Question

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

Answer 1

xy= log y+C
Differentiating both sides of this equation with respect to x, we get:
$\frac{d}{dx}(xy)=\frac{d}{dx}\,(log \,y)$

$\Rightarrow y.\frac{d}{dx}(x)+x.\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}$

$\Rightarrow y+xy'=\frac{1}{yy'}$

$\Rightarrow y^2+xyy'=y'$

$\Rightarrow y'=\frac{y^2}{1-xy}$

∴ L.H.S. = R.H.S.

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