Mathematics · 2024
JEE Main · 4 April 2024, Shift 2 · Q12
Let y=y(x) be the solution of the differential equation (x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0. If y(0)=0, then y(2) is equal to
Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \left(x^2+4\right)^2 d y+\left(2 x^3 y+8 x y-2\right) d x=0$. If $\displaystyle y(0)=0$, then $\displaystyle y(2)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{\pi}{32}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.