If x log x frac{dy}{dx} + y = log x^2 and y(e | vedclass

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

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$\frac{3}{2}$

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(D) The given differential equation is $x \log x \frac{dy}{dx} + y = 2 \log x$. Dividing by $x \log x$,we get $\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$. The integrating factor $IF = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$. The general solution is $y \cdot IF = \int Q \cdot IF dx + C$. $y \log x = \int \frac{2}{x} \cdot \log x dx + C$. Let $u = \log x$,then $du = \frac{1}{x} dx$. $y \log x = \int 2u du + C = u^2 + C = (\log x)^2 + C$. Given $y(e) = 0$,we have $0 \cdot \log e = (\log e)^2 + C$,so $0 = 1 + C$,which means $C = -1$. Thus,$y \log x = (\log x)^2 - 1$. For $x = e^2$,$y \log(e^2) = (\log e^2)^2 - 1$. $y(2) = (2)^2 - 1 = 4 - 1 = 3$. $2y = 3$,so $y = \frac{3}{2}$.

If $x \log x \frac{dy}{dx} + y = \log x^2$ and $y(e) = 0$,then $y(e^2) = $

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