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(C) The given differential equation is $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$.Dividing by $(2x \ln x)$,we get $\frac{dy}{dx} + \frac{y}{x

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$-\frac{3}{e}$

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(C) The given differential equation is $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$.Dividing by $(2x \ln x)$,we get $\frac{dy}{dx} + \frac{y}{x \ln x} = \frac{3}{2x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \ln x}$ and $Q = \frac{3}{2x^2}$.The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{1}{x \ln x} dx} = e^{\ln(\ln x)} = \ln x$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y \ln x = \int \frac{3}{2x^2} \ln x dx$.Using integration by parts,$\int \ln x \cdot (\frac{3}{2} x^{-2}) dx = \ln x \cdot (-\frac{3}{2x}) - \int \frac{1}{x} (-\frac{3}{2x}) dx = -\frac{3 \ln x}{2x} + \frac{3}{2} \int x^{-2} dx = -\frac{3 \ln x}{2x} - \frac{3}{2x} + C$.Given $y(e^{-1}) = 0$,so $0 \cdot \ln(e^{-1}) = -\frac{3 \ln(e^{-1})}{2e^{-1}} - \frac{3}{2e^{-1}} + C$.$0 = -\frac{3(-1)}{2e^{-1}} - \frac{3}{2e^{-1}} + C \Rightarrow 0 = \frac{3e}{2} - \frac{3e}{2} + C \Rightarrow C = 0$.Thus,$y \ln x = -\frac{3 \ln x}{2x} - \frac{3}{2x} \Rightarrow y = -\frac{3}{2x} - \frac{3}{2x \ln x}$.At $x = e$,$y(e) = -\frac{3}{2e} - \frac{3}{2e \ln e} = -\frac{3}{2e} - \frac{3}{2e} = -\frac{6}{2e} = -\frac{3}{e}$.

Let $y=y(x)$ be the solution of the differential equation $(2x \ln x) \frac{dy}{dx} + 2y = \frac{3}{x} \ln x$,$x > 0$ and $y(e^{-1}) = 0$. Then,$y(e)$ is equal to

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