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  • Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 3

Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 3

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Answer: 58 years

Given:  Present Population = 1,00,000

To find: When the city will have a population of 5,00,000

Hint: The population of city increase in time i.e. \frac{d P}{d t} \propto P and then find the equation using integration.

Solution: The present population is 1,00,000 and the population of a city doubled in the past 25 years

Let P be the population at any time t

Then, \frac{d P}{d t} \propto P

        \begin{aligned} &\frac{d P}{d t}=k P \\\\ &\frac{d P}{d t}=k d t \end{aligned}

Integrating on both sides we get,

        \begin{aligned} &=>20 \int \frac{d P}{P}=\int k d t \\\\ &=>\log P=k t+C \ldots(i) \end{aligned}

At t=0 we take P=P_{0}

Then,

        =>\log P_{0}=k \times 0+C                [Putting t=0  and P=P_{0}  in equation (i)]

        =>C=\log P_{0}

Putting C=\log P_{0}  in equation (i) we get

        \begin{aligned} &=>\log P=k t+\log P_{0} \\\\ &=>\log P-\log P_{0}=k t \\\\ &=>\log \frac{P}{p_{0}}=k t \ldots(i i) \end{aligned}

When P=2P_{0}  at t=25 we have

        \begin{aligned} &\log \left(\frac{2 P_{o}}{P}\right)=k t \\\\ &=>\log 2=25 k \end{aligned}

Putting k=\frac{\log 2}{25}  in equation (ii) we get

        =\log \left(\frac{p}{p_{0}}\right)=\left(\frac{\log 2}{25}\right) t

We assume that t_{1} be the time take for the population to become 5,00,000 from 1,00,000

Then,\log g\left(\frac{5,00,000}{1,00,000}\right)=\left(\frac{\log 2}{25}\right) t_{1}

        \begin{aligned} &=\log 5=\left(\frac{\log 2}{25}\right) t_{1} \\\\ &=25 \log 5=\log 2 t_{1} \\\\ &=t_{1}=25\left(\frac{\log \log 5}{\log \log 2}\right) \\\\ &=t_{1}=25\left(\frac{1.609}{0.6931}\right) \end{aligned}        [\log \log 5=1.609, \log \log 2=0.6931]

        =t_{1}=58.08 \text { Years }

Therefore the required time is 58 years (approximate).

 

Posted by

infoexpert26

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