$2xy + {y^2} - 2{x^2}\cfrac{{dy}}{{dx}} = 0;y = 2$ when $x = 1$

: The given equation is $2xy + {y^2} - 2{x^2}\cfrac{{dy}}{{dx}} = 0$

$\Rightarrow 2{x^2}\cfrac{{dy}}{{dx}} = 2xy + {y^2} \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{y}{x} + \cfrac{1}{2}{\left( {\cfrac{y}{x}} \right)^2}$

…(1)
Put $y = vx \Rightarrow \cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

$\therefore$ (1) becomes $v + x\cfrac{{dv}}{{dx}} = v + \cfrac{1}{2}{v^2}$

$\Rightarrow x\cfrac{{dv}}{{dx}} = \cfrac{1}{2}{v^2} \Rightarrow \cfrac{2}{{{v^2}}}dv = \cfrac{{dx}}{x}$

Integrating both sides,
we get $2\int {{v^{ - 2}}} dv = \int {\cfrac{{dx}}{x}}$

$\Rightarrow \cfrac{{2{v^{ - 1}}}}{{ - 1}} = \log |x| + C$

$\Rightarrow \cfrac{{ - 2}}{v} = \log \left| x \right| + C \Rightarrow - 2\cfrac{x}{y} = \log |x| + C$ 2

…(2)
When $x = 1,y = 2,\,\,\,\therefore \,\,\, - 2 \cdot \cfrac{1}{2} = \log |1| + C \Rightarrow C = - 1$

Putting ${\rm{C}} = - 1$ in (2),

we get
$- \cfrac{{2x}}{y} = \log |x| - 1 \Rightarrow \cfrac{{2x}}{y} = 1 - \log |x|$

$\Rightarrow y = \cfrac{{2x}}{{1 - \log |x|}}(x \ne 0)$ ,

which is the required solution.