Mathematics · 2026
JEE Main · 2 April 2026, Shift 2 · Q20
Let x=x(y) be the solution of the differential equation 2 y^2 (d x)/(d y)-2 x y+x^2=0, y>1, x( e )= e. Then x( e^2) is equal to:
Let $\displaystyle x=x(y)$ be the solution of the differential equation $\displaystyle 2 y^2 \frac{\mathrm{~d} x}{\mathrm{~d} y}-2 x y+x^2=0, y>1, x(\mathrm{e})=\mathrm{e}$. Then $\displaystyle x\left(\mathrm{e}^2\right)$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{2}{3} \mathrm{e}^2$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.