Let a curve $$y = y(x)$$ be given by the solution of the differential equation $$\cos\left(\frac{1}{2}\cos^{-1}(e^{-x})\right)dx = \left(\sqrt{e^{2x}-1}\right)dy$$. 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 ___.

Let

$$\theta=\cos^{-1}(e^{-x})$$

Then,

$$\cos\theta=e^{-x}$$

Using the half-angle formula,

$$\cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}}$$

Therefore,

$$\cos\left(\frac12\cos^{-1}(e^{-x})\right)=\sqrt{\frac{1+e^{-x}}{2}}$$

The differential equation becomes

$$\sqrt{\frac{1+e^{-x}}{2}}\,dx=\sqrt{e^{2x}-1}\,dy$$

Now,

$$e^{2x}-1=e^{2x}(1-e^{-2x})$$

So,

$$\sqrt{e^{2x}-1}=e^x\sqrt{1-e^{-2x}}$$

$$=e^x\sqrt{(1-e^{-x})(1+e^{-x})}$$

Substituting,

$$\sqrt{\frac{1+e^{-x}}{2}}\,dx=e^x\sqrt{(1-e^{-x})(1+e^{-x})}\,dy$$

Cancelling

$$\sqrt{1+e^{-x}},$$

$$\frac1{\sqrt2}\,dx=e^x\sqrt{1-e^{-x}}\,dy$$

Hence,

$$dy=\frac1{\sqrt2}\cdot\frac{e^{-x}}{\sqrt{1-e^{-x}}}\,dx$$

Integrating,

$$y=\frac1{\sqrt2}\int\frac{e^{-x}}{\sqrt{1-e^{-x}}}\,dx+C$$

Put

$$u=1-e^{-x}$$

Then,

$$du=e^{-x}dx$$

Therefore,

$$y=\frac1{\sqrt2}\int\frac{du}{\sqrt u}+C$$

$$=\frac1{\sqrt2}(2\sqrt u)+C$$

$$=\sqrt2\sqrt{1-e^{-x}}+C$$

The curve passes through

$$(0,-1)$$

Substituting

$$x=0,\qquad y=-1,$$

$$-1=\sqrt2\sqrt{1-e^0}+C$$

$$-1=C$$

So,

$$y=\sqrt2\sqrt{1-e^{-x}}-1$$

At the x-axis,

$$y=0$$

Therefore,

$$0=\sqrt2\sqrt{1-e^{-\alpha}}-1$$

$$1=\sqrt2\sqrt{1-e^{-\alpha}}$$

$$\frac1{\sqrt2}=\sqrt{1-e^{-\alpha}}$$

Squaring,

$$\frac12=1-e^{-\alpha}$$

$$e^{-\alpha}=\frac12$$

Hence,

$$e^\alpha=2$$