The solution of $y^{\mathrm{\prime }}+xy=x{y}^{-1},y\left(0\right)=2$ is 

given by

$y^{\mathrm{\prime }}+xy=x{y}^{-1}\Rightarrow y{y}^{\mathrm{\prime }}+x{y}^2=x$

This is reducible to linear equation, so put $y^2=z\Rightarrow y\frac{\mathrm{d}y}{\mathrm{d}x}$

$=\frac{1}{2}\frac{\mathrm{d}z}{\mathrm{d}x}$

$\begin{array}{r}\frac{1}{2}\frac{\mathrm{d}z}{\mathrm{d}x}+xz=x\\ \frac{\mathrm{d}z}{\mathrm{d}x}+2xz=2x\end{array}$

$\text{ I.F. }=e^{2\int xdx}=e^{x^2}$ Multiplying with I.F., we have

$\begin{array}{l}\frac{d}{dx}\left(z{e}^{x^2}\right)=2x{e}^{x^2}\\ \Rightarrow z{e}^{x^2}=\int 2x{e}^{x^2}dx+C=e^{x^2}+C\\ \Rightarrow y^2{e}^{x^2}=e^{x^2}+C\end{array}$

since $y\left(0\right)=1\text{ so }C=3$

$\Rightarrow y^2=3+e^{-x^2}$.