Mathematics · 2024
JEE Main · 6 April 2024, Shift 1 · Q13
Let y=y(x) be the solution of the differential equation (1+x^2) (d y)/(d x)+y=e^tan^-1 x, y(1)=0. Then y(0) is
Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$, $\displaystyle y(1)=0$. Then $\displaystyle y(0)$ is
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \frac{1}{2}\left(1-e^{\pi / 2}\right)$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.