Given differential equation is
$\left(y^{2}+2 x\right) \frac{d y}{d x}=y
\Rightarrow \frac{d x}{d y}=\left(\frac{y^{2}+2 x}{y}\right)$
$\Rightarrow \frac{d x}{d y}=y+\frac{2 x}{y}$
$\Rightarrow \frac{d x}{d y}-\frac{2}{y} \cdot x=y$
$IF =e^{\int \frac{2}{y} d y}=e^{-2 \log y}=y^{-2}=\frac{1}{y^{2}}$
- Complete solution is
$ x \cdot \frac{1}{y^{2}} =\int y \cdot \frac{1}{y^{2}} d y+C$
$\Rightarrow \frac{x}{y^{2}} =\int \frac{d y}{y}+C=\log _{e} y+C $
$\Rightarrow x =y^{2} \log _{e} y+c y^{2}$
$A t+x=1, y=1$ then from Eq. (i), we get
$1=0+C \Rightarrow C=1$
$\therefore $ From Eq. (i), we get
$x=y^{2}\left(\log _{e} y+1\right)$
Which is the required solution.