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MCQMediumJEE Main 2023 · 29 January, Shift 2Linear Differential Equations

Mathematics Question from JEE Main 2023 · 29 January, Shift 2

Let y=y(x)y = y(x) be the solution of the differential equation

xlog⁡exdydx+y=x2log⁡ex,(x>1).x \log_e x \frac{dy}{dx} + y = x^2 \log_e x, \quad (x > 1).

If y(2)=2y(2) = 2, then y(e)y(e) is equal to:

  • A

    4+e24\frac{4 + e^2}{4}

  • B

    1+e24\frac{1 + e^2}{4}

  • C

    2+e22\frac{2 + e^2}{2}

  • D

    1+e22\frac{1 + e^2}{2}

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