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

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 $P$

Then,

\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.