Consider the differential equation,

Consider the differential equation,

$y^{2} d x+\left(x-\frac{1}{y}\right) d y=0$

$\Rightarrow \frac{d x}{d y}+\left(\frac{1}{y^{2}}\right) x=\frac{1}{y^{3}}$

$\mathrm{I} \mathrm{F} .=e^{\int \frac{1}{y^{2}} d y}=e^{-\frac{1}{y}}$

$\therefore \quad x . e^{-\frac{1}{y}}=\int e^{-\frac{1}{y}} \frac{1}{y^{3}} d y+c$

Put $-\frac{1}{y}=u \Rightarrow \frac{1}{y^{2}} d y=d u$

$\Rightarrow x . e^{-\frac{1}{y}}=-\int u e^{u} d u+c=-u e^{u}+e^{u}+c$

$\Rightarrow x . e^{-\frac{1}{y}}=e^{-\frac{1}{y}}\left(\frac{1}{y}+1\right)+c$

At $y=1, x=1$

$1=2+c e \Rightarrow c=-\frac{1}{e} \Rightarrow x=\left(1+\frac{1}{y}\right)-\frac{1}{e} e^{\frac{1}{y}}$

On putting $y=2$, we get $x=\frac{3}{2}-\frac{1}{\sqrt{e}}$