JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $y = y(t)$ be a solution of the differential equation $\frac{dy}{dt} + \alpha y = \gamma e^{-\beta t}$ where, $\alpha > 0, \beta > 0$ and $\gamma > 0$ . Then $\lim_{t \to \infty} y(t)$

Choose the correct answer:

Explanation

Solving

1. Identify the Equation Type

This is a Linear Differential Equation of the form $\frac{dy}{dt} + Py = Q$ .

$P = \alpha$
$Q = \gamma e^{-\beta t}$

2. Integrating Factor ( $IF$ )

IF = e^{\int \alpha \, dt} = e^{\alpha t}

3. General Solution

y(t) \cdot e^{\alpha t} = \int (\gamma e^{-\beta t} \cdot e^{\alpha t}) \, dt + C

y(t) \cdot e^{\alpha t} = \gamma \int e^{(\alpha - \beta)t} \, dt + C

4. Evaluate for $\alpha \neq \beta$

y(t) \cdot e^{\alpha t} = \gamma \left[ \frac{e^{(\alpha - \beta)t}}{\alpha - \beta} \right] + C

Divide by $e^{\alpha t}$ :

y(t) = \frac{\gamma e^{-\beta t}}{\alpha - \beta} + Ce^{-\alpha t}

5. Apply Limit ( $t \to \infty$ )

Given $\alpha > 0$ and $\beta > 0$ :

\lim_{t \to \infty} y(t) = \lim_{t \to \infty} \left( \frac{\gamma}{(\alpha - \beta)e^{\beta t}} + \frac{C}{e^{\alpha t}} \right)

= 0 + 0 = 0

6. Evaluate for $\alpha = \beta$

y(t) \cdot e^{\alpha t} = \int \gamma \, dt + C = \gamma t + C

y(t) = \frac{\gamma t}{e^{\alpha t}} + \frac{C}{e^{\alpha t}}

Using L'Hôpital's Rule for the first term:

\lim_{t \to \infty} \frac{\gamma t}{e^{\alpha t}} = \lim_{t \to \infty} \frac{\gamma}{\alpha e^{\alpha t}} = 0

Thus, $\lim_{t \to \infty} y(t) = 0$ .

Correct Option: (4)

Choose the correct answer:

Explore More