Let x = x ( t ) and y = y ( t ) be solutions of the differential...

Given: $\dfrac{dx}{dt} + ax = 0$

$⇒ \dfrac{dx}{x} = - adt$

Integrating both sides,

$⇒ ∫ \dfrac{dx}{x} = - a ∫ dt$

$⇒ log | x | = - at + c$

At $t = 0, x = 2$

$⇒ \log 2 = 0 + c$

$⇒ \log x = - at + \log 2$

$⇒ \log x - \log 2 = - at$

$⇒ \log \left(\dfrac{x}{2}\right) = - at$

$⇒ \dfrac{x}{2} = {e}^{- at}$

$⇒ x = 2{e}^{- at} . . . ( i )$

It is also given that, $\dfrac{dy}{dt} + by = 0$

$⇒ \dfrac{dy}{y} = - bdt$

$⇒ log | y | = - bt + \lambda$

At, $t = 0, y = 1$

$⇒ 0 = 0 + \lambda$

$⇒ y = {e}^{- bt} . . . ( ii )$

According to question,

$3y ( 1 ) = 2x ( 1 )$

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

$⇒ {e}^{a - b} = \dfrac{4}{3}$

$\text{For}x\left(t\right) = y\left(t\right)$

$⇒ 2{e}^{- at} = {e}^{- bt}$

$⇒ 2 = {e}^{( a - b ) t}$

$⇒ 2 = {\left(\dfrac{4}{3}\right)}^{t}$

$⇒ \log \left(2\right) = t \times \log \left(\dfrac{4}{3}\right)$

$⇒ log\dfrac{4}{3}2 = t$