QJEEM24Ma4b12: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4b12 Solution: Let and be solutions of the differential equations and…

$\begin{aligned}& \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0 \\& \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt} \\& \int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt} \\& \ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c} \\& \mathrm{at} \mathrm{t}=0, \mathrm{x}=2 \\& \ln 2=0+\mathrm{c} \\& \ln \mathrm{x}=-\mathrm{at}+\ln 2 \\& \frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}} \\& \mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad \text{.... (i)}\end{aligned}$

$\begin{aligned}& \frac{d y}{d t}+b y=0 \\& \frac{d y}{y}=-b d t \\& \ln |y|=-b t+\lambda \\& t=0, y=1 \\& 0=0+\lambda \\& y=e^{-b t} \quad \text{..... (ii)}\end{aligned}$

According to question

$\begin{aligned}& 3 \mathrm{y}(1)=2 \mathrm{x}(1) \\& 3 \mathrm{e}^{-\mathrm{b}}=2\left(2 \mathrm{e}^{-\mathrm{a}}\right) \\& \mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3}\end{aligned}$

$\begin{aligned}& \text { For } x(t)=y(t) \\& \begin{array}{r}2 \mathrm{e}^{-a t}=e^{-b t} \\2=e^{(a-b) t} \\2=\left(\frac{4}{3}\right)^t \\\log _{\frac{4}{3}} 2=t\end{array}\end{aligned}$