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, $a, b \in \mathbb{R}$. 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 :

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Differential Equations

Let

x = x (t)

and

y = y (t)

be solutions of the differential equations

\frac{d x}{d t} + a x = 0

and

\frac{d y}{d t} + b y = 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 :

Answer: (D)

Solving

\frac{d x}{d t} + a x = 0

with

x (0) = 2

gives

x (t) = 2 e^{- a t}

; solving

\frac{d y}{d t} + b y = 0

with

y (0) = 1

gives

y (t) = e^{- b t}

.

From

3 y (1) = 2 x (1)

3 e^{- b} = 4 e^{- a}

, so

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

.

Setting

x (t) = y (t)

2 e^{- a t} = e^{- b t} \Rightarrow 2 = e^{(a - b) t} = (\frac{4}{3})^{t}

, so

t = lo g_{4/3} 2

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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