I have made some modifications to the shapes of the arrays in your code and changed the representation of the system of ODEs as a cell array of function handles to an array of function handles. The code below also demonstrates how "ode45" can be used to solve the problem. Both methods give the same result!
function testODE
f = @(t,y) [y(1)-4.*y(2);-y(1)+y(2)];
y0 = [1;0];
t0 = 0;
T = 1;
N = 11;
[Y,t] = RK42d(f, t0, T, y0, N);
figure;
hold on;
plot(t,Y(:,1),'r');
plot(t,Y(:,2),'b');
sol = ode45(f,[0,1],y0);
plot(sol.x,sol.y(1,:),'go');
plot(sol.x,sol.y(2,:),'ko');
legend('RK42d - y1','RK42d - y2','ode45 - y1','ode45 - y2');
end
function [Y, t] = RK42d(f, t0, T, y0, N)
%the input of f and y0 must both be vectors
%composed of two fuctions and their respective
%intial conditions
%the vector Y will return a vector as well
h = (T - t0)/(N - 1); %Calculate and store the step size
Y = zeros(N,2); %Initialize the X and Y vector
t = linspace(t0,T,N); % A vector to store the time values
Y(1,:) = y0; % Start Y vector at the intial values.
for i = 1:(N-1)
y = Y(i,:)';
k1 = f(t(i),y);
k2 = f(t(i) +0.5*h, y +0.5*h*k1);
k3 = f(t(i) +0.5*h, y +0.5*h*k2);
k4 = f(t(i) +h , y +h*k3);
Y(i+1,:) = y + (h/6)*(k1+ 2.*k2 + 2*k3 + k4);
%Update approximation
end
end
For more information on "ode45", please see
