Let x = x(t) and y = y(t) be solutions of the differential equations d x d t + a x = 0 and d y d t + b y = 0 respectively, a , b ? 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: [2024]

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Solution

Q.
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 3y(1) = 2x(1), the value of t, for which x(t) = y(t), is: [2024]

1 $\log_{\frac{4}{3}} 2$

2 $\log_{\frac{2}{3}} 2$

3 $\log_{4} 3$

4 $\log_{3} 4$

Ans.
(1)

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

$\int \frac{d x}{a x} = - \int d t \Rightarrow \frac{1}{a} \log x = - t + c_{1}$

$∵$ x(0) = 2

$c_{1} = \frac{1}{a} \log 2 \Rightarrow \frac{1}{a} \log x = - t + \frac{1}{a} \log 2 \Rightarrow x = 2 e^{- a t}$

Now, $\frac{d y}{d t} + b y = 0 \Rightarrow \frac{1}{b} \log y = - t + c_{2}$

$∵ y (0) = 1$

$∴ \frac{1}{b} \log (1) = - 0 + c_{2} \Rightarrow c_{2} = 0 \Rightarrow y = e^{- b t}$

Now, $3 y (1) = 2 x (1) \Rightarrow 3 e^{- b} = 4 e^{- a}$

$\Rightarrow e^{a - b} = \frac{4}{3} \Rightarrow a - b = \log (\frac{4}{3})$ ... (i)

For $x (t) = y (t), 2 e^{- a t} = e^{- b t} \Rightarrow e^{(a - b) t} = 2 \Rightarrow (a - b) t = \log 2$

$\Rightarrow \log (\frac{4}{3}) t = \log 2$ [From (i)]

$\Rightarrow t = \frac{\log 2}{\log \frac{4}{3}} \Rightarrow t = \log_{\frac{4}{3}} 2$

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