Let ​​x = x(t) and y = y(t) be solutions of the differential equations dx/dt + ax = 0

Question

Let \(x=x(t)\) and \(y=y(t)\) be solutions of the differential equations \(\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0\) and \(\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{by}=0\) respectively, \(\mathrm{a}, \mathrm{b} \in \mathrm{R}\). Given that \(x(0)=2 ; y(0)=1\) and \(3 y(1)=2 x(1)\), the value of \(t\), for which \(x(t)=y(t)\), is :

(1) \(\log _{\frac{2}{3}} 2\)

(2) \(\log _4 3\)

(3) \(\log _3 4\)

(4) \(\log _{\frac{4}{3}} 2\)

Answer 1

Correct option is (4) \(\log _{\frac{4}{3}} 2\) 

\(\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0\)

\(\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt}\)

\(\int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt}\)

\(\ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c}\)

at \(\mathrm{t}=0, \mathrm{x}=2\)

\(\ln 2=0+\mathrm{c}\)

\(\ln \mathrm{x}=-\mathrm{at}+\ln 2\)

\(\frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}}\)

\(\mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad....(i)\)

\(\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{by}=0\) 

\(\frac{\mathrm{dy}}{\mathrm{y}}=-\mathrm{bdt}\)

\(\ln |y|=-b t+\lambda\)

\(\mathrm{t}=0, \mathrm{y}=1\)

\(0=0+\lambda\)

\(\mathrm{y}=\mathrm{e}^{-\mathrm{bt}} \quad ....(ii)\)

According to question

\(3 \mathrm{y}(1)=2 \mathrm{x}(1)\)

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

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

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

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

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

\(2=\left(\frac{4}{3}\right)^{t}\)

\(\log _{\frac 43} 2=\mathrm{t}\)