JEE 2024 — Mathematics PYQ

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Question

JEE 2024 — Mathematics PYQ

JEE | Mathematics | 2024

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 $t$ ,

Choose the correct answer:

Explanation

We have, $\frac{d x}{d t} + a x = 0 ⟹ \frac{d x}{d t} = - a x$

$⟹ \int \frac{d x}{x} = - a \int d t$

$⟹ ln x = - a t + C_{1}$

$⟹ x = C e^{- a t}$

As, $x (0) = 2$

So, $2 = C e^{0} ⟹ C = 2$

$⟹ x = 2 e^{- a t}$

Also, $\frac{d y}{d t} + b y = 0 ⟹ \frac{d y}{d t} = - b y$

$⟹ \int \frac{d y}{y} = - b \int d t$

$⟹ ln y = - b t + C_{2}$

$⟹ y = K e^{- b t}$

As, $y (0) = 1$

So, $1 = K e^{0} ⟹ K = 1 ⟹ y = e^{- b t}$

Given, $3 y (1) = 2 x (1)$

$⟹ 3 e^{- b} = 2 (2 e^{- a})$

$⟹ e^{a - b} = \frac{4}{3}$

If $x (t) = y (t)$

$⟹ 2 e^{- a t} = e^{- b t}$

$⟹ 2 \frac{e ^{- a t}}{e ^{- b t}} = 1$

$⟹ 2 e^{- (a - b) t} = 1$

$⟹ 2 (\frac{4}{3})^{- t} = 1 ⟹ (\frac{4}{3})^{t} = 2$

$⟹ t = lo g_{\frac{4}{3}} 2$

Choose the correct answer:

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