#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 4 maths

Answer: $\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \text { hours }$

Given:  Present bacteria count$=1,00,000$

To find: The amount of hours taken for the count to reach $2,00,000$

Hint: Use $\frac{d C}{d t}=\lambda C$ and then find the value of lambda using integration.

Solution: The bacteria count is $1,00,000$ and rate of increase $=10%$ in $2$ hours

Let $C$ be the bacteria count at any time $t$,

Then $\frac{d C}{d t} \propto C$

$=>\frac{d C}{d t}=\lambda C$           [Where $\lambda$ is the constant]

Integrating on both sides

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

Where $\log \log k$  is integral constant

At $t=0$ we have $C=1,00,000$

Then $\log (1,00,000)=\lambda \times 0+\log k$         [putting $t=0$ and $C=1,00,000$ in equation (i)]

$=\log (1,00,000)=\log k \ldots(i i)$

\begin{aligned} &\text { At } t=2 \\ &\qquad \begin{aligned} C &=1,00,000+1,00,000\left(\frac{10}{100}\right) \\ &=1,10,000 \end{aligned} \end{aligned}

From equation (i) we have,

\begin{aligned} &=\log 1,10,000=\lambda \times 2+\log k \quad \text { [Putting } C=1,10,000 \text { and } t=2] \\\\ &=\log 1,10,000=2 \lambda+\log k \ldots(i i i) \end{aligned}

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

\begin{aligned} &=\log 1,10,000-\log 1,00,000=2 \lambda+\log k-\log k \\\\ &=\log \left(\frac{1,10,000}{1,00,000}\right)=2 \lambda \\\\ &=\log \frac{11}{10}=2 \lambda \\\\ &=\lambda=\frac{1}{2} \log \left(\frac{11}{10}\right) \end{aligned}

Now, we have to find time $t$ in which the count reaches $2,00,000$

\begin{aligned} &\text { Now } C=2,00,000 \text { in equation }(i)\\\\ &=\log (2,00,000)=\lambda t+\log k \end{aligned}

Putting value of lambda and  $\log k$, we get

\begin{aligned} &=\log (2,00,000)=\frac{1}{2} \log \left(\frac{11}{10}\right) t+\log 1,00,000 \\\\ &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log 2,00,000-\log 1,00,000 \end{aligned}

\begin{aligned} &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log \left(\frac{2,00,000}{1,00,000}\right) \\\\ &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log 2 \\\\ &=t=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \end{aligned}

Hence, the time required $=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}$  hours