Mathematics · 2023
JEE Main · 29 January 2023, Shift 2 · Q70
Let y=y(x) be the solution of the differential equation x log_e x (d y)/(d x)+y=x^2 log_e x,(x>1). If y(2)=2, then y(e) is equal to
Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle x \log _e x \frac{d y}{d x}+y=x^2 \log _e x,(x>1)$.
If $\displaystyle y(2)=2$, then $\displaystyle y(e)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{4+e^2}{4}$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.