Exponential Growth & Decay Equations & Examples

Recent questions in Exponential growth and decay
Algebra IAnswered question
Marilyn Cameron Marilyn Cameron 2022-10-26

Consider the modification to the Malthusian equation
d N d t = r S ( N ) N ,
where r>0 is the per capita growth rate, and S(N) is a survival fraction. For some organisms, finding a mate at low population densities may be difficult. In such cases, the survival fraction can take the form S ( N ) = N A + N , where A>0 is a constant.
i. By examining what happens to S(N) for N≫A and N≪A explain why this fraction models the situation outlined above.
When N≫A (my assumption is that the sign '≫'means significantly greater than ) so A is much smaller than N then the constant A.This would effect the survival fraction in a way which it becomes roughly equal to 1 since A is really small.
When N≪A (my assumption is that the sign '≪' means significantly smaller than) so N is much smaller than A will impact the survival fraction in a way which will make it become small.
The survival fraction models matches with the situation above since:
For a low population finding a mate is difficult (a.k.a N≪A )
For a large population finding a mate is more likely (a.k.a N≫A )
ii. By examining the form of the equation, determine the long-term behaviour of a population for an initial condition N(0)>0.
Attempt: For the initial conditions stated above the Malthusian equation has the form:
d N d t = r N A + N N ,
which has the solution.
N ( t ) = N 0 e λ t
If λ > 0 then exponential growth
If λ < 0 then exponential decay
Please could you check if this is correct. If it is please can you suggest any improvement I could add to my attempts.

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