# Let a curve

Question:

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 $\mathrm{y}$-axis at $\mathrm{y}=-1$, and the intersection point of the curve with $x$-axis is $(\alpha, 0)$, then $e^{\alpha}$ is equal to

Solution:

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

Put $\cos ^{-1}\left(\mathrm{e}^{-\mathrm{x}}\right) \theta, \theta \in[0, \pi]$

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

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

$\sqrt{\frac{e^{x}+1}{2 c^{x}}} d x=\sqrt{e^{2 x}-1} d y$

$\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{e^{x}} \sqrt{e^{x}-1}}=\int d y$

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

$\frac{1}{\sqrt{2}} \int \frac{d t}{e^{x} \sqrt{e^{x}} \sqrt{e^{x}-1}}=\int d y$

$\int \frac{d t}{t \sqrt{t^{2}-t}}=\sqrt{2} y$

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

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

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

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

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

$2\left(1-\mathrm{e}^{-\mathrm{x}}\right)^{1 / 2}=\sqrt{2 \mathrm{y}}+\mathrm{c} \stackrel{(0,-1)}{\longrightarrow} \Rightarrow \mathrm{c}=\sqrt{2}$

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

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

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

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