Question

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

Answer 1

1.
`(dy)/(dt)=ky`

2.543.6

Explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

`(dy)/(dt)`=Number of bacteria per unit time

`(dy)/(dt)\proportional y`

`(dy)/(dt)=ky`

Where k=Proportionality constant

2.
`(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(y)/(200)=kt`

`(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=(100)/(200)=(1)/(2)`

`y=200e^{(t)/(2)}`

Substitute t=2

Then, we get
`y=200e^{(2)/(2)}=200e`

`y=200(2.718)=543.6=543.6`

e=2.718

Hence, the number of bacteria after 2 hours=543.6