with exit code 0 Press ENTER to exit console.

So, depending upon the no of iteration the approximate value of y changes.