JEE Main 2024 — 27 January, Shift 1 — Question 3

$\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0$

$\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt}$

$\int \frac{d x}{x}=-a \int d t$

$\ln |x|=-$ at $+c$ at $\mathrm{t}=0, \mathrm{x}=2$

$\ln 2=0+c$

$\ln x=-$ at $+\ln 2$

$\frac{x}{2}=e^{-a t}$

$\mathrm{x}=2 \mathrm{e}^{-\mathrm{at}}$

$\frac{d y}{d t}+b y=0$

$\frac{d y}{y}=-b d t$

$\ln |\mathrm{y}|=-\mathrm{bt}+\lambda$

$\mathrm{t}=0, \mathrm{y}=1$

$0=0+\lambda$

$y=e^{-b t}$

According to question

$$\begin{aligned}&3\mathrm{y}(1)=2\mathrm{x}(1)& \end{aligned}$$

$3\mathrm{e}^{\mathrm{b}}=2\left(2\mathrm{e}^{-\mathrm{a}}\right)$

$\mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3}$

For $\mathrm{x}(\mathrm{t})=\mathrm{y}(\mathrm{t})$

$\Rightarrow 2 \mathrm{e}^{-\mathrm{at}}=\mathrm{e}^{-\mathrm{bt}}$

$2=e^{(a-b) t}$

$$\begin{aligned}& 2=\left(\frac{4}{3}\right)^{t} &\end{aligned}$$

$\log _{\frac{4}{3}} 2=t$