Suppose a bee population model is such that dB/dt = 0.06y - 2.3t, Where B is mea

Question

Suppose a bee population model is such that dB/dt = 0.06y - 2.3t, Where B is measured in bees (approximately scaled e.g., millions, etc.). Due to the life expectancy, time is considered a natural enemy to bees. If there aren't enough in the population, time proves to be a cause for possible extinction.

a) Using the method of integrating factors, solve the differential.

I did this and he said it is correct and when I use b instead of y in my calculator for the DE it says it is correct. It is y(t) = Ce^0.06t + (2.3/0.06)t +(2.3/0.0036)

b) Under what mathematical condition will the bee population grow forever? Under what mathematical condition will the bee population eventually die off? Use the idea of limits and long-run analysis of the solution you have found. Explain your reasoning clearly.

I am under the assumption that we can't use negative time (unless we have a time machine). And it doesn't make sense to have negative bees (unless you count zombie bees). Yet not a single one of those terms converge to 0. And the sum of them will never converge to 0 b/c we can't start off with negative bees. Yet we still have bees even if we start off with 0 bees. What is it that I am not seeing and causing me to retard this problem and want to rip my hair out?

Thanks in advanced.

Explanation / Answer

y(t) = Ce^0.06t + (2.3/0.06)t +(2.3/0.0036) looks great

Simplifying this :
y = Ce^(0.06t) + 38.33t + 638.89

As t ---> 0, y = C + 638.89
so clearly C >= -638.89

Clearly when C is a positive constant
y = Ce^(0.06t) + 38.33t + 638.89
when derived becomes
y' = 0.06Ce^(0.06t) + 38.33, both positive terms, as in
function keeps increasing, as in the bee population keeps on growing

So, for bee pop to keep on growing,
we just need C>= 0

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To eventually die off,
we just neee constant C < 0

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