Official solution
\[\left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){{dy} \over {dx}} = 1 + \left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){y \over x}\]
Putting y = tx
\[\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)\left( {t + x{{dt} \over {dx}}} \right) = 1 + \left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)t\]
\[ \Rightarrow x\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right){{dt} \over {dx}} = 1\]
\[ \Rightarrow {\sin ^{ - 1}}t + {e^t} = \ln x + C\]
\[ \Rightarrow {\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + C\]
at x = 1, y = 0
So, \[0 + {e^0} = 0 + C \Rightarrow C = 1\]
at \[(2\alpha ,\alpha )\]
\[{\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + 1\]
\[ \Rightarrow {\pi \over 6} + {e^{{1 \over 2}}} - 1 = \ln (2\alpha )\]
\[ \Rightarrow \alpha = {1 \over 2}{e^{\left( {{\pi \over 6} + {e^{{1 \over 2}}} - 1} \right)}}\]