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(D) Given the differential equation: $y(1+\log x) = (\log x^x) \frac{dy}{dx}$.Since $\log x^x = x \log x$,the equation becomes $y(1+\log x) = (x \log

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$ex \log x - y = 0$

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(D) Given the differential equation: $y(1+\log x) = (\log x^x) \frac{dy}{dx}$.Since $\log x^x = x \log x$,the equation becomes $y(1+\log x) = (x \log x) \frac{dy}{dx}$.Separating the variables,we get: $\frac{1+\log x}{x \log x} dx = \frac{dy}{y}$.Integrating both sides: $\int \frac{1+\log x}{x \log x} dx = \int \frac{dy}{y}$.Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1+u}{u} du = \int (\frac{1}{u} + 1) du = \log |u| + u = \log |\log x| + \log x$.Thus,$\log |\log x| + \log x = \log |y| + C$.Using the condition $y(e) = e^2$: $\log |\log e| + \log e = \log |e^2| + C$.Since $\log e = 1$,we have $\log |1| + 1 = 2 + C$,which gives $0 + 1 = 2 + C$,so $C = -1$.Substituting $C$ back: $\log |\log x| + \log x = \log |y| - 1$.Since $1 = \log e$,we have $\log |\log x| + \log x + \log e = \log |y|$.Using $\log a + \log b = \log(ab)$,we get $\log |e \cdot x \log x| = \log |y|$.Therefore,$y = ex \log x$,or $ex \log x - y = 0$.

The particular solution of the differential equation $y(1+\log x) = (\log x^x) \frac{dy}{dx}$,given $y(e) = e^2$,is

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