The number of bacteria in a flask grows according to the differential equation d

Question

The number of bacteria in a flask grows according to the differential equation dy/dt = 0.08y In this question, time is measured in hours and the number of bacteria, y, is measured in millions. The number of bacteria at time t = 0 is 20 million. Enter a formula for the number of bacteria at time t Y= What is the value of the growth constant? Growth constant : per hour. How long does it take for the number of bacteria to double? (Enter your answer correct to two decimal places.) Doubling time : hours. How many million bacteria will be present after 8 hours have passed? (Enter your answer correct to one decimal place.) Number present after 8 hours : million.

Explanation / Answer

a)dy/dt =0.08y

dy/y=0.08 dt

integrate on both sides

dy/y=0.08 dt

lny =0.08t +c

y=e0.08t +c

y=Ce0.08t

at =0 ,y =20

20=Ce0

C=20

y=20e0.08t

b)growth constant =e0.08 per hour

growth constant =1.0833

c)to double =>y=2082=40

40=20e0.08t

0.08t =ln(40/20)

t =(1/0.08)ln2

t=8.66 hours

d)after 8 hours,t=8

y=20e0.08*8

y=37.9 million

number after 8 hours =37.9million

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