Let x = x(t) and y = y(t) be solutions of the differential equations dx/dt + ax = 0 and dy/dt + by =

Q: Wait, why do we reject the negative root here? Or do we even have one?

In this problem, we are dealing with exponential growth or decay. Since $x(t)$ and $y(t)$ are solutions to $\frac{dx}{dt} + ax = 0$, they are inherently exponential. When we set $x(t) = y(t)$, we are solving $2e^{-at} = e^{-bt}$. This leads to $2 = e^{(a-b)t}$. Since $e^{(a-b)t}$ is always positive for any real $t$, and $2$ is positive, we don't have to worry about negative roots or complex solutions here. The math is straightforward!

Q: Why can't we just solve for $a$ and $b$ individually?

That is a fantastic question! You have two unknowns, $a$ and $b$, but you only have one constraint: $3y(1) = 2x(1)$. With only one equation, you cannot uniquely determine both $a$ and $b$. However, notice that the final expression for $t$ only depends on the difference $(a-b)$. We don't need the individual values of $a$ and $b$, just their relationship. This is a common trick in JEE problems—look for what you actually need, not what you think you need!

Q: How do I know the solution to $\frac{dx}{dt} + ax = 0$ is $x(t) = Ce^{-at}$?

Great catch! This is a standard result for first-order linear differential equations. If you separate the variables, you get $\frac{dx}{x} = -a dt$. Integrating both sides gives $\ln|x| = -at + C'$. Exponentiating both sides, we get $|x| = e^{-at+C'} = e^{C'}e^{-at}$. Letting $C = \pm e^{C'}$, we arrive at $x(t) = Ce^{-at}$. It is a fundamental tool in your physics and math toolkit, so keep it handy!

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, $a, b \in R$. Given that $x(0) = 2; y(0) = 1$ and $3y(1) = 2x(1)$, the value of $t$, for which $x(t) = y(t)$, is :

Options:

(A) $\log_{2/3} 2$

(B) $\log_4 3$

(C) $\log_3 4$

(D) $\log_{4/3} 2$

Sigma Percentile · Accepted Answer

### The Dance of Exponential Decay

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a differential equation; we are witnessing a beautiful dance between two variables, $x(t)$ and $y(t)$.

These variables are governed by the laws of change, specifically the first-order linear differential equations:

$$
\frac{dx}{dt} + ax = 0 \quad 	ext{and} \quad \frac{dy}{dt} + by = 0
$$

These equations describe systems that are constantly fading away, losing their magnitude at a rate proportional to their current value. It is the classic signature of exponential decay.

### Unveiling the Functions

Let us start by solving for $x(t)$. We begin with the equation $\frac{dx}{dt} = -ax$.

By separating the variables, we bring the $x$ terms to one side and the $t$ terms to the other:

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

Integrating both sides is a moment of pure elegance: the integral of $\frac{1}{x}$ is $\ln x$, and the integral of $-a$ with respect to $t$ is $-at$. Adding our constant of integration, $C_1$, we get $\ln x = -at + C_1$.

Exponentiating both sides, we find the general solution:

$$
x(t) = C_1 e^{-at}
$$

We are given the initial condition $x(0) = 2$. Substituting $t=0$ into our general solution, we get $2 = C_1 e^0$, which simplifies beautifully to $C_1 = 2$.

Thus, our specific function is $x(t) = 2e^{-at}$. We apply the exact same logic to $y(t)$. With $y(0) = 1$, we find:

$$
y(t) = e^{-bt}
$$

### The Bridge of Constraints

We are given a crucial piece of information: $3y(1) = 2x(1)$. This is our bridge, connecting the two separate worlds of $x$ and $y$ at the specific moment $t=1$.

Let us substitute our functions into this constraint:

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

Now, we need to find the relationship between $a$ and $b$. By rearranging the terms, we get:

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

Using the laws of exponents, we know that $\frac{e^{-b}}{e^{-a}} = e^{-b - (-a)} = e^{a-b}$. So, we have discovered the hidden key to the entire problem:

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

### The Grand Intersection

Finally, we ask: at what time $t$ do these two curves intersect? Geometrically, this means $x(t) = y(t)$.

Setting our functions equal, we have:

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

To solve for $t$, we divide both sides by $e^{-at}$, yielding $2 = \frac{e^{-bt}}{e^{-at}}$. Again, using exponent rules, this becomes:

$$
2 = e^{(a-b)t}
$$

We can rewrite $e^{(a-b)t}$ as $(e^{a-b})^t$. Since we already know that $e^{a-b} = \frac{4}{3}$, our equation becomes:

$$
2 = \left( \frac{4}{3} ight)^t
$$

To isolate $t$, we take the logarithm base $\frac{4}{3}$ on both sides. The final result is:

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

This is the exact moment when the two paths cross. You have navigated the differential equations, applied the initial conditions, bridged the gap with the constraint, and found the intersection. **Well done!**

Select Answer:

Visualized Solution

Analyze the Differential Equations

Solve for $x (t)$

Apply Initial Condition for $x (t)$

Solve for $y (t)$

Apply Initial Condition for $y (t)$

Use the Constraint at $t = 1$

Find the Ratio $e^{a - b}$

Equate $x (t)$ and $y (t)$

Isolate the Exponential Term

Substitute and Solve for $t$

Final Conclusion

The Sigma Insight: Variable Separable Method

The Dance of Exponential Decay

Unveiling the Functions

The Bridge of Constraints

The Grand Intersection

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Select Answer:

Visualized Solution

Analyze the Differential Equations

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