Differential Equations -Population Models, guaranteed and verified 100% Pass

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Population Models


We have already seen one model for population growth. In that model, one
assumes a constant rate of growth which is proportional to the population size at
time 𝑡:


𝑑𝑃
= 𝑘𝑃 (𝑡 ); 𝑘 > 0.
𝑑𝑡



The general solution is:

𝑃(𝑡) = 𝑃0 𝑒 𝑘𝑡 .


More generally, we could assume a birth rate, 𝛽(𝑡), and a death rate, 𝛿(𝑡),
where the rate of change in the population is given by:


𝑑𝑃
= (𝛽(𝑡) − 𝛿(𝑡))𝑃(𝑡).
𝑑𝑡



If 𝛽 (𝑡 ) − 𝛿 (𝑡 ) = 𝑘, a constant, then we get the first population model. Birth
and death rates can also depend on the size of the population 𝑃(𝑡). Notice that
the birth and death rates are percentages of the population. For example,
suppose the population is 500 at time 𝑡 and the birth rate is 0.02 = 2% per
year. Then over the next year there will be . 02(500) = 10 births. So the
absolute birth rate for that year is 10, but the (relative) birth rate is 0.02.

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Ex. Suppose a certain lake is stocked with fish and the birth and death rates are
inversely proportional to √𝑃(𝑡 ). At the time 𝑡 = 0, 𝑃0 = 100 and after 6
months there are 169 fish. a) Find a formula for 𝑃 (𝑡 ). b) How many fish are
there after 1 year?


𝑎 𝑏
a) 𝛽 = , 𝛿= , where 𝑎, 𝑏 are constants.
√𝑃 √𝑃


𝑑𝑃 𝑎 𝑏
= (𝛽 − 𝛿 )𝑃 = ( − ) 𝑃
𝑑𝑡 √𝑃 √𝑃
𝑑𝑃
= (𝑎 − 𝑏)√𝑃 Now separate variables.
𝑑𝑡
1
𝑑𝑃 = (𝑎 − 𝑏)𝑑𝑡
√𝑃
1
∫ 𝑑𝑃 = ∫(𝑎 − 𝑏)𝑑𝑡
√𝑃
2√𝑃 = (𝑎 − 𝑏)𝑡 + 𝑐1
1
√𝑃 = (𝑎 − 𝑏)𝑡 + 𝑐2
2
(∗)
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𝑃 = ( (𝑎 − 𝑏)𝑡 + 𝑐2 )2 ; 𝑙𝑒𝑡 𝑘 = 𝑎 − 𝑏
2
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𝑃(𝑡) = ( 𝑘𝑡 + 𝑐2 )2 Now find 𝑐2 .
2


100 = 𝑃(0) = (𝑐2 )2 ⟹ 𝑐2 = 10, since 𝑐2 ≥ 0 from (∗).
1
𝑃(𝑡 ) = ( 𝑘𝑡 + 10)2 Now find 𝑘.
2
1
169 = 𝑃(6) = ( 𝑘(6) + 10)2
2
169 = (3𝑘 + 10)2
13 = 3𝑘 + 10; 3𝑘 + 10 ≠ −13, else 𝑘 < 0 and 𝑃(𝑡) decreases
𝑘=1

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So 𝑃(𝑡 ) = ( 𝑡 + 10)2 .
2

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b. 𝑃(12) = ( (12) + 10)2 = 162 = 256 fish after 1 year.
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