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(A) The given differential equation is $x \ln x \frac{dy}{dx} + y = x^2 \ln x$.Dividing by $x \ln x$,we get $\frac{dy}{dx} + \frac{1}{x \ln x} y = x$.

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

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(A) The given differential equation is $x \ln x \frac{dy}{dx} + y = x^2 \ln x$.Dividing by $x \ln x$,we get $\frac{dy}{dx} + \frac{1}{x \ln x} y = x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x \ln x}$ and $Q(x) = x$.The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{1}{x \ln x} dx} = e^{\ln(\ln x)} = \ln x$.The solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y \ln x = \int x \ln x dx + C$.Using integration by parts,$\int x \ln x dx = \ln x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$.So,$y \ln x = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$.Given $y(2) = 2$,we have $2 \ln 2 = \frac{4}{2} \ln 2 - \frac{4}{4} + C \Rightarrow 2 \ln 2 = 2 \ln 2 - 1 + C \Rightarrow C = 1$.Thus,$y \ln x = \frac{x^2}{2} \ln x - \frac{x^2}{4} + 1$.For $x = e$,$y \ln e = \frac{e^2}{2} \ln e - \frac{e^2}{4} + 1$.Since $\ln e = 1$,$y(e) = \frac{e^2}{2} - \frac{e^2}{4} + 1 = \frac{e^2}{4} + 1 = \frac{e^2 + 4}{4}$.

Let $y=y(x)$ be the solution of the differential equation $x \log _e x \frac{d y}{d x}+y=x^2 \log _e x, (x > 1)$. If $y(2)=2$,then $y(e)$ is equal to

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