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

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

According to question

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

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