Solving an ODE using ode45 in Octave

I am trying to solve the following ODE using Octave, and in particular the function ode45.

dx/dt = x(1-x/2), 0<= t <= 10

with the initial condition x(0) = 0.5

But the graphs I get are not what I expect.

I think that the graph with red crosses represents x' vs x and not x vs t.

The code is the following:

clear all
close all
% Differential Equation: x' = x(1-x/2)
function dx = f(x,t)
   dx = x*(1-x./2);
endfunction 
% Exacte Solution: 2*e^t/(3+e^t) 
function xexac =solexac(t)
xexac =  (2*exp(t))./(3+exp(t));
endfunction
x0=0.5;   %%Initial condition
T=10;            %% maximum time T
t=[0:0.1:T];              %% we choose the times t(k) where is calculated 'y'  
sol=ode45(@f,[0,T],x0);   %% numerical solution of (E)
tt=sol.x;y=sol.y;         %% extraction of the results
clf;hold on  ;            %% plot the exact and numerical solutionss
plot(tt,y,'xr')
plot(t,solexac(t),'-b') 
xlabel('t')
ylabel('x(t)')
title('Chemostat Model')
legend("Numerical Solution","Exacte Solution ")

It would we great that any of you could help me with this code.

Solution

ode45 expects the ODE function to have arguments in the order (time, state), so exactly the other way around. What you effectively did was integrate t-t^2/2, and the resulting function 0.5+t^2/2-t^3/6 is what you got in the plot.