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Q.
If $y = y ( x )$ is the solution curve of the differential equation $x^{2} d y+\left(y-\frac{1}{x}\right) d x=0 \quad ; x>0$ and $y (1)=1$, then $y \left(\frac{1}{2}\right)$ is equal to :
$x^{2} d y+\left(y-\frac{1}{x}\right) d x=0: x>0, y(1)=1$
$x^{2} d y+\frac{(x y-1)}{x} d x=0$
$x^{2} d y=\frac{(x y-1)}{x} d x$
$\frac{d y}{d x}=\frac{1-x y}{x^{3}}$
$\frac{d y}{d x}=\frac{1}{x^{3}}-\frac{y}{x^{2}}$
$\frac{d y}{d x}=\frac{1}{x^{2}} \cdot y=\frac{1}{x^{3}}$
If $e^{\int \frac{1}{x^{2}} d x}=e^{\frac{1}{x}}$
$y e^{-\frac{1}{x}}=\int \frac{1}{x^{3}} \cdot e^{-\frac{1}{x}}$
$ye ^{-\frac{1}{x}}= e ^{- x }\left(1+\frac{1}{ x }\right)+ C$
$1 . e ^{-1}= e ^{-1}(2)+ C$
$C =- e ^{-1}=-\frac{1}{ e }$
$y e^{-\frac{1}{x}}=e^{-\frac{1}{x}}\left(1+\frac{1}{x}\right)-\frac{1}{e}$
$y\left(\frac{1}{2}\right)=3-\frac{1}{e} \times e^{2}$
$y\left(\frac{1}{2}\right)=3-e$