Mathematics · Differential Equations

JEE Main 2026 — 24 January, Evening Shift — Question 16

Let y=y(x)\mathrm{y}=\mathrm{y}(\mathrm{x}) be a differentiable function in the interval (0,∞)(0, \infty) such that y(1)=2\mathrm{y}(1)=2. and lim⁡t→x(t2y(x)−x2y(t)x−t)=3\lim _{\mathrm{t} \rightarrow \mathrm{x}}\left(\frac{\mathrm{t}^{2} \mathrm{y}(\mathrm{x})-\mathrm{x}^{2} \mathrm{y}(\mathrm{t})}{\mathrm{x}-\mathrm{t}}\right)=3 for each x>0\mathrm{x}>0. Then 2y(2)2 \mathrm{y}(2) is equal to

  1. Option A:

    18

  2. Option B:

    23

    Correct
  3. Option C:

    27

  4. Option D:

    12

Answer: B

Step-by-step solution

lim⁡t→x2tf(x)−x2f′(t)−1=3\lim _{\mathrm{t} \rightarrow \mathrm{x}} \frac{2 \mathrm{t} \mathrm{f}(\mathrm{x})-\mathrm{x}^{2} \mathrm{f}^{\prime}(\mathrm{t})}{-1}=3

x2f′(x)−2xf(x)=3x^{2} f^{\prime}(x)-2 x f(x)=3

dydx−2yx=3x2\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{2 \mathrm{y}}{\mathrm{x}}=\frac{3}{\mathrm{x}^{2}}

I.F. =e−∫2xdx=e−2log⁡ex=1/x2=\mathrm{e}^{-\int \frac{2}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-2 \log _{\mathrm{e}} \mathrm{x}}=1 / \mathrm{x}^{2}

y⋅1x2=∫3x4dx\mathrm{y} \cdot \frac{1}{\mathrm{x}^{2}}=\int \frac{3}{\mathrm{x}^{4}} \mathrm{dx}

yx2=−1x3+c⇒y=cx2−1x=f(x)\frac{y}{x^{2}}=-\frac{1}{x^{3}}+c \Rightarrow y=c x^{2}-\frac{1}{x}=f(x)

f(1)=2=c−1⇒c=3\mathrm{f}(1)=2=\mathrm{c}-1 \Rightarrow \mathrm{c}=3

f(x)=3x2−1xf(x)=3 x^{2}-\frac{1}{x}

f(2)=12−12⇒2f(2)=23f(2)=12-\frac{1}{2} \Rightarrow 2 f(2)=23

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Mathematics
Chapter
Differential Equations
Topic
Introduction to Differential Equations
Let y = y ( x ) be a differentiable function in the interval (0, ∞)… | JEE Main 2026 PYQ with Solution · DhiX AI