Let x = x(t) and y = y(t) be solutions of the different equations $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}+\mathrm{a}\mathrm{x}=0$ and $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}+\mathrm{b}\mathrm{y}=0$ 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:

The correct answer is log_4/3 2 (Option 4).

We are given two differential equations and need to find the value of t for which x(t) = y(t).

Step 1: Solve the differential equations.

The first equation is:

$\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0$

This is a separable linear ODE. Rearranging:

$\frac{\mathrm{dx}}{x}=-a\mathrm{dt}$

Integrating both sides gives: $\ln \left|x\right|=-\mathrm{at}+C$, so

$x\left(t\right)=A{e}^{-\mathrm{at}}$

Applying x(0) = 2:   A = 2, so $x\left(t\right)=2e^{-\mathrm{at}}$

Similarly, the second equation $\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{by}=0$ gives:

$y\left(t\right)=B{e}^{-\mathrm{bt}}$

Applying y(0) = 1:   B = 1, so $y\left(t\right)=e^{-\mathrm{bt}}$

Step 2: Use the condition 3y(1) = 2x(1) to relate a and b.

At t = 1:

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

Substituting into 3y(1) = 2x(1):

$3e^{-b}=2·2e^{-a}=4e^{-a}$

Therefore:

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

$e^{a-b}=\frac{4}{3}$  ...(★)

Step 3: Find t such that x(t) = y(t).

We set x(t) = y(t):

$2e^{-\mathrm{at}}=e^{-\mathrm{bt}}$

Rearranging:

$2=\frac{e^{-\mathrm{bt}}}{e^{-\mathrm{at}}}=e^{(a-b)t}$

Taking natural log on both sides:

$\ln 2=(a-b)t$

$t=\frac{\ln 2}{a-b}$

Step 4: Substitute the value of (a - b) from condition (★).

From (★): $a-b=\ln \frac{4}{3}$

Therefore:

$t=\frac{\ln 2}{\ln \frac{4}{3}}$

By the change-of-base formula, this is precisely:

$t={\log }_{\frac{4}{3}}2$

Conclusion: The value of t for which x(t) = y(t) is:

$t={\log }_{\frac{4}{3}}2$