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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

58. In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present.

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Vishal Baghel | Contributor-Level 10

4 months ago

Let ‘x’ be the number of bacteria present in instantaneous time t.

Then, $\frac{d x}{d t} \propto x$

$\Rightarrow \frac{d x}{d t} = k x, w h e r e, k =$ constant of proportionality.

$\Rightarrow \frac{d x}{x} = k d t$

Integrating both sides,

$\begin{array}{l} \int \frac{d x}{x} = \int k d t \\ \Rightarrow l o g x = k t + c \end{array}$

Given, at $t = 0, x = x_{0} (s a y) t h e n,$

$l o g x_{0} = c (I n i t i a l, x_{0} = 100000)$

So, the differential equation is

$\begin{array}{l} l o g x = k t + l o g x_{0} \\ \Rightarrow l o g x - l o g x_{0} = k t \\ \Rightarrow l o g \frac{x}{x_{0}} = k t \end{array}$

As the bacteria number increased by 10% in 2 hours.

The number of bacteria increased in 2hours $= 10 % \times 100000 = 10000$

Hence, at t=2,

$x = 100000 + 10000 = 110000$

So, $\begin{array}{l} l o g \frac{110000}{100000} = 2 k \\ \Rightarrow k = \frac{1}{2} l o g (\frac{11}{10}) \end{array}$

Hence, $l o g \frac{x}{x_{0}} = [\frac{1}{2} l o g \frac{11}{10}] \times t$

$w h e n, x = 200000,$ then we get,

$l o g \frac{200000}{100000} = \frac{1}{2} l o g \frac{11}{10} \times t$

$\begin{array}{l} \Rightarrow 2 l o g 2 = l o g (\frac{11}{10}) \times t \\ \Rightarrow t = \frac{2 l o g 2}{l o g (\frac{11}{10})} h o u r s \end{array}$

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58. In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present.

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