Consider the differential equation, y 2 d x + x - 1 y d y = 0 . If value of y is 1 when x = 1 , then the…

Given differential equation is
$y^{2}dx=\left(\frac{1}{y}-x\right)dy$

$\Rightarrow y^2\frac{dx}{dy}+x=\frac{1}{y}$
$\Rightarrow \frac{dx}{dy}+\left(\frac{1}{y^2}\right)x=\frac{1}{y^3}$
It is a linear differential equation whose integrating factor $I.F. =e^{\int \frac{dy}{y^2}}=e^{-\frac{1}{y}}$
Solution of a given differential equation can be written as
$x{e}^{-\frac{1}{y}}=\int e^{-\frac{1}{y}}\frac{1}{y^3}dy=I$
Let $\frac{-1}{y}=t\Rightarrow \frac{dy}{y^2}=dt\Rightarrow I=\int \left(-t\right)e^t dt$
$=e^t\left(1-t\right)+C$, (Integrating by parts)
$\Rightarrow$ Solution of differential equation is $x{e}^{-\frac{1}{y}}=e^{-\frac{1}{y}}\left(1+\frac{1}{y}\right)+C$
Since for $x=1,$we have $y=1$,
$\Rightarrow 1.e^{-1}=e^{-1}\left(1+1\right)+C\Rightarrow C=\frac{-1}{e}$
$\Rightarrow$ Solution with given condition is $x{e}^{\frac{-1}{y}}=e^{\frac{-1}{y}}\left(1+\frac{1}{y}\right)-e^{-1}$
or $x=1+\frac{1}{y}-e^{\left(\frac{1}{y}-1\right)}$
So, ${\left.x\right|}_{y=2}=1+\frac{1}{2}-e^{\frac{1}{2}-1}=\frac{3}{2}-\frac{1}{\sqrt{e}}$