"dN\/dt=aN(k-N)" is the logistic growth equation, where dN/dt is rate of growth of bacteria population, a is an undetermined constant and k is the saturation population or equilibrium population.
Solving the above equation we get;
"dN\/N(k-N)=adt=(1\/k)dN(1\/N+1\/(k-N))"
Integrating both sides we get;
"\\implies \\int dN(1\/N+1\/(k-N))=k\\int adt"
"ln(N\/(k-N))=(ka)t+c"
Using the given conditions :
"k=10000; N(0)=1000; N(1)=2000"
We get;
"c=ln(1000\/(10000-1000))=-ln9=-2.197"
and
"ln(2000\/(10000-2000))=ka+c"
"\\implies ka=(-ln4-c)=-1.386+2.197=0.811"
Thus, the population as a function of time is given by;
"ln(N\/(10000-N)))=0.811t-2.197"
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