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Assume there is a certain population of fish in a pond whose growth is described by the logistic equation.
It is estimated that the carrying capacity for the pond is 1600 fish. Absent constraints,
the population would grow by 120% per year.
If the starting population is given by p0=200, then estimate population of fish in the pond
after one and two years.
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One good Internet source to read and to learn about a logistic equation is Libre text
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/08%3A_Introduction_to_Differential_Equations/8.04%3A_The_Logistic_Equation
The general solution for the standard logistic differential equation is so called logistic function
P(t) = .
Here is the initial population, P(t) is current population, K is the carrying capacity
and "r" is growth rate. Number "e" = 2.71828 is the base of natural logarithm
(shown here as an approximate value).
In this current problem in the post, we are given
= 200, K = 1600, r = 1.2.
They want you determine P(1) and P(2), the population after 1 and 2 years.
(a) For t = 1 year
P(1) = = 514.74, or 514 (rounded).
(b) For t = 2 years
P(2) = = 978.58, or 979 (rounded).
Solved.
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Post-solution notice
In my opinion, the question in the post is posed/worded INCORRECTLY.
It asks about the breeding season, but we are not given information
for a breading season - we are given information, related to years - so,
the question should be about the times of years, not the breeding seasons.
The life of fish includes not only breeding - it includes also struggle for food,
survival from predators and diseases, natural death, that is, processes with a period of a year.