Question

1) Explain the runge-kutta method

2) Produce an example that estimates a differential equation with this technique and the necessary code to run your iterations.

Answer 1

The runge-kutta method / Runge kutta method of fourth order :-

It is the general method , the most widely known member of the Runge–Kutta family is generally referred to as "RK4", the "classic Runge–Kutta method" or simply as "the Runge–Kutta method".

Procedure:-

let the given differential equation be dy/dx = f(x,y) with the initial conditions ,

X = X0, y = y0 .

To find the value of y = y0 + k, say at x = x0 + h , we calculate,

K1 = hf (x0,y0)

K2 = hf (x0 + h/2 , y0 + k1/2)

K3 = hf (x0 + h/2 , y0 + k2/2)

K4= hf (x0 + h, y0 + k3)

Lastlty,

K = 1/6 [K1 + 2 K2 + 2 K3 + K4]

Required value: y = y0 + k

For example:

{Question} Solve dy/dx with initial conditions y(1)=2 and find y at x=1.2,1.4 by Runge –kutta method of fourth order.

Solution:-

K1 = hf (x0,y0) = 0.2(1*2) = 0.4

K2 = hf (x0 + h/2 , y0 + k1/2) = 0.2 [(1+0.1)(2+0.2)] = 0.484

K3 = hf (x0 + h/2 , y0 + k2/2) = 0.2[(1+0.1)(2+2.242)] = 0.49324

K4= hf (x0 + h, y0 + k3) = 0.2 [(1+0.2)(2+0.49324)] = 0.598378

=>K = 1/6[k1 + 2k2 + 2k3 + k4]

    K = 1/6[0.4+2*0.484+2*0.4932+0.492143]

=>y=y0+k = 2+0.492143

At x=1.2, y=2.492143

b) x0 = 1.2

y0 = 2.492143

h=0.2

f(x,y) = xy  

hence, f(x,y) = xy  .

2) //Eulers Method to solve a differential equation in c++ #include<iostream> #include<iomanip> #include<cmath> using namespace std; double df(double x, double y) //function for defining dy/dx { double a=x+y; //dy/dx=x+y return a; } int main() { int n; double x0,y0,x,y,h; //for initial values, width, etc. cout.precision(5); //for precision cout.setf(ios::fixed); cout<<"\nEnter the initial values of x and y respectively:\n"; //Initial values cin>>x0>>y0; cout<<"\nFor what value of x do you want to find the value of y\n"; cin>>x; cout<<"\nEnter the width of the sub-interval:\n"; //input width cin>>h; cout<<"x"<<setw(19)<<"y"<<setw(19)<<"dy/dx"<<setw(16)<<"y_new\n"; cout<<"----------------------------------------------------------\n"; while(fabs(x-x0)>0.0000001) //I couldn't just write "while(x0<x)" as they both are floating point nos. It is dangerous to compare two floating point nos. as they are not the same in binary as they are in decimal. For instance, a computer cannot exactly represent 0.1 or 0.7 in binary just like decimal can't represent 1/3 exactly without recurring digits. { y=y0+(h*df(x0,y0)); //calculate new y, which is y0+h*dy/dx cout<<x0<<setw(16)<<y0<<setw(16)<<df(x0,y0)<<setw(16)<<y<<endl; y0=y; //pass this new y as y0 in the next iteration. x0=x0+h; //calculate new x. } cout<<x0<<setw(16)<<y<<endl; cout<<"The approximate value of y at x=0 is "<<y<<endl; //print the solution. return 0; }