Let the solution curve $y = y(x)$ of the differential equation

$$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$$

pass through the points (1, 0) and (2$\alpha$, $\alpha$), $\alpha$ > 0. Then $\alpha$ is equal to

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