Differential Equations Question 5
Question 5 - 2024 (27 Jan Shift 1)
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 $3 y(1)=2 x(1)$, 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) $\log _{\frac{4}{3}} 2$
Show Answer
Answer (4)
Solution
$\frac{d x}{d t}+a x=0$
$\frac{d x}{x}=-a d t$
$\int \frac{d x}{x}=-a \int d t$
$\ln |x|=-a t+c$
at $t=0, x=2$
$\ln 2=0+c$
$\ln x=-a t+\ln 2$
$\frac{x}{2}=e^{-a t}$
$x=2 e^{-at}$.
$\frac{d y}{d t}+b y=0$
$\frac{d y}{y}=-b d t$
$\ln |y|=-b t+\lambda$
$t=0, y=1$
$0=0+\lambda$
$y=e^{-b t}$.
According to question
$3 y(1)=2 x(1)$
$3 e^{-b}=2\left(2 e^{-a}\right)$
$e^{a-b}=\frac{4}{3}$
For $x(t)=y(t)$
$\Rightarrow 2 e^{-at}=e^{-bt}$
$2=e^{(a-b) t}$
$ \begin{aligned} & 2=\left(\frac{4}{3}\right)^{t} \\ & \log _{\frac{4}{3}} 2=t \end{aligned} $
