Let a curve y = y(x) be given by the solution of the differential equation $$\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$$. If it intersects y-axis at y = $-$1, and the intersection point of the curve with x-axis is ($\alpha$, 0), then e^$\alpha$ is equal to __________________.

$$\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$$<br><br>Put cos^$-$1(e^$-$x) $\theta$, $\theta$ $\in$ [0, $\pi$]<br><br>$$\cos \theta = {e^{ - x}} \Rightarrow 2{\cos ^2}{\theta \over 2} - 1 = {e^{ - x}}$$<br><br>$$\cos {\theta \over 2} = \sqrt {{{{e^{ - x}} + 1} \over 2}} = \sqrt {{{{e^x} + 1} \over {2{c^x}}}} $$<br><br>$\sqrt {{{{e^x} + 1} \over {2{c^x}}}} dx = \sqrt {{e^{2x}} - 1} dy$<br><br>$${1 \over {\sqrt 2 }}\int {{{dx} \over {\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $$<br><br>Put ${e^x} = t,{{dt} \over {dx}} = {e^x}$<br><br>$${1 \over {\sqrt 2 }}\int {{{dx} \over {{e^x}\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $$<br><br>$\int {{{dt} \over {t\sqrt {{t^2} - t} }} = \sqrt 2 y}$<br><br>Put $t = {1 \over z},{{dt} \over {dz}} = - {1 \over {{z^2}}}$<br><br>$$\int {{{ - {{dz} \over {{z^2}}}} \over {{1 \over z}\sqrt {{1 \over {{z^2}}} - {1 \over z}} }} = \sqrt {2y} } $$<br><br>$- \int {{{dz} \over {\sqrt {1 - z} }} = \sqrt 2 y}$<br><br>${{ - 2{{(1 - z)}^{1/2}}} \over { - 1}} = \sqrt 2 y + c$<br><br>$2{\left( {1 - {1 \over t}} \right)^{1/2}} = \sqrt 2 y + c$<br><br>$$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 y + c\buildrel {(0, - 1)} \over \longrightarrow \Rightarrow c = \sqrt 2 $$<br><br>$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 (y + 1)$, passes through ($\alpha$, 0)<br><br>$2{(1 - {e^{ - \alpha }})^{1/2}} = \sqrt 2$<br><br>$$\sqrt {1 - {e^{ - \alpha }}} = {1 \over {\sqrt 2 }} \Rightarrow 1 - {e^{ - \alpha }} = {1 \over 2}$$<br><br>${e^{ - \alpha }} = {1 \over 2} \Rightarrow {e^\alpha } = 2$