Let x = x(y) be the solution of the differential equation ${\text{y}}^2\mathrm{dx}+\left(\mathrm{x}-\frac{1}{\mathrm{y}}\right)\mathrm{dy}=0$. If x(1) = 1, then $\mathrm{x}\left(\frac{1}{2}\right)$ is :

${\text{y}}^2\mathrm{dx}+\left(\mathrm{x}-\frac{1}{\mathrm{y}}\right)\mathrm{dy}=0$

$\frac{dx}{dy}=\frac{-x}{y^2}+\frac{1}{y^3}$

$\frac{dx}{dy}+\frac{x}{y^2}=\frac{1}{y^3}$

 $I.F\begin{array}{r}=e^{\int \frac{1}{y^2}\mathrm{dy}}\end{array}$

$={\mathrm{e}}^{-\frac{1}{\mathrm{y}}}$

The general solution is

$x\cdot {\mathrm{e}}^{-\frac{1}{y}}=\int \left({\mathrm{e}}^{-\frac{1}{\mathrm{t}}}\right)\cdot \frac{1}{{\mathrm{y}}^3}\mathrm{dy}$

Put $\frac{-1}{y}=t$

$\begin{array}{r}+\frac{1}{y^2}dy=dt\end{array}$

$\begin{array}{r}x\cdot e^{-\frac{1}{y}}=-\int t\cdot e^{t}dt\end{array}$

$\begin{array}{r}x\cdot e^{-\frac{1}{y}}=-t{e}^t+e^t+C\end{array}$

$\begin{array}{r}x\cdot e^{-\frac{1}{y}}=\frac{+1}{y}{e}^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C\end{array}$

$\begin{array}{r}x=1,y=1\end{array}$

$\begin{array}{r}\frac{1}{\mathrm{e}}=\frac{1}{\mathrm{e}}+\frac{1}{\mathrm{e}}+\mathrm{C}\end{array}$

$\Rightarrow \mathrm{C}=-\frac{1}{\mathrm{e}}$

Put $y=\frac{1}{2}$

$\begin{array}{r}\frac{\mathrm{x}}{{\mathrm{e}}^2}=\frac{2}{{\mathrm{e}}^2}+\frac{1}{{\mathrm{e}}^2}-\frac{1}{\mathrm{e}}\end{array}$

$\mathrm{x}=3-\mathrm{e}$