Course:MATH102/Question Challenge/2002 December Q4

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Question

The population of two species of rodents is found to satisfy

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

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

(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?

Hints

Hint
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Solutions

Solution

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