The particular solution to the differential equation which describes the number of fish over time will be y = 40(0.99)^t.
Consider the function:
y = a (1 ± r) ˣ
Where x is the number of times this growth/decay occurs, a = initial amount, and r = fraction by which this growth/decay occurs.
If there is a plus sign, then there is exponential growth happening by r fraction or 100r %
If there is a minus sign, then there is exponential decay happening by r fraction or 100r %
A population of endangered salmon, S(t), starts out with 40 fish and decreases proportionally at a rate of 0.01 fish per year. Let t be measured in years.
Let y be the number of fish over time (t).
Then we have
[tex]\rm y = 40(1 - 0.01)^t\\\\y = 40(0.99)^t[/tex]
Then the particular solution to the differential equation which describes the number of fish over time will be y = 40(0.99)^t.
More about the exponent link is given below.
https://brainly.com/question/5497425
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