Let a solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0$ satisfy $y(2)=\frac{2}{\sqrt{3}}$
Statement $1 y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$
Statement $2 y(x)$ is given by $\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}$

Question

Aicharya · Accepted Answer

$\because \frac{d y}{d x}=\frac{y \sqrt{y^2-1}}{x \sqrt{x^2-1}} \Rightarrow \int \frac{d y}{y \sqrt{y^2-1}}=\int \frac{d x}{x \sqrt{x^2-1}}$ $\Rightarrow \quad \sec ^{-1} y=\sec ^{-1} x+C$ At $\quad x=2, y=\frac{2}{\sqrt{3}}$ $ \frac{\pi}{6}=\frac{\pi}{3}+C \Rightarrow C=-\frac{\pi}{6} $ Now, $y=\sec \left(\sec ^{-1} x-\frac{\pi}{6}ight)$ $\begin{aligned} & =\cos \left[\cos ^{-1} \left(\frac{1}{x}ight)-\cos ^{-1} \left(\frac{\sqrt{3}}{2}ight)ight] \ & =\cos \left[\cos ^{-1}\left(\frac{\sqrt{3}}{2 x}+\sqrt{1-\frac{1}{x^2}} \sqrt{1-\frac{3}{4}}ight)ight] \ \Rightarrow \quad \frac{1}{y} & =\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^2}} \end{aligned}$

Let a solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0$ satisfy…

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