Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$xy=\log y+C:y^{\prime }=\frac{y^2}{1-xy}(xy\ne 1)$

Answer 1

Let $xy=\log y+C$
Differentiating both sides w.r.t. $x$ we get,
$\frac{d\left(xy\right)}{dx}=\frac{d}{dx}\left[\log \right(y)+C]$
$\frac{d\left(x\right)}{dx}.y+x\frac{d\left(y\right)}{dx}=\frac{d\left[\log y\right]}{dx}+\frac{dC}{dx}$
$1.y+x\frac{dy}{dx}=\frac{1}{y}.\frac{dy}{dx}+0$
$y+x\frac{dy}{dx}=\frac{1}{y}.\frac{dy}{dx}$
$y=\frac{1}{y}.\frac{dy}{dx}-x.\frac{dy}{dx}$
$y=\frac{dy}{dx}[\frac{1}{y}-x]$
$y=\frac{dy}{dx}\left[\frac{1-xy}{y}\right]$
$\frac{dy}{dx}=\frac{y^2}{1-xy}$
$\frac{dy}{dx}=y^{\prime }$
$y^{\prime }=\frac{y^2}{1-xy}$
Thus $LHS=RHS$
Hece verified.

Final answer:
Hence, the function $xy=\log y+C$ is a solution of the differential equation $y^{\prime }=\frac{y^2}{1-xy}$.