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

Question

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

Accepted Answer

$\begin{gathered} \frac{dx}{dt}+ax=0 \ \frac{dx}{x}=-adt \ \int \frac{dx}{x}=-a\int dt \ \ln |x|=-at+c \ att=0,x=2 \ \ln 2=0+c \ \ln x=-at+\ln 2 \ \frac{x}{2}={e}^{-at} \ x=2{e}^{-at}\, \end{gathered}$

$\begin{gathered} \frac{dy}{dt}+by=0 \ \frac{dy}{y}=-bdt \ \ln |y|=-bt+\lambda \ t=0,y=1 \ 0=0+\lambda \ y={e}^{-bt}\, \end{gathered}$

According to question

$\begin{gathered} 3y(1)=2x(1) \ 3{e}^{-b}=2(2{e}^{-a}) \ {e}^{a-b}=\frac{4}{3} \end{gathered}$

$\begin{gathered} x(t)=y(t) \ \begin{gathered} 2{e}^{-at}={e}^{-bt} \ 2={e}^{(a-b)t} \ 2={(\frac{4}{3})}^{t} \ {\log }_{\frac{4}{3}}2=t \end{gathered} \end{gathered}$

Solutions · 1