Question

Let a solution $y=y\left(x\right)$ of the differential equation $x\sqrt{x^2-1}dy-y\sqrt{y^2-1}dx=0$ satisfy $y\left(2\right)=\frac{2}{\sqrt{3}}$

STATEMENT-1: $y\left(x\right)=\sec (se{c}^{-1}x-\frac{\pi }{6})$

STATEMENT-2 : $y\left(x\right)$ is given by $\frac{1}{y}=\frac{2\sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}$

• A. STATEMENT1 is True, STATEMENT2 is True; STATEMENT2 is a correct explanation for STATEMENT1
• B. STATEMENT1 is True, STATEMENT2 is True; STATEMENT2 is NOT a correct explanation for STATEMENT1.
• C. STATEMENT1 is True, STATEMENT2 is False
• D. STATEMENT1 is False, STATEMENT2 is True

Answer 1

The correct option is C STATEMENT1 is True, STATEMENT2 is False

Given differential equation is $x\sqrt{x^2-1}dy-y\sqrt{y^2-1}dx=0$

$\therefore \frac{dx}{x\sqrt{x^2-1}}=\frac{dy}{y\sqrt{y^2-1}}$

integrating both sides

$\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{dy}{y\sqrt{y^2-1}}$

$\Rightarrow$${\sec }^{-1}x={\sec }^{-1}y+c$

$\Rightarrow$${\sec }^{-1}2={\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)+c$

$\Rightarrow$ $c=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$

$\Rightarrow$ ${\sec }^{-1}x={\sec }^{-1}y+\frac{\pi }{6}$

$\Rightarrow$ $y=\sec (se{c}^{-1}x-\frac{\pi }{6})$

$\Rightarrow$ ${\cos }^{-1}\frac{1}{x}={\cos }^{-1}\frac{1}{y}+\frac{\pi }{6}$

$\Rightarrow$ ${\cos }^{-1}\frac{1}{y}={\cos }^{-1}\frac{1}{x}-{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\Rightarrow$ $\frac{1}{y}=\frac{\sqrt{3}}{2x}-\sqrt{1-\frac{1}{x^2}}\left(\frac{1}{2}\right)$

$\Rightarrow$$\frac{2}{y}=\frac{\sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}$