$\text{ The solution of }\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}\text{ satisfying y}\left(1\right)=1\text{ is given by }$

$$\begin{gathered} \frac{dy}{dx}=\frac{x^2+y^2+1}{2xy} \\ 2xy dy=\left(x^2+y^2+1\right)dx \end{gathered}$$

$2xydy-y^{2}dx=\left(x^2+1\right)dx$

$\int \frac{{\text{2xy}}^{\text{dy}}{\text{-y}}^{\text{2}}\text{dx}}{{\text{x}}^{\text{2}}}=\int \left(1+\frac{1}{{\text{x}}^2}\right)\text{dx}$

$\begin{array}{l}\to \int \mathrm{d}\left({\mathrm{y}}^2/\mathrm{x}\right)=\mathrm{x}-\frac{1}{\mathrm{x}}+\mathrm{c}\\ \to \frac{{\mathrm{y}}^2}{\mathrm{x}}=\mathrm{x}-\frac{1}{\mathrm{x}}+\mathrm{c}\\ \to {\mathrm{y}}^2=\left({\mathrm{x}}^2-1\right)+\mathrm{cx} \mathrm{pass}es through(1,1)\\ 1=\mathrm{c}\\ {\mathrm{y}}^2={\mathrm{x}}^2-1+\mathrm{x}\end{array}$