Example: Let P(t) be the population of a certain animal species. Assume that P(t) satisfies the logistic growth equation
Find A and B.
Hint: Use the initial condition and the result of 5.
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,
200 being the carrying capacity.
Then, the non-constant solutions can be generated by separating the variables
,
and the integration
.
Next, the left hand-side can be handled by using the technique of integration of rational functions. We get
,
which gives
Hence, we have
.
Easy algebraic manipulations give
,
where .
Therefore, all the solutions are
where C is a constant parameter.
Remark: We may rewrite the non-constant solutions as
,
where a and B are two parameters. If we use the conditions
we will be able to get the desired solution. Indeed, we have
Thus, and consequently
.
,
where we used the chain rule and the fact that . Since our solution is not one of the two constant solutions we are only left with the equation
This simplifies to
,
which gives P=100 (half-way between 0 and 200).
Remark: You still need to convince yourself that t=100 is indeed the inflection point, that is the second derivative changes sign at that point.
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Author: Helmut Knaust