For the differential equation $xy\cfrac{{dy}}{{dx}} = (x + 2)(y + 2)$, find the solution curve passing through the point $\left( {1, - 1} \right)$.

.: We have $xy\cfrac{{dy}}{{dx}} = (x + 2)(y + 2)$ … (1)

Integrating (1) both sides,

we get
$\int {\cfrac{{ydy}}{{y + 2}}} = \int {\cfrac{{x + 2}}{x}} dx \Rightarrow \;\int {\cfrac{{y + 2 - 2}}{{y + 2}}} dy = \int {\left( {1 + \cfrac{2}{x}} \right)} dx$

$\Rightarrow \int {\left( {1 - \cfrac{2}{{y + 2}}} \right)} dy = x + 2\log x + C$

$\Rightarrow y - 2\log (y + 2) = x + 2\log x + C$

As point $(1,\; - 1)$ lies on it, i.e., $x = 1,y = - {1^ \cdot }$

$\therefore$ $= - 1 - 2\log 1 = 1 + 2\log 1 + C \Rightarrow C = - 2$

So, particular solution is:

$y - 2\log (y + 2) = x + 2\log x - 2$
$\Rightarrow y - x + 2 = \log [{x^2}{(y + 2)^2}]$