Mathematics · 2019
JEE Main · 12 January 2019, Shift 1 · Q77
Let y=y(x) be the solution of the differential equation, x (d y)/(d x)+y=x log_e x,(x>1). If 2 y(2)= log_e 4-1, then y(e) is equal to:
Let $\displaystyle y=y(x)$ be the solution of the differential equation, $\displaystyle x \frac{d y}{d x}+y=x \log _e x,(x>1)$. If $\displaystyle 2 y(2)=\log _e 4-1$, then $\displaystyle y(e)$ is equal to :
Official answer
(1)
$\displaystyle \frac{e}{4}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.