Let x = x ( y ) be the solution of the differential equation 2 y 2 frac d x d y - 2 x y + x 2 = 0 , y 1 , x ( e ) = e . Then x ( e 2) is equal to:

Mathematics · Differential Equations

JEE Main 2026 — 2 April, Evening Shift — Question 42

Let $\mathbf{x} = \mathbf{x}(\mathbf{y})$ be the solution of the differential equation $2\mathbf{y}^2 \frac{\mathrm{d}\mathbf{x}}{\mathrm{d}\mathbf{y}} - 2\mathbf{x}\mathbf{y} + \mathbf{x}^2 = 0$ , $\mathbf{y} > 1$ , $\mathbf{x}(\mathbf{e}) = \mathbf{e}$ . Then $\mathbf{x}(\mathbf{e}^2)$ is equal to:

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

Answer: B

Step-by-step solution

$2 y(y d x-x d y)+x^{2} d y=0$ $\Rightarrow-2 \mathrm{yx}^{2} \mathrm{~d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{x}^{2} \mathrm{dy}=0$ $\Rightarrow-2 y d\left(\frac{y}{x}\right)+d y=0$ $\Rightarrow-2 \mathrm{~d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\frac{1}{\mathrm{y}} \mathrm{dy}=0$ $\Rightarrow \frac{-2 \mathrm{y}}{\mathrm{x}}+\log _{\mathrm{e}} \mathrm{y}=\mathrm{C}$ Given $\mathrm{x}(\mathrm{e})=\mathrm{e}$ $\Rightarrow \mathrm{C}=-1$ $\therefore \frac{-2 y}{x}+\log _{e} y=-1$ $\Rightarrow \frac{2 \mathrm{y}}{\mathrm{x}}-\log _{\mathrm{e}} \mathrm{y}=1$ Put $y=e^{2}$ ,

we get, $\Rightarrow \frac{2 \mathrm{e}^{2}}{\mathrm{x}}-2=1 \Rightarrow \frac{2 \mathrm{e}^{2}}{\mathrm{x}}=3 \Rightarrow \mathrm{x}=\frac{2 \mathrm{e}^{2}}{3}$ $\Rightarrow \mathrm{x}\left(\mathrm{e}^{2}\right)=\frac{2 \mathrm{e}^{2}}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Differential Equations
Topic: Methods of solving a First Order,First Degree Differential