From the question it is given that,

The bacteria count is 1,00,000

The number is increased by 10% in 2 hours.

So, let us assume C be the surface area of balloon,

\(( \dfrac{dC}{dt})\) ∝ C

Then,

\( \dfrac{dC}{dt}\) = λC

\( \dfrac{dC}{dt}\) = λ dt

Integrating on both side we get,

∫\( \dfrac{dC}{dt}\) = λ ∫dt

Log C = λt + log k … [equation (i)]

From question, t = 0 when c = 1,00,000,

log (1,00,000) = λ × 0 + log k … [equation (ii)]

log (1,00,000) = log k … [equation (iii)]

And also form question, given t = 2,

c = 1,00,000 + 1,00,000 × (\( \dfrac{10}{100}\))

= 110000

So, from equation (i) we have,

log 110000 = λ × 2 + log K … [equation (iv)]

Now, subtracting equation (ii) from equation (iv), we have,

Log 110000 – log 100000 = 2 λ

Then, log 11 × 10000 – log 10 × 10000 = 2 λ

Log \(\dfrac{ ((11 × 10000)}{(10 × 10000))}\) = 2λ

Log \(\dfrac{11}{10}\) = 2λ

So, λ = \( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\)) … [equation (v)]

Now we need to find the time ‘t’ in which the count reaches 200000.

Then, substituting the values of λ and k from equation (iii) and (v) in equation (i),

Log 200000 = \( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\))t + log 100000

\( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\))t = log 200000 – log 100000

\( \dfrac{1}{2}\)log (\( \dfrac{11}{10}\))t = log\(\dfrac{ 200000}{100000}\)

\( \dfrac{1}{2}\)log(\( \dfrac{11}{10}\))t = log 2

Therefore, the required time t = \(\dfrac{ 2log2}{log}(\dfrac{11}{10}) hours\)

Answered by Aaryan | 1 year agoFind the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)

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