Consider the differential equation  

$y^2 dx+\left(x-\frac{1}{y}\right) dy=0$. If $y\left(1\right)=1$ then $x$is given by:

The given differential equation can be written as

$\frac{dx}{dy}+\frac{1}{y^2} x=\frac{1}{y^3}$ which is a linear equation with I.F. =

$e^{\int \left(1/y^2\right)dy}=e^{-1/y}$ . Multiplying with the integrating factor.

             $\frac{d}{dx}\left(x{e}^{-1/y}\right)=\frac{1}{y^3}{e}^{-1/y}$

$\begin{array}{r}\Rightarrow x{e}^{-1/y}=\int \frac{1}{y^3}{e}^{-1/y}dy+C\\ =\int -u e^{u}du+C(u=-1/y)\\ =-u e^u+\int e^u du+C\\ =e^{-1/y}(1+1/y)+C\end{array}$

Thus,       $x=1+\frac{1}{y}+C{e}^{1/y}$

When     $\begin{array}{l}x=1,y=1\\ 1=1+1+Ce=C=-1/e.\\ x=1+\frac{1}{y}-\frac{e^{1/y}}{e}.\end{array}$