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(D) The given differential equation is $x \frac{dy}{dx}+y=x \log x$. Dividing by $x$,we get $\frac{dy}{dx}+\frac{1}{x} y=\log x$. This is a linear dif

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$\frac{e}{4}$

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(D) The given differential equation is $x \frac{dy}{dx}+y=x \log x$. Dividing by $x$,we get $\frac{dy}{dx}+\frac{1}{x} y=\log x$. This is a linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$,where $P(x)=\frac{1}{x}$ and $Q(x)=\log x$. The integrating factor $(I.F.)$ is $e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$. The general solution is given by $y(I.F.) = \int Q(I.F.) dx + c$. Substituting the values,$xy = \int x \log x dx + c$. Using integration by parts,$\int x \log x dx = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + c$. So,$xy = \frac{x^2}{2} \log x - \frac{x^2}{4} + c$. Given $2(y(2)) = \log 4 - 1$,which implies $y(2) = \frac{1}{2} \log 4 - \frac{1}{2} = \log 2 - \frac{1}{2}$. Substituting $x=2$ in the general solution: $2(\log 2 - \frac{1}{2}) = \frac{4}{2} \log 2 - \frac{4}{4} + c$. $2 \log 2 - 1 = 2 \log 2 - 1 + c$,which gives $c=0$. Thus,$y = \frac{x}{2} \log x - \frac{x}{4}$. For $x=e$,$y(e) = \frac{e}{2} \log e - \frac{e}{4} = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}$.

Let $y=y(x)$ be the solution of the differential equation $x \frac{dy}{dx}+y=x \log x, (x > 1)$. If $2(y(2))=\log 4-1$,then the value of $y(e)$ is:

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