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(B) Given the differential equation: $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$.Rearranging the terms, we get: $\frac{1 - y \log x}{y^2} dy = \frac{

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$y = x^y$

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(B) Given the differential equation: $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$.Rearranging the terms, we get: $\frac{1 - y \log x}{y^2} dy = \frac{1}{x} dx$.Integrating both sides: $\int (y^{-2} - \frac{\log x}{y}) dy = \int \frac{1}{x} dx$ is not direct. Let us check the options.If $y = x^y$, then taking $\log$ on both sides: $\log y = y \log x$.Differentiating w.r.t. $x$: $\frac{1}{y} \frac{dy}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}$.Rearranging for $\frac{dy}{dx}$: $\frac{dy}{dx} (\frac{1}{y} - \log x) = \frac{y}{x}$.$\frac{dy}{dx} (\frac{1 - y \log x}{y}) = \frac{y}{x}$.$x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$.This matches the given differential equation. Also, for $x=1$, $y=1^y=1$, which satisfies the condition $y(1)=1$.

Let $y$ be the solution of the differential equation $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$ satisfying $y(1) = 1$. Then, $y$ satisfies:

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