Let a solution y=y(x) of the differential equation xx2−1 dy−yy2−1dx=0 satisfies y(2)=23Statement-I: y(x)=sec(sec−1x−π6)Statement-II: 1y=23x−1−1x2

∫dxxx2−1=∫dyyy2−1⇒sec−1x=sec−1y+c so, sec−1(23)+c⇒c=π6∴ y = sec(sec−1x−π6)=1cos[cos−11x−π6]=11x32+1−1x212⇒1y=32x+1−1x22

Let a solution y=y(x) of the differential equation xx2−1 dy−yy2−1dx=0 satisfies y(2)=23Statement-I: y(x)=sec(sec−1x−π6)Statement-II: 1y=23x−1−1x2

∫dxxx2−1=∫dyyy2−1⇒sec−1x=sec−1y+c so, sec−1(23)+c⇒c=π6∴ y = sec(sec−1x−π6)=1cos[cos−11x−π6]=11x32+1−1x212⇒1y=32x+1−1x22

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Let a solution $y = y (x)$ of the differential equation $x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0$ satisfies $y (2) = \frac{2}{\sqrt{3}}$
Statement-I: $y (x) = \sec (\sec^{- 1} x - \frac{π}{6})$
Statement-II: $\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}$

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Statement-I is true, Statement-II is true, Statement-II is the correct explanation of Statement-I

Statement-I is true, Statement-II is true, statement-II is not the correct explanation of Statement-I

Statement-I is true, Statement-II is false

Statement-I is false, Statement-II is true.

answer is C.

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Detailed Solution

$\int \frac{d x}{x \sqrt{x^{2} - 1}} = \int \frac{d y}{y \sqrt{y^{2} - 1}} \Rightarrow \sec^{- 1} x = \sec^{- 1} y + c$ so, $\sec^{- 1} (\frac{2}{\sqrt{3}}) + c \Rightarrow c = \frac{π}{6}$

$∴ y = sec (\sec^{- 1} x - \frac{π}{6}) = \frac{1}{\cos [\cos^{- 1} \frac{1}{x} - \frac{π}{6}]}$

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

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

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Let a solution y=y(x) of the differential equation xx2−1 dy−yy2−1dx=0 satisfies y(2)=23Statement-I: y(x)=sec(sec−1x−π6)Statement-II: 1y=23x−1−1x2

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Let a solution $y = y (x)$ of the differential equation $x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0$ satisfies $y (2) = \frac{2}{\sqrt{3}}$
Statement-I: $y (x) = \sec (\sec^{- 1} x - \frac{π}{6})$
Statement-II: $\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}$