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White raven [17]
3 years ago
12

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

te down a differential equation connecting y and the time t. 2. If the rate of increase of the number is initially 100 per hour, how many bacteria are there after 2 hours? Solution:
Mathematics
1 answer:
GuDViN [60]3 years ago
4 0

Answer:

1.\frac{dy}{dt}=ky

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

\frac{dy}{dt}=Number of bacteria per unit time

\frac{dy}{dt}\proportional y

\frac{dy}{dt}=ky

Where k=Proportionality constant

2.\frac{dy}{y}=kdt,y'(0)=100

Integrating on both sides then, we get

lny=kt+C

We have y(0)=200

Substitute the values then , we get

ln 200=k(0)+C

C=ln 200

Substitute the value of C then we get

ln y=kt+ln 200

ln y-ln200=kt

ln\frac{y}{200}=kt

\frac{y}{200}=e^{kt}

y=200e^{kt}

Differentiate w.r.t

y'=200ke^{kt}

Substitute the given condition then, we get

100=200ke^{0}=200 \;because \;e^0=1

k=\frac{100}{200}=\frac{1}{2}

y=200e^{\frac{t}{2}}

Substitute t=2

Then, we get y=200e^{\frac{2}{2}}=200e

y=200(2.718)=543.6=543.6

e=2.718

Hence, the number of bacteria after 2 hours=543.6

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