Let x = x ( y ) be the solution of the differential equation y 2 d x+ (x-1/y ) d y=0 . If x(1)=1 , then x (1/2 ) is :

Mathematics · Differential Equations

JEE Main 2025 — 22 January, Morning Shift — Question 12

Let $\mathrm{x}=\mathrm{x}(\mathrm{y})$ be the solution of the differential equation $y^{2} d x+\left(x-\frac{1}{y}\right) d y=0$ . If $x(1)=1$ , then

$x\left(\frac{1}{2}\right)$ is :

Option A:
$\frac{1}{2}+e$
Option B:
$\frac{3}{2}+e$
Option C:
$3-\mathrm{e}$
Correct
Option D:
$3+\mathrm{e}$

Answer: C

Step-by-step solution

\frac{dx}{dy} + \left( \frac{1}{y^2} \right) x = \frac{1}{y^3}

I.F. = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}

\Rightarrow x e^{-\frac{1}{y}} = \int \left( e^{-\frac{1}{y}} \right) \frac{1}{y^3} dy

\text{Put } -\frac{1}{y} = t

\frac{1}{y^2} dy = dt

x e^{-\frac{1}{y}} = - \int t e^t dt

x e^{-\frac{1}{y}} = -te^t + e^t + C

x e^{-\frac{1}{y}} = \frac{1}{y} e^{-\frac{1}{y}} + e^{-\frac{1}{y}} + C

x = 1, y = 1

\frac{1}{e} = \frac{1}{e} + \frac{1}{e} + C

\Rightarrow C = -\frac{1}{e}

\text{Put } y = \frac{1}{2}

\frac{x}{e^2} = \frac{2}{e^2} + \frac{1}{e^2} - \frac{1}{e}

x = 3 - e

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Differential Equations
Topic: Methods of solving a First Order,First Degree Differential