Matlab (Runge-Kutta Method):

Could someone explain what Y = [0 1; -1 ...] (why 0 1; -1) and u = [0; ...] (why 0 and only one column) mean in the following code?

(refference: https://www.mathworks.com/matlabcentral/answers/705963-evaluating-a-2nd-order-ode-using-the-runge-kutta-method)

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% ODE: y''(t) = -e^(3t)*y'(t) - y(t) + (5-2e^(-3t))*e^(-2t) +1

% With initial conditions y(0) = 2 ; y'(0) = -2 where 0

a = 0;

b = 1;

%n = input('Enter the number of intervals:');

n = 100;

h = (b-a)/n;

y = [2; -2];

t = 0:h:n*h;

for i=1:n

k1 = f(t(i),y(:,i));

k2 = f(t(i)+0.5*h,y(:,i)+0.5*k1*h);

k3 = f(t(i)+h,y(:,i)-k1*h+2*k2*h);

y(:,i+1)= y(:,i)+h/6*(k1+4*k2+k3); %approximated value of the ODE using RK3

end

exact = exp(-2*t)+1;

disp(max(abs(exact-y(1,:))))

plot(t,y(1,:),'r',t,exact,'b--'),grid

xlabel('t'),ylabel('y')

legend('RK3','Exact')

function fty = f(t,y) %matrix form of the 2nd order ODE

Y = [0 1; -1 -exp(-3*t)]*y; % HERE

u = [0; (5-2*exp(-3*t))*exp(-2*t)+1]; % HERE

fty = Y + u;

end