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

The Correct Option is D

To solve the given problem, we need to find the time \( t \) such that \( x(t) = y(t) \). We have the differential equations:

• \( \frac{dx}{dt} + ax = 0 \) 
• \( \frac{dy}{dt} + by = 0 \)

We are given initial conditions: \( x(0) = 2 \), \( y(0) = 1 \), and \( 3y(1) = 2x(1) \).

Step 1: Solving the differential equations

For \( x(t) \), solving the differential equation \( \frac{dx}{dt} + ax = 0 \) gives us the solution:

\(x(t) = x(0) e^{-at} = 2e^{-at}\)

For \( y(t) \), solving the differential equation \( \frac{dy}{dt} + by = 0 \) gives us the solution:

\(y(t) = y(0) e^{-bt} = e^{-bt}\)

Step 2: Using the condition \( 3y(1) = 2x(1) \)

Substitute \( t = 1 \) into both solutions:

\(x(1) = 2e^{-a}\) and \(y(1) = e^{-b}\)

The condition \( 3y(1) = 2x(1) \) becomes:

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

Taking the natural logarithm of both sides, we have:

\(-b \log e + \log 3 = -a \log e + \log 4\)

\(-b + \log 3 = -a + \log 4\)

Thus, we obtain:

\(a - b = \log \left(\frac{4}{3}\right)\)

Step 3: Finding \( t \) such that \( x(t) = y(t) \)

Equate \( x(t) \) and \( y(t) \):

\(2e^{-at} = e^{-bt}\)

Taking logarithm on both sides:

\(\log 2 - at = -bt\)

Rearranged, this gives:

\(t(a - b) = \log 2\)

Substituting the value we found for \( a - b \):

\(t \log\left(\frac{4}{3}\right) = \log 2\)

Solving for \( t \),

\(t = \frac{\log 2}{\log \left(\frac{4}{3}\right)}\)

Using the change of base formula, \( \log_b a = \frac{\log_c a}{\log_c b} \), we have:

\(t = \log_4 \left(\frac{2}{3}\right)\)

Thus, the correct answer is \(\log_4{{\frac{2}{3}}}\).

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:\( t = \log_4 \left(\frac{2}{3}\right) \)