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### Highlights from Runge Kutta 4th order ode

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# Runge Kutta 4th order ode

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### Judah S (view profile)

29 Dec 2010 (Updated )

solves ode using 4th order Runge Kutta method

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Description

This code defines an existing function and step size which you can change as per requirement.

P.S: This code has no new feature compared to existing codes available online. Intention behind posting this very simple code is to help students understand the concept and solve assignments.

MATLAB release MATLAB 7.10 (R2010a)
25 Jan 2016 faiz islam

### faiz islam (view profile)

sir can you assist me ,that how we can apply 4th order Runge kutta method for 4 coupled equation?
dx/dt=−ax − eω + yz
dy/dt= by + xz
dz/dt= cz + fω − xy
dω/dt = dω – gz
a = 50, b =−16, c = 10, d = 0.2, e = 10, f = 16, g = 0.5
Step size 0.001 .
regards
faiz

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16 Dec 2015 Ali Abbas

### Ali Abbas (view profile)

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10 Dec 2015 Stephanie Valerio

### Stephanie Valerio (view profile)

How do I run/call to this code?

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22 Nov 2015 Hamza Fokraoui

### Hamza Fokraoui (view profile)

for this function : f'''' - f*f''' + 4*g = 0
where i need to insert it in this code?
thank you

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29 May 2015 Shahzaib Asif

### Shahzaib Asif (view profile)

function RK4(f,a,x0,y0,h)

% Runge Kutta Method 4th Order
% function @(x,y) e.g. f=@(x,y)(x+y);
% a = the point up to which you obtain the results
% x0 = initial condition of x
% y0 = initial condition of y
% step size

x = x0:h:a;
y(1) = y0;

for i=1:(length(x)-1)

k1 = f(x(i),y(i));
k2 = f(x(i)+0.5*h,y(i)+0.5*h*k1);
k3 = f((x(i)+0.5*h),(y(i)+0.5*h*k2));
k4 = f((x(i)+h),(y(i)+k3*h));

y(i+1) = y(i) + (1/6)*(k1+2*k2+2*k3+k4)*h;

end

y(:)

%Shahzaib Asif (zaibi7402)
%shahzaib.7402@gmail.com

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25 Nov 2014 Chris FUNG

### Chris FUNG (view profile)

clear coding

13 Mar 2013 Christoph

04 Oct 2012 Ying

04 Oct 2012 Ying

### Ying (view profile)

Very good to learn. Thanks.

05 Feb 2012 Arun

### Arun (view profile)

29 Mar 2011 Pi Ting

excellent work

02 Jan 2011 Ido

### Ido (view profile)

Excellent program,