Mathematics · 2020
JEE Main · 3 September 2020, Shift 2 · Q62
If x^3 d y+x y d x=x^2 d y+2 y d x; y(2)= e and x>1, then y(4) is equal to:
If $\displaystyle x^3 \mathrm{~d} y+x y \mathrm{~d} x=x^2 \mathrm{~d} y+2 y \mathrm{~d} x ; y(2)=\mathrm{e}$ and $\displaystyle x>1$, then $\displaystyle y(4)$ is equal to :
Official answer
(4)
$\displaystyle \frac{3}{2} \sqrt{e}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.