JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Asymptotic behavior of a first-order linear differential equation

Q: What happens if α = β?

If α = β, the integral becomes integral of γ dt = γ t + C, so y(t) = (γ t + C)e^(-α t). By L'Hôpital's rule, t e^(-α t) → 0 as t → ∞ since α > 0, giving the exact same limit of 0.

Let y = y(t) be a solution of the differential equation dy/dt + α y = γ e^(-β t) where, α > 0, β > 0 and γ > 0. Then lim as t → ∞ of y(t)

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Ais -1Bis 1Cdoes not existDis 0

Step-by-step solutionView

Correct answer

For positive constants $α, β, γ$ , both the transient complementary solution and the particular solution decay exponentially to zero as $t \to \infty$ , so $\lim_{t \to \infty} y (t) = 0$ .

Option analysis

Why each option works or fails

A · is -1

Believing that evaluating the limit involves dividing $γ e^{- β t}$ by the coefficient of $y$ with an inverted sign, or confusing the limit with $- 1$ due to a sign error during integration. Solve for the general solution $y (t) = C e^{- α t} + \frac{γ}{α - β} e^{- β t}$ (for $α \neq β$ ) and note that both terms contain decaying exponentials with positive rates $α$ and $β$ , which approach 0 as $t \to \infty$ .

B · is 1

Confusing the asymptotic limit $\lim_{t \to \infty} y (t)$ with the normalized initial condition or misinterpreting the ratio of matching coefficients in the limit. Evaluate $\lim_{t \to \infty} e^{- k t}$ where $k > 0$ ; because exponential decay yields $0$ rather than $1$ , the entire function approaches $0$ .

C · does not exist

Assuming that when $α = β$ , the secular term $t e^{- α t}$ diverges as $t \to \infty$ , leading to a nonexistent limit. Recall that $\lim_{t \to \infty} t e^{- α t} = 0$ for any $α > 0$ by L'Hôpital's Rule, so the limit exists and equals 0 in all cases.

D · is 0

The solution is correct. Both the homogeneous solution $C e^{- α t}$ and the particular solution decay to zero as $t \to \infty$ because $α > 0$ and $β > 0$ .

Reviewed route

Solution

StepWorking

01Given

The differential equation is $\frac{d y}{d t} + α y = γ e^{- β t}$ with constants $α > 0, β > 0, γ > 0$ .

02Goal

Determine the value of $\lim_{t \to \infty} y (t)$ .

03Approach

Recognize this as a first-order linear differential equation $\frac{d y}{d t} + P (t) y = Q (t)$ . Find the integrating factor $I.F. = e^{\int α d t} = e^{α t}$ , solve for $y (t)$ , and evaluate the limit as $t \to \infty$ .

04Execute

Multiply across by the integrating factor: $y (t) e^{α t} = \int γ e^{(α - β) t} d t$ . Assuming $α \neq β$ , we integrate to get $y (t) e^{α t} = \frac{γ}{α - β} e^{(α - β) t} + C$ . Dividing by $e^{α t}$ gives $y (t) = \frac{γ}{α - β} e^{- β t} + C e^{- α t}$ .

✓Verify

Since $α > 0$ and $β > 0$ , both $e^{- β t} \to 0$ and $e^{- α t} \to 0$ as $t \to \infty$ . Thus, $\lim_{t \to \infty} y (t) = 0 + 0 = 0$ . In the edge case $α = β$ , $y (t) = γ t e^{- α t} + C e^{- α t}$ , which also approaches $0$ as $t \to \infty$ .

Hints that build this answer step by step

What is the integrating factor for the first-order linear differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ ?
$I (t) = e^{α t}$
Assuming $α \neq β$ , what is the general solution $y (t)$ obtained by multiplying by the integrating factor and integrating?
$y (t) = C e^{- α t} + \frac{γ}{α - β} e^{- β t}$
Given $α > 0$ and $β > 0$ , what is the limit of each term in $y (t) = C e^{- α t} + \frac{γ}{α - β} e^{- β t}$ as $t \to \infty$ ?
Both $e^{- α t} \to 0$ and $e^{- β t} \to 0$ , so $\lim_{t \to \infty} y (t) = 0$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

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What happens if

α = β

If $α = β$ , the integral becomes $\int γ d t = γ t + C$ , so $y (t) = (γ t + C) e^{- α t}$ . By L'Hôpital's rule, $t e^{- α t} \to 0$ as $t \to \infty$ since $α > 0$ , giving the exact same limit of 0.

Answer

For positive constants $α, β, γ$ , both the transient complementary solution and the particular solution decay exponentially to zero as $t \to \infty$ , so $\lim_{t \to \infty} y (t) = 0$ .

Why each option works or fails

A: is -1 - Believing that evaluating the limit involves dividing $γ e^{- β t}$ by the coefficient of $y$ with an inverted sign, or confusing the limit with $- 1$ due to a sign error during integration. Solve for the general solution $y (t) = C e^{- α t} + \frac{γ}{α - β} e^{- β t}$ (for $α \neq β$ ) and note that both terms contain decaying exponentials with positive rates $α$ and $β$ , which approach 0 as $t \to \infty$ .
B: is 1 - Confusing the asymptotic limit $\lim_{t \to \infty} y (t)$ with the normalized initial condition or misinterpreting the ratio of matching coefficients in the limit. Evaluate $\lim_{t \to \infty} e^{- k t}$ where $k > 0$ ; because exponential decay yields $0$ rather than $1$ , the entire function approaches $0$ .
C: does not exist - Assuming that when $α = β$ , the secular term $t e^{- α t}$ diverges as $t \to \infty$ , leading to a nonexistent limit. Recall that $\lim_{t \to \infty} t e^{- α t} = 0$ for any $α > 0$ by L'Hôpital's Rule, so the limit exists and equals 0 in all cases.
D · correct: is 0 - The solution is correct. Both the homogeneous solution $C e^{- α t}$ and the particular solution decay to zero as $t \to \infty$ because $α \neq β$ and $β > 0$ .

Step-by-step solution

Given: The differential equation is $\frac{d y}{d t} + α y = γ e^{- β t}$ with constants $α > 0, β > 0, γ > 0$ .
Goal: Determine the value of $\lim_{t \to \infty} y (t)$ .
Approach: Recognize this as a first-order linear differential equation $\frac{d y}{d t} + P (t) y = Q (t)$ . Find the integrating factor $I.F. = e^{\int α d t} = e^{α t}$ , solve for $y (t)$ , and evaluate the limit as $t \to \infty$ .
Execute: Multiply across by the integrating factor: $y (t) e^{α t} = \int γ e^{(α - β) t} d t$ . Assuming $α \neq β$ , we integrate to get $y (t) e^{α t} = \frac{γ}{α - β} e^{(α - β) t} + C$ . Dividing by $e^{α t}$ gives $y (t) = \frac{γ}{α - β} e^{- β t} + C e^{- α t}$ .
Verify: Since $α > 0$ and $β > 0$ , both $e^{- β t} \to 0$ and $e^{- α t} \to 0$ as $t \to \infty$ . Thus, $\lim_{t \to \infty} y (t) = 0 + 0 = 0$ . In the edge case $α = β$ , $y (t) = γ t e^{- α t} + C e^{- α t}$ , which also approaches $0$ as $t \to \infty$ .

Asymptotic behavior of a first-order linear differential equation

What feels right?

For positive constants $α, β, γ$ , both the transient complementary solution and the particular solution decay exponentially to zero as $t \to \infty$ , so $\lim_{t \to \infty} y (t) = 0$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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