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Question:

$x y=\log y+\mathrm{C} \quad: y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)$

Solution:

$x y=\log y+\mathrm{C}$

Differentiating both sides of this equation with respect to x, we get:

$\frac{d}{d x}(x y)=\frac{d}{d x}(\log y)$

$\Rightarrow y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}$

$\Rightarrow y+x y^{\prime}=\frac{1}{y} y^{\prime}$

$\Rightarrow y^{2}+x y y^{\prime}=y^{\prime}$

$\Rightarrow(x y-1) y^{\prime}=-y^{2}$

$\Rightarrow y^{\prime}=\frac{y^{2}}{1-x y}$

L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.