Question

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 $x(0) = 2$, $y(0) = 1$, and $3y(1) = 2x(1)$, the value of t for which $x(t) = y(t)$, is:

• A. $\log_{ \frac{2}{3}}2$
• B. $\log_{4} 3$
• C. $\log_{3} 4$
• D. $\log_4{{\frac{2}{3}}}$

Answer 1

Answer: (D)

$\log_4{{\frac{2}{3}}}$

Answer 2

Given differential equations are:
$\frac{dx}{dt} + ax = 0 \Rightarrow x(t) = x(0)e^{-at}$
$\frac{dy}{dt} + by = 0 \Rightarrow y(t) = y(0)e^{-bt}$
From the initial conditions, we are provided:
$x(0) = 2, \; y(0) = 1$
Thus, the solutions for $x(t)$ and $y(t)$ become:
$x(t) = 2e^{-at}$, $y(t) = e^{-bt}$

Step 1: Relation at $t = 1$

We are given:
$3y(1) = 2x(1)$
Substituting the values of $x(1)$ and $y(1)$:
$3e^{-b} = 2 \times 2e^{-a} \Rightarrow 3e^{-b} = 4e^{-a}$
Taking the natural logarithm on both sides:
$-b = -a + \ln\left(\frac{4}{3}\right)$
Rearranging terms, we get:
$b = a + \ln\left(\frac{4}{3}\right)$

Step 2: Finding $t$ for which $x(t) = y(t)$

We need to find the value of $t$ such that:
$2e^{-at} = e^{-bt}$
Dividing both sides by $e^{-bt}$:
$2 = e^{(b-a)t}$
Taking the natural logarithm of both sides:
$\ln 2 = (b - a)t$
Substituting the expression for $b - a$ from earlier:
$b - a = \ln\left(\frac{4}{3}\right)$
Thus:$t = \frac{\ln 2}{\ln \left(\frac{4}{3}\right)}$

Step 3: Simplifying the Expression

To simplify
$\frac{\ln 2}{\ln \left(\frac{4}{3}\right)}$, we recognize that:
$\log_4 \left(\frac{2}{3}\right) = \frac{\ln \left(\frac{2}{3}\right)}{\ln 4} = \frac{\ln 2}{\ln \left(\frac{4}{3}\right)}$
Thus, the value of $t$ is:$t = \log_4 \left(\frac{2}{3}\right)$