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Question

A population of bacteria given by y(t) grows according to the equation dy/dt=ky, where k is a constant and t is measured in minutes. If y(10)=10 and y(30)=25, what is the value of K?

Answer 1

dy/dt=ky
dy/y = kdt
integral(dy/y) = k integral dt
ln(y) = kt + c
ln(10) = 10k + c
ln(25) = 30k + c
20 k = ln(25) - ln(10)
= ln(2.5)
k = ln(2.5)/20 = 0.045815

Answer 2

So dy/y = kdt.
(This is a separable differential equation.)
Integrating both sides
ln y = kt + ln C = ln(e^kt) + ln C.
y = Ce^kt.
Next, we plug in our points:
10 = Ce^10k
25 = Ce^30k.
So, dividing the second equation by the first,
2.5 = e^20k
20k = ln 2.5
k = ln 2.5/20 = 0.0458(approx)