differential equation...

\hspace{-16}$Let $\bf{\frac{dy}{dx}=p\Leftrightarrow dy=p.dx.........(1)}$\\\\\\ Now our equation is Convert into $\bf{y=p.x+\frac{1}{p}}$\\\\\\ Now Diff. both side w.r.to $\bf{x}$\\\\\\ $\bf{dy = p.dx+x.dp-\frac{dp}{p^2}}$\\\\\\ Using $\bf{(1)}$ Relation......\\\\\\ $\bf{x.dp-\frac{dp}{p^2}=0\Leftrightarrow dp\left(x-\frac{1}{p^2}\right)=0}$\\\\\\ So either $\bf{dp=0\Leftrightarrow p=\mathbb{C}.}$\\\\\\ So Solution of Diff. eqn. is $\bf{y=x.\mathbb{C}+\frac{1}{\mathbb{C}}}$\\\\\\ or $\bf{x-\frac{1}{p^2}=0}$. put this value in $\bf{y=p.x+\frac{1}{p}}$\\\\\\ We Get $\bf{y=\frac{2}{p}\Leftrightarrow y.dy = 2.dx}$\\\\\\ $\bf{\frac{y^2}{2}=2x+\mathbb{C^{'}}\Leftrightarrow y^2=4x+\mathbb{C}}$