Question

For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

xy=logy +C and $y^{\prime }=\frac{y}{1-xy}(xy\ne 1).$

Answer 1

Given, xy=logy+C
On differentiating both sides w.r.t. x, we get
$\frac{d}{dx}\left(xy\right)=\frac{d}{dx}(logy+C)\Rightarrow y\frac{d}{dx}x+x{y}^{\prime }=\frac{1}{y}{y}^{\prime }\Rightarrow x{y}^{\prime }+y=\frac{1}{y}{y}^{\prime }\Rightarrow x{y}^{\prime }y+y^2=y^{\prime }...\left(i\right)$
But we have to verify, $y^{\prime }=\frac{y}{1-xy}$
From Eq. (i), $y^2=y^{\prime }(1-xy)\Rightarrow y^{\prime }=\frac{y^2}{1-xy}Henceproved.$
Hence, xy=logy+C is a solution of the given differential equation.