Course:MATH102/Question Challenge/2002 December Q4

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Question

The population of two species of rodents is found to satisfy

$y_{1} (t) = 1000 e^{- 0.2 t}$

$y_{2} (t) = 50 e^{0.5 t}$

(a) What differential equation and initial conditions govern the populations of each species?

(b) At what time is the population of species 1 twice as large as that of species 2?

Hint
add your hints here

Solution
Part A) $y_{1} (t) = 1000 e^{0.2 t}$ $y_{2} (t) = 50 e^{0.5 t}$ To find the differential equation for each species we take the first derivative: $y_{1} (t) = - 200 e^{- 0.2 t}$ $y_{2} (t) = 25 e^{0.5 t}$ To find the initial conditions we set t equal to zero for the equations that govern the population of both species: $y_{1} (0) = 1000 e^{0}$ $y_{1} (0) = 1000$ $y_{2} (0) = 50 e^{0}$ $y_{2} (0) = 50$ Part B) To find the time at which which the population of the first species is double that of the second, we set $y_{1} = 2 y_{2}$ and solve for $t$ : $1000 e^{- 0.2 t} = 2 (50 e^{0.5 t})$ $1000 e^{- 0.2 t} = 100 e^{0.5 t}$ $10 e^{- 0.2 t} = e^{0.5 t}$ $10 = e^{0.5 t + 0.2 t}$ $l n (10) = 0.7 t$ $t = l n (10) / 0.7$