#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 77 textbook solution.

Answer : $\frac{\log 2}{\lambda}$

Given:    The decay rate of radium at any time t is proportional to its mass at that time.

Hint:       Using variable separable method

Explanation:   Acc to given,

\begin{aligned} &\frac{d m}{d t} \alpha m \\ &\Rightarrow \frac{d m}{d t}=-d m \end{aligned}

\begin{aligned} &\text { where } \lambda>0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { [-ve sign } \therefore \text { of decay rate }] \\ &\Rightarrow \frac{d m}{m}=-d t \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{\mathrm{dm}}{\mathrm{m}}=-\int \lambda \mathrm{dt}\\ &\Rightarrow \log \mathrm{m}=-\lambda+\mathrm{c}-(1)\\ &\text { Let } \mathrm{m}_{0} \text { be the initial mass at } \mathrm{t}=0\\ &\log m_{0}=-\lambda \cdot 0+c\\ &\Rightarrow \mathrm{c}=\log \mathrm{m}_{0} \end{aligned}

Put in (1)

\begin{aligned} &\Rightarrow \log \mathrm{m}=-\lambda \mathrm{t}+\log \mathrm{m}_{0} \\ &\Rightarrow \log \mathrm{m}-\log \mathrm{m}_{0} \\ &=-\lambda \mathrm{t} \\ &\Rightarrow \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)=-\lambda \mathrm{t}-2 \end{aligned}                                                 $\left[\because \log \mathbf{m}-\log \mathbf{x}=\log \mathbf{m}_{\mathrm{h}}\right]$

We have to find the time when $m=\frac{1}{2}m_{0}$

$\Rightarrow 2m=m_{0}$

Put in (2)

\begin{aligned} &\Rightarrow \log \frac{\mathrm{m}}{2 \mathrm{~m}}=-\lambda \mathrm{t} \\ &\Rightarrow \log \frac{1}{2}=-\lambda \mathrm{t} \\ &\Rightarrow \log 1-\log 2=-\lambda \mathrm{t} \\ &\Rightarrow 0-\log 2=-\lambda \mathrm{t} \end{aligned}                                        $[\because \log 1=0]$

\begin{aligned} &\Rightarrow \lambda t=\log 2 \\ &\Rightarrow \mathrm{t}=\frac{\log 2}{\lambda} \end{aligned}