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Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 2

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Answer: 20\log 2 years

Given:  Population growth rate =5% per year.

To find: Time taken for increase in population.

Hint: The rate of population increases with increase in time i.e. \frac{d P}{d t}=r \% \times P

Solution: Population growth rate =5% per year

Let initial population be P_{0}  and the population after time t be P


        \begin{aligned} &\frac{d P}{d t}=5 \% \times P \\\\ &=\frac{d P}{d t}=\frac{5}{100} \times P \end{aligned}

        =>\frac{d P}{d t}=\frac{P}{20}

By cross multiplication we get,

        =>20 \frac{d P}{p}=d t

Integrating on both sides

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

At t=0 we have P=P_{0}                                 [Putting t=0 and P=P_{0} in equation i]       

        \begin{aligned} &=>20 \log \left(P_{0}\right)=0+C \\\\ &=>C=20 \log P_{0} \end{aligned}        

Putting C=20 \log P_{0}  in equation (i)

        \begin{aligned} &=>20 \log P=t+20 \log \left(P_{0}\right) \\\\ &=>20 \log P-20 \log \left(P_{0}\right)=t \\\\ &=>20 \log \frac{p}{p_{0}}=t \end{aligned}

When P=2 P_{0} we get

        \begin{aligned} &=20 \log \left(\frac{2 P_{0}}{P_{0}}\right) \\\\ &=t=20 \log 2 \end{aligned}

Hence, the required time period is 20\log 2  years.


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