Mathematics · 2021
JEE Main · 26 August 2021, Shift 2 · Q68
Let y(x) be the solution of the differential equation 2 x^2 d y+( e^y-2 x) d x=0, x>0. If y( e )=1, then y(1) is equal to:
Let $\displaystyle y(x)$ be the solution of the differential equation $\displaystyle 2 x^2 \mathrm{~d} y+\left(\mathrm{e}^y-2 x\right) \mathrm{d} x=0, x>0$. If $\displaystyle y(\mathrm{e})=1$, then $\displaystyle y(1)$ is equal to :
Official answer
(3)
$\displaystyle \log _{\mathrm{e}} 2$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.