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

Given equation,

$\cos \left(\frac{1}{2}{\cos }^{-1}\left({\mathrm{e}}^{-\mathrm{x}}\right)\right)\mathrm{dx}=\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}\mathrm{dy}$        ...(i)

Let    ${\cos }^{-1}\left({\mathrm{e}}^{-\mathrm{x}}\right)=\mathrm{\theta },\text{ then }{\mathrm{e}}^{-\mathrm{x}}=\cos \mathrm{\theta }$

$\begin{array}{l}\Rightarrow 2{\cos }^2\frac{\mathrm{\theta }}{2}-1={\mathrm{e}}^{-\mathrm{x}}\\ \Rightarrow \cos \frac{\mathrm{\theta }}{2}=\sqrt{\frac{{\mathrm{e}}^{-\mathrm{x}}+1}{2}} ..\left(\mathrm{ii}\right)\end{array}$

From Eq. (i),

$\cos \left(\frac{1}{2}\cdot \mathrm{\theta }\right)\mathrm{dx}=\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}\mathrm{dy}$

From Eqs. (i) and (ii),

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

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

$\begin{array}{l}\mathrm{or} \frac{\sqrt{{\mathrm{e}}^{-\mathrm{x}}+1}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{dx}=\sqrt{2}\mathrm{dy}\\ \mathrm{or} \frac{\sqrt{1+{\mathrm{e}}^{\mathrm{x}}}}{\sqrt{{\mathrm{e}}^{\mathrm{x}}}\cdot \sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}}\mathrm{dx}=\sqrt{2}\mathrm{dy}\end{array}$

Put ${\mathrm{e}}^{\mathrm{x}}=\mathrm{t},\text{ then }{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=\mathrm{dt}\text{ or }\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{t}}$,

$\begin{array}{l}\frac{\sqrt{1+\mathrm{t}}}{\sqrt{\mathrm{t}}\sqrt{{\mathrm{t}}^2-1}}\frac{\mathrm{dt}}{\mathrm{t}}=\sqrt{2}\mathrm{dy}\\ \Rightarrow \frac{\sqrt{1+\mathrm{t}}}{\sqrt{\mathrm{t}}\sqrt{\mathrm{t}+1}\cdot \sqrt{\mathrm{t}-1}}\frac{\mathrm{dt}}{\mathrm{t}}=\sqrt{2}\mathrm{dy}\\ \Rightarrow \frac{\mathrm{dt}}{\mathrm{t}\sqrt{\mathrm{t}}\sqrt{\mathrm{t}-1}}=\mathrm{dy}\cdot \sqrt{2}\end{array}$

Put $\mathrm{t}=\frac{1}{\mathrm{z}},\text{ then }\mathrm{dt}=-\frac{1}{{\mathrm{z}}^2}\mathrm{dz}$

$\begin{array}{c}\frac{-\frac{\mathrm{dz}}{{\mathrm{z}}^2}}{\frac{1}{\mathrm{z}}\sqrt{\frac{1}{{\mathrm{z}}^2}-\frac{1}{\mathrm{z}}}}=\sqrt{2}\mathrm{dy}\\ -\frac{\mathrm{dz}}{\sqrt{1-\mathrm{z}}}=\sqrt{2}\mathrm{dy}\end{array}$

Integrating both sides,

$\begin{array}{l}-\int \frac{\mathrm{dz}}{\sqrt{1-\mathrm{z}}}=\sqrt{2}\int \mathrm{dy}\\ \Rightarrow -2\frac{(1-\mathrm{z}{)}^{\frac{1}{2}}}{-1}=\sqrt{2}\mathrm{y}+\mathrm{c}\\ \Rightarrow 2{\left(1-\frac{1}{\mathrm{t}}\right)}^{\frac{1}{2}}=\sqrt{2}\mathrm{y}+\mathrm{c}\\ \Rightarrow 2{\left(1-{\mathrm{e}}^{-\mathrm{x}}\right)}^{\frac{1}{2}}=\sqrt{2}\mathrm{y}+\mathrm{c} ...\left(\mathrm{iii}\right)\end{array}$

Given condition y(0) = -1

From Eq. (iii),

$\begin{array}{l}2(1-1{)}^{\frac{1}{2}}=\sqrt{2}(-1)+\mathrm{C}\\ \Rightarrow \mathrm{C}=\sqrt{2}\end{array}$

From Eq. (iii),

$2\sqrt{1-{\mathrm{e}}^{-\mathrm{x}}}=\sqrt{2}(\mathrm{y}+1)$         ...(iv)

It passes through $(\mathrm{\alpha }, 0)$, putting in Eq. (iv),

$\begin{array}{l}2\sqrt{1-{\mathrm{e}}^{-\mathrm{\alpha }}}=\sqrt{2}(0+1)\\ \Rightarrow \sqrt{1-{\mathrm{e}}^{-\mathrm{\alpha }}}=\frac{1}{\sqrt{2}}\Rightarrow 1-{\mathrm{e}}^{-\mathrm{\alpha }}=\frac{1}{2}\\ \Rightarrow {\mathrm{e}}^{-\mathrm{\alpha }}=\frac{1}{2}\Rightarrow {\mathrm{e}}^{\mathrm{\alpha }}=2\end{array}$