Asolution 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:

1.

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

=Number of bacteria per unit time

Where k=Proportionality constant

2.,y'(0)=100

Integrating on both sides then, we get

We have y(0)=200

Substitute the values then , we get

Substitute the value of C then we get

Differentiate w.r.t

Substitute the given condition then, we get

Substitute t=2

Then, we get

e=2.718

Hence, the number of bacteria after 2 hours=543.6

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