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 $\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0$ 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 $\mathrm{x}(\mathrm{t})=\mathrm{y}(\mathrm{t})$, 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 $t=0, x=2$
$\ln 2=0+c$
$\ln x=-a t+\ln 2$
$\frac{x}{2}=e^{-a t}$
$\mathrm{x}=2 \mathrm{e}^{-\mathrm{at}}$.
$\frac{d y}{d t}+b y=0$
$\frac{d y}{y}=-b d t$
$\ln |y|=-b t+\lambda$
$\mathrm{t}=0, \mathrm{y}=1$
$0=0+\lambda$
$y=e^{-b t}$.
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 $x(t)=y(t)$
$\Rightarrow 2 \mathrm{e}^{-\mathrm{at}}=\mathrm{e}^{-\mathrm{bt}}$
$2=\mathrm{e}^{(\mathrm{a}-\mathrm{b}) \mathrm{t}}$
$$ \begin{aligned} & 2=\left(\frac{4}{3}\right)^{\mathrm{t}} \ & \log _{\frac{4}{3}} 2=\mathrm{t} \end{aligned} $$
