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(B) Given the differential equation: $y \log y \, dx - x \, dy = 0$ Rearranging the terms,we get: $y \log y \, dx = x \, dy$ Separating the variables,

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$y = e^{cx}$

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(B) Given the differential equation: $y \log y \, dx - x \, dy = 0$ Rearranging the terms,we get: $y \log y \, dx = x \, dy$ Separating the variables,we have: $\frac{dx}{x} = \frac{dy}{y \log y}$ Integrating both sides: $\int \frac{dx}{x} = \int \frac{dy}{y \log y}$ Let $u = \log y$,then $du = \frac{1}{y} \, dy$. The integral becomes: $\int \frac{dx}{x} = \int \frac{du}{u}$ $\log |x| = \log |u| + C_1$ $\log |x| = \log |\log y| + C_1$ Taking the exponential of both sides: $|x| = e^{C_1} |\log y|$ Let $e^{C_1} = k$,then $x = k \log y$ or $\log y = \frac{x}{k} = cx$ (where $c = 1/k$). Therefore,$y = e^{cx}$.

The general solution of the differential equation $y \log y \, dx - x \, dy = 0$ is . . . . . . .

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