Let $y=y\left(x\right)$ be the solution of the differential equation  $x^{3}dy+\left(xy-1\right)dx=0,$ $x>0,$ $y\left(\frac{1}{2}\right)=3-e$then y(1) =______

$\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}\to \left(1\right)$

$P=\frac{1}{x^2},Q=\frac{1}{x^3}$

$\mathrm{I}\cdot \mathrm{F}={\mathit{e}}^{\int pdx}={\mathit{e}}^{\frac{-1}{x}}$

Solution is  $y\left(e^{-1/x}\right)=\int \left(\frac{1}{x^3}\right)e^{-1/x}dx+c$

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

Given  $y\left(\frac{1}{2}\right)=3-e\Rightarrow y\left(1\right)=1$