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(D) Given the differential equation: $y(1+\log x) \frac{dx}{dy} = x \log x$. Rearranging the terms to separate the variables: $\frac{dy}{y} = \frac{1+

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$y=e x \log x$

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(D) Given the differential equation: $y(1+\log x) \frac{dx}{dy} = x \log x$. Rearranging the terms to separate the variables: $\frac{dy}{y} = \frac{1+\log x}{x \log x} dx$. Integrating both sides: $\int \frac{1}{y} dy = \int \frac{1+\log x}{x \log x} dx$. Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1+u}{u} du = \int (\frac{1}{u} + 1) du = \log |u| + u + C$. Substituting back: $\log |y| = \log |\log x| + \log x + C$. Given $x=e, y=e^2$: $\log |e^2| = \log |\log e| + \log e + C \Rightarrow 2 = \log(1) + 1 + C \Rightarrow 2 = 0 + 1 + C \Rightarrow C = 1$. Thus,$\log |y| = \log |\log x| + \log x + 1$. Since $1 = \log e$,we have $\log |y| = \log |\log x| + \log x + \log e = \log |e \log x \cdot x|$. Therefore,$y = ex \log x$.

The particular solution of the differential equation $y(1+\log x) \frac{dx}{dy} - x \log x = 0$ when $x=e, y=e^2$ is

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