MATLAB: Runge-kutta

mathematicsrunge kutta

hello i have this equation y''+3y'+5y=1 how can i solve it by programming a runge kutta 4'th order method ? i know how to solve it by using a pen and paper but i can not understand how to programe it please any one can solve to me this problem ? i dont have any idea about how to use ODE and i read the help in matlab but did not understand how to solve this equation please any one can solve this and help me with it ? thanx

Best Answer

Writing a new answer so that I can use markup (MathWorks, please fix this!!). You need a couple of things - first, the time values

t = 0:h:100

You can now work out how many steps you need:

n = numel(t)

Now, we need an array in which to store your results: each column should correspond to a value from your t vector

x = zeros(2, n)

We know the initial condition, so we'll pop that in

x(:, 1) = [0; 0]

Now we need a loop to store the subsequent x values. The basic structure will look like this

for i = 2:n % We've already filled in the first column
  k1 = h*f(x(:, i-1));
  k2 = ...
  k3 = ...
  k4 = ...
  % Now define the new x vector based on k1, ... , k4
  x(:, i) = x(:, i-1) + ...
end
% Plot your results
plot(t, x)

And that should pretty much be all you need

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MATLAB: Coupled Second order differential eq. solution

%Suppose
x1=x
x2=x'
x3=y
%we get the system
dx1=x2
dx2=x2/a*(f/c - b*(1-x2*x3/d)/x3)  % You have to check your brackets 
dx3=1-x2*x3/d

MATLAB: Runge-Kutta function with a second order ODE

I'm still confused about your initial conditions and what y0 and y1 are. But let's back up a bit.

For a 2nd order ODE, your state vector will have 2 elements. E.g., if y is your state then y(1) could be position and y(2) could be velocity. This looks like the setup you have in your FunctionC( ) above. So for that case, the states in your RK4 solver have to reflect this and be 2-element vectors, not scalars. So that y pre-allocation needs to have 2 rows, not 1. And all the downstream indexing needs to account for this as well. E.g., something like this if we assume that those y0 and y1 values are the initial values at the x0 time (CAUTION: Not checked):

function [x, y] = FunctionBeta_Executor(F,h,x0,x1,y0,y1)
% Note that F function expression is defined via Function Handle: F = @(x, y)(x+y)               
% change the function as you desire
                                             % step size (smaller step size gives more accurate solutions)
x = x0:h:x1;                                          % x space
y = zeros(2,length(x)); % CHANGED                            % Memory allocation
y(1,1) = y0;  % CHANGED                                       % initial condition
y(2,1) = y1;  % NEW
for i=1:(length(x)-1)                             
% i=1:(length(x)-1)                              % calculation loop
    k1 = F( x(i)         , y(:,i)            ); % CHANGED



    k2 = F( x(i) + 0.5*h , y(:,i) + 0.5*h*k1 ); % CHANGED
    k3 = F( x(i) + 0.5*h , y(:,i) + 0.5*h*k2 ); % CHANGED
    k4 = F( x(i) +     h , y(:,i) +     h*k3 ); % CHANGED
    y(:,i+1) = y(:,i) + (1/6)*( k1 + 2*k2 + 2*k3 + k4 )*h;  % main equation % CHANGED
end
figure, plot(x, y(1,:))  % To see the solution results % CHANGED
end

Note also how much easier it is to read (and debug) when you use spacing!

Best Answer

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MATLAB: Coupled Second order differential eq. solution

MATLAB: Runge-Kutta function with a second order ODE

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