JEE Main 2021 — Differential Equations Question with Solution

Given $\frac{dB}{dt}\propto B\Rightarrow \frac{dB}{dt}=\lambda B, \left(\lambda \text{is proportionility constant}\right)$

$\Rightarrow \frac{dB}{B}=\lambda dt$

Integrate both sides with proper limit we get, 

${\int }_{B_1}^{B_2}\frac{dB}{B}={\int }_{t_1}^{t_2}\lambda dt$

$\Rightarrow \ln \frac{B_2}{B_1}=\lambda \left(t_2-t_1\right) ...\left(i\right)$

Given that, at $t=0, B=1000 \& t=20\% \text{increase}=1200$

$\therefore \ln \frac{1200}{1000}=2\lambda \Rightarrow \lambda =\frac{1}{2}\ln \frac{6}{5}$

Now an equation $\left(i\right)$becomes, 

$\ln \frac{B_2}{B_1}=\frac{1}{2}\ln \frac{6}{5}\left(t_2-t_1\right) ...\left(ii\right)$

Given that, at $t_2=\frac{k}{\ln 2}, B_2=2000 \& t=0, B_1=1000$

From $\left(ii\right), \ln \frac{2000}{1000}=\frac{1}{2}\ln \frac{6}{5}\left(\frac{k}{\ln {\displaystyle \frac{6}{5}}}-0\right)=\frac{1}{2}k$

$\Rightarrow k=2\ln 2$

$\therefore {\left(\frac{k}{\ln 2}\right)}^2={\left(\frac{2\ln 2}{\ln 2}\right)}^2=4$