dy$$

Put cos^$$-$$1(e^$$-$$x) $$\theta$$, $$\theta$$ $$\in$$ [0, $$\pi$$]

$$\cos \theta = {e^{ - x}} \Rightarrow 2{\cos ^2}{\theta \over 2} - 1 = {e^{ - x}}$$

$$\cos {\theta \over 2} = \sqrt {{{{e^{ - x}} + 1} \over 2}} = \sqrt {{{{e^x} + 1} \over {2{c^x}}}} $$

$$\sqrt {{{{e^x} + 1} \over {2{c^x}}}} dx = \sqrt {{e^{2x}} - 1} dy$$

$${1 \over {\sqrt 2 }}\int {{{dx} \over {\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $$

Put $${e^x} = t,{{dt} \over {dx}} = {e^x}$$

$${1 \over {\sqrt 2 }}\int {{{dx} \over {{e^x}\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} } $$

$$\int {{{dt} \over {t\sqrt {{t^2} - t} }} = \sqrt 2 y} $$

Put $$t = {1 \over z},{{dt} \over {dz}} = - {1 \over {{z^2}}}$$

$$\int {{{ - {{dz} \over {{z^2}}}} \over {{1 \over z}\sqrt {{1 \over {{z^2}}} - {1 \over z}} }} = \sqrt {2y} } $$

$$ - \int {{{dz} \over {\sqrt {1 - z} }} = \sqrt 2 y} $$

$${{ - 2{{(1 - z)}^{1/2}}} \over { - 1}} = \sqrt 2 y + c$$

$$2{\left( {1 - {1 \over t}} \right)^{1/2}} = \sqrt 2 y + c$$

$$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 y + c\buildrel {(0, - 1)} \over \longrightarrow \Rightarrow c = \sqrt 2 $$

$$2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 (y + 1)$$, passes through ($$\alpha$$, 0)

$$2{(1 - {e^{ - \alpha }})^{1/2}} = \sqrt 2 $$

$$\sqrt {1 - {e^{ - \alpha }}} = {1 \over {\sqrt 2 }} \Rightarrow 1 - {e^{ - \alpha }} = {1 \over 2}$$

$${e^{ - \alpha }} = {1 \over 2} \Rightarrow {e^\alpha } = 2$$

Differential Equations

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