Official solution
\[\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0\]
\[ \Rightarrow {{dx} \over {dy}} + {x \over {\left( {1 + {y^2}} \right)}} = {{{e^{{{\tan }^{ - 1}}y}}} \over {\left( {1 + {y^2}} \right)}}\]
\[I.F = e{}^{\int {{1 \over {\left( {1 + {y^2}} \right)}}dy} } = {e^{{{\tan }^{ - 1}}y}}\]
\[x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \int {{{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}}} {e^{{{\tan }^{ - 1}}y}}\,dy\]
\[x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {{{e^{2{{\tan }^{ - 1}}y}}} \over 2} + C\]
\[\therefore\] \[\,\,\,\,\,2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k\]