Let a solution y=y(x) of the differential xx2−1dy−yy2−1dx=0 satisfy y(2)=23 Statement 1: y(x)=sec sec−1 x−π/6Statement 2: y(x) is given by 1y=23x−1−1x2

(c) STATEMENT-1 is True, STATEMENT-2 is False

(c) STATEMENT-1 is True, STATEMENT-2 is False

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

(d) STATEMENT-1 is False, STATEMENT-2 is True

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

The given equation can be written as

$\frac{d y}{y \sqrt{y^{2} - 1}} - \frac{d x}{x \sqrt{x^{2} - 1}} = 0$

Integrating $\sec^{- 1} y - \sec^{- 1} x = Const$

Putting $x = 2, Const = \sec^{- 1} 2 / \sqrt{3} - \sec^{- 1} 2 = \frac{π}{6} - \frac{π}{3} = - \frac{π}{6}$

so $y = \sec (\sec^{- 1} x - π / 6)$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE