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

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(D) Given the linear differential equation: $x \log x \frac{dy}{dx} + y = \log x^2$.Dividing by $x \log x$, we get: $\frac{dy}{dx} + \frac{y}{x \log x

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$\frac{3}{2}$

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(D) Given the linear differential equation: $x \log x \frac{dy}{dx} + y = \log x^2$.Dividing by $x \log x$, we get: $\frac{dy}{dx} + \frac{y}{x \log x} = \frac{2 \log x}{x \log x} = \frac{2}{x}$.This is of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x \log x}$ and $Q = \frac{2}{x}$.The Integrating Factor $(IF)$ is $e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$.The general solution is $y \cdot IF = \int (Q \cdot IF) dx + c$.$y \log x = \int \left(\frac{2}{x} \cdot \log x\right) dx + c$.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 $y(e) = 0$, substitute $x = e$ and $y = 0$:$0 \cdot \log e = (\log e)^2 + c \Rightarrow 0 = 1 + c \Rightarrow c = -1$.Thus, $y \log x = (\log x)^2 - 1$.For $x = e^2$, $y \log(e^2) = (\log e^2)^2 - 1$.$y(2) = (2)^2 - 1 = 4 - 1 = 3$.$2y = 3 \Rightarrow y = \frac{3}{2}$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

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

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