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

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$\frac{e}{4}$

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(C) The given differential equation is $x\frac{dy}{dx} + y = x \ln x$. Dividing by $x$,we get $\frac{dy}{dx} + \frac{1}{x}y = \ln x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \ln x$. The integrating factor $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is $y \cdot IF = \int Q \cdot IF dx + C$. $y \cdot x = \int (\ln x) \cdot x dx + C$. 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,$xy = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$. Given $2y(2) = \ln 4 - 1$,we have $y(2) = \frac{\ln 4 - 1}{2}$. Substituting $x=2$ in the general solution: $2y(2) = \frac{2^2}{2} \ln 2 - \frac{2^2}{4} + C \implies \ln 4 - 1 = 2 \ln 2 - 1 + C$. Since $\ln 4 = 2 \ln 2$,we get $2 \ln 2 - 1 = 2 \ln 2 - 1 + C$,which implies $C = 0$. Thus,$y = \frac{x}{2} \ln x - \frac{x}{4}$. For $x = e$,$y(e) = \frac{e}{2} \ln e - \frac{e}{4} = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}$.

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

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