Question

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

Answer 1

$\int dy=\int \frac{\cos \frac{1}{2}{\cos }^{-1}\left(e^{-x}\right)}{\sqrt{e^{2x}-1}}dx$
Let $\frac{1}{2}{\cos }^{-1}\left(e^{-x}\right)=\theta$
$\Rightarrow e^{-x}=\cos 2\theta$
$\Rightarrow x=\ln \sec 2\theta$
$\Rightarrow dx=2\tan 2\theta d\theta$
$\Rightarrow y=\int 2\cos \theta d\theta =2\sin \theta +C=\sqrt{2}\sqrt{1-\cos 2\theta }+C$
$\Rightarrow y=\sqrt{2}\sqrt{1-e^{-x}}+C$
$y\left(0\right)=-1\Rightarrow C=-1$
$y=\sqrt{2(1-e^{-x})}-1$
$y=0\Rightarrow e^{\alpha }=2$