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

Given x³dy + (xy - 1)dx = 0
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{{\mathrm{x}}^2}=\frac{1}{{\mathrm{x}}^3}\\ \text{ I.F }={\mathrm{e}}^{-\frac{1}{\mathrm{x}}}\end{array}$
Solution is $\mathrm{y}\left({\mathrm{e}}^{-\frac{1}{\mathrm{x}}}\right)=\int \frac{1}{{\mathrm{x}}^3}\left({\mathrm{e}}^{-\frac{1}{\mathrm{x}}}\right)+\mathrm{c}$
Put $-\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow \frac{1}{{\mathrm{x}}^2}\mathrm{dx}=\mathrm{dt}$
$\mathrm{y}\left({\mathrm{e}}^{-\frac{1}{\mathrm{x}}}\right)=\int -{\mathrm{te}}^{\mathrm{t}}+\mathrm{c}={\mathrm{e}}^{-\frac{1}{\mathrm{x}}}\left(\frac{1}{\mathrm{x}}+1\right)+\mathrm{c}$

$y=\frac{1}{x}+1+C{e}^{\frac{1}{x}}$

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

$\begin{array}{r}\text{ Put }x=\frac{1}{2}\\ 3-\mathrm{e}=2+1+{\mathrm{Ce}}^2\\ \mathrm{C}=-\frac{1}{\mathrm{e}}\end{array}$

$y=\frac{1}{x}+1-e^{\frac{1}{x}-1}$

$\Rightarrow y\left(1\right)=1$