If , the integral becomes , so . By L'Hôpital's rule, as since , giving the exact same limit of 0.

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Differential Equations: Mathematics | JEE Main

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)

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Ais

-1

Bis 1Cdoes not existDis 0

Step-by-step solutionView

Correct answer

For positive constants $\alpha, \beta, \gamma$ , 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 $\gamma e^{-\beta t}$ 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 $y(t) = C e^{-\alpha t} + \frac{\gamma}{\alpha - \beta} e^{-\beta t}$ (for $\alpha \neq \beta$ ) and note that both terms contain decaying exponentials with positive rates $\alpha$ and $\beta$ , 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 $\alpha = \beta$ , the secular term $t e^{-\alpha t}$ diverges as $t \to \infty$ , leading to a nonexistent limit. Recall that $\lim_{t \to \infty} t e^{-\alpha t} = 0$ for any $\alpha > 0$ by L'Hôpital's Rule, so the limit exists and equals 0 in all cases.

D · is 0

None. The solution is correct. Both the homogeneous solution $C e^{-\alpha t}$ and the particular solution decay to zero as $t \to \infty$ because $\alpha > 0$ and $\beta > 0$ .

Reviewed route

Solution

StepWorking

01given

The differential equation is $\frac{dy}{dt} + \alpha y = \gamma e^{-\beta t}$ with constants $\alpha > 0, \beta > 0, \gamma > 0$ .

02goal

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

03approach

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

04execute

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

✓verify

Since $\alpha > 0$ and $\beta > 0$ , both $e^{-\beta t} \to 0$ and $e^{-\alpha t} \to 0$ as $t \to \infty$ . Thus, $\lim_{t \to \infty} y(t) = 0 + 0 = 0$ . In the edge case $\alpha = \beta$ , $y(t) = \gamma t e^{-\alpha t} + C e^{-\alpha t}$ , 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 $\frac{dy}{dt} + \alpha y = \gamma e^{-\beta t}$ ?
$I(t) = e^{\alpha t}$
Assuming $\alpha \neq \beta$ , what is the general solution $y(t)$ obtained by multiplying by the integrating factor and integrating?
$y(t) = C e^{-\alpha t} + \frac{\gamma}{\alpha - \beta} e^{-\beta t}$
Given $\alpha > 0$ and $\beta > 0$ , what is the limit of each term in $y(t) = C e^{-\alpha t} + \frac{\gamma}{\alpha - \beta} e^{-\beta t}$ as $t \to \infty$ ?
Both $e^{-\alpha t} \to 0$ and $e^{-\beta t} \to 0$ , so $\lim_{t \to \infty} y(t) = 0$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

What happens if

\alpha = \beta

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

Answer

For positive constants $\alpha, \beta, \gamma$ , 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 $\gamma e^{-\beta t}$ 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 $y(t) = C e^{-\alpha t} + \frac{\gamma}{\alpha - \beta} e^{-\beta t}$ (for $\alpha \neq \beta$ ) and note that both terms contain decaying exponentials with positive rates $\alpha$ and $\beta$ , 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 $\alpha = \beta$ , the secular term $t e^{-\alpha t}$ diverges as $t \to \infty$ , leading to a nonexistent limit. Recall that $\lim_{t \to \infty} t e^{-\alpha t} = 0$ for any $\alpha > 0$ by L'Hôpital's Rule, so the limit exists and equals 0 in all cases.
D · correct: is 0 - None. The solution is correct. Both the homogeneous solution $C e^{-\alpha t}$ and the particular solution decay to zero as $t \to \infty$ because $\alpha > 0$ and $\beta > 0$ .

Step-by-step solution

given: The differential equation is $y$ with constants $\alpha > 0, \beta > 0, \gamma > 0$ .
goal: Determine the value of $\lim_{t \to \infty} y(t)$ .
approach: Recognize this as a first-order linear differential equation $\frac{dy}{dt} + P(t)y = Q(t)$ . Find the integrating factor $\text{I.F.} = e^{\int \alpha dt} = e^{\alpha t}$ , solve for $0$ , and evaluate the limit as $t \to \infty$ .
execute: Multiply across by the integrating factor: $y(t) e^{\alpha t} = \int \gamma e^{(\alpha - \beta)t} dt$ . Assuming $\alpha \neq \beta$ , we integrate to get $y(t) e^{\alpha t} = \frac{\gamma}{\alpha - \beta} e^{(\alpha - \beta)t} + C$ . Dividing by $e^{\alpha t}$ gives $y(t) = \frac{\gamma}{\alpha - \beta} e^{-\beta t} + C e^{-\alpha t}$ .
verify: Since $\alpha > 0$ and $\beta > 0$ , both $e^{-\beta t} \to 0$ and $e^{-\alpha t} \to 0$ as $t \to \infty$ . Thus, $\lim_{t \to \infty} y(t) = 0 + 0 = 0$ . In the edge case $\alpha = \beta$ , $y(t) = \gamma t e^{-\alpha t} + C e^{-\alpha t}$ , which also approaches $0$ as $t \to \infty$ .

Differential Equations: Mathematics | JEE Main

What feels right?

For positive constants $\alpha, \beta, \gamma$ , 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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