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x log x dy/dx + y = f(x) log x: find y at a point

The same question, asked 2 times: 29 Jan 2023*, 6 Apr 2024. Every wording below, oldest first.

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  1. Question 1

    29 Jan 2023

    165
    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,(x > 1). If y(2)=2y(2) = 2, then y(e)y(e) is equal to

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  2. Question 2

    6 Apr 2024

    78
    Let y=y(x)y = y(x) be the solution of the differential equation (2xlog⁡ex)dydx+2y=3xlog⁡ex,x>0\left( 2x\log_{e}x \right)\frac{dy}{dx}+ 2y =\frac{3}{x}\log_{e}x,x > 0 and y(e−1)=0y\left( e^{- 1} \right)= 0. Then, y(e)y(e) is equal to

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