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JEE Advanced 2025
PAPER-2 2025
Question
Let $y(x)$ be the solution of the differential equation
$ x^2 \frac{d y}{d x}+x y=x^2+y^2, \quad x>\frac{1}{e} $
satisfying $y(1)=0$. Then the value of $2 \frac{(y(e))^2}{y\left(e^2\right)}$ is $\_\_\_\_$ "
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Question Tags
JEE Advance
Mathematics
Easy
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