Consider the differential equation $y^{2}dx+\left(x-\frac{1}{y}\right)dy=0$ if value of $y$ is 1 when $x = 1,$ then the value of$x$for which $y = 2$, is  

Given differential equation is

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

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

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

$\therefore$Required solution is $x\cdot e^{-\frac{1}{y}}=\int e^{-\frac{1}{y}}\frac{1}{y^3}dy+C$

Let $I=\int e^{-\frac{1}{y}}\frac{1}{y^3}dy$ Putting $-\frac{1}{y}=t\Rightarrow \frac{dy}{y^2}=dt$

$\therefore I=\int e^t(-t)dt=-\left[t{e}^t-\int e^{t}dt\right]=-t{e}^t+e^t=e^{-1/y}+\frac{1}{y}{e}^{-1/y}$

$\because y\left(1\right)=1\Rightarrow 1=1+1+Ce\therefore C=\frac{-1}{e}$

$\therefore$ Equation of curve is $x=1+\frac{1}{y}-e^{\frac{1}{y}-1}$

At $y=2,x=1+\frac{1}{2}-e^{\frac{1}{2}-1}=\frac{3}{2}-\frac{1}{\sqrt{e}}$