Mathematics - Differential Equation Question with Solution | TestHub

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