Official solution
\[{x^2}dy + \left( {y - {1 \over x}} \right)dx = 0\] ; x > 0, y(1) = 1
\[{x^2}dy + {{(xy - 1)} \over x}dx = 0\]
\[{x^2}dy = {{(xy - 1)} \over x}dx\]
\[{{dy} \over {dx}} = {{1 - xy} \over {{x^3}}}\]
\[{{dy} \over {dx}} = {1 \over {{x^3}}} - {y \over {{x^2}}}\]
\[{{dy} \over {dx}} = {1 \over {{x^2}}}.y = {1 \over {{x^3}}}\]
If \[{e^{\int {{1 \over {{x^2}}}dx} }} = {e^{ - {1 \over x}}}\]
\[y{e^{ - {1 \over x}}} = \int {{1 \over {{x^3}}}.{e^{ - {1 \over x}}}} \]
\[y{e^{ - {1 \over x}}} = {e^{ - x}}\left( {1 + {1 \over x}} \right) + C\]
\[1.\,{e^{ - 1}} = {e^{ - 1}}(2) + C\]
\[C = - {e^{ - 1}} = - {1 \over e}\]
\[y{e^{ - {1 \over x}}} = {e^{ - {1 \over x}}}\left( {1 + {1 \over x}} \right) - {1 \over e}\]
\[y\left( {{1 \over 2}} \right) = 3 - {1 \over e} \times {e^2}\]
\[y\left( {{1 \over 2}} \right) = 3 - e\]