r=1;   % rate constant in the logistic equation
K=1;   % saturation population size


% choose a step that is much smaller than the smallest time scale in the
% problem

step=0.1;

% Total Integration time
TotalTime=10;

%  The initial conditions is
N0=0.01;


% Now, do it for the actual integration
% We'll have our time information in the vector N
N(1)=N0;
% m is a counter that we will use in the for loop
m=2;

for i=step:step:TotalTime
    N(m)=N(m-1)+r*step*N(m-1);  % 
    m=m+1;  % increment the counter
end

Times=0:step:TotalTime;  % create a vector of times for x-axis

plot(Times,N,'-b')
xlabel('time')
ylabel('number of cells')
title('Exponential Growth')



