Let a solution y=y(x) of the differential equ | vedclass

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(C) Given differential equation: $x \sqrt{x^2-1} dy = y \sqrt{y^2-1} dx$.Separating variables: $\int \frac{dy}{y \sqrt{y^2-1}} = \int \frac{dx}{x \sqr

Vedclass · Accepted Answer

$STATEMENT-1$ is True,$STATEMENT-2$ is False

Vedclass · Answer

(C) Given differential equation: $x \sqrt{x^2-1} dy = y \sqrt{y^2-1} dx$.Separating variables: $\int \frac{dy}{y \sqrt{y^2-1}} = \int \frac{dx}{x \sqrt{x^2-1}}$.Integrating both sides: $\sec^{-1} y = \sec^{-1} x + C$.Using the condition $y(2) = \frac{2}{\sqrt{3}}$: $\sec^{-1} \left(\frac{2}{\sqrt{3}}\right) = \sec^{-1} (2) + C$.$\frac{\pi}{6} = \frac{\pi}{3} + C \implies C = -\frac{\pi}{6}$.Thus,$\sec^{-1} y = \sec^{-1} x - \frac{\pi}{6}$,which gives $y(x) = \sec \left(\sec^{-1} x - \frac{\pi}{6}\right)$. So,$STATEMENT-1$ is True.Now,$\cos^{-1} \left(\frac{1}{y}\right) = \cos^{-1} \left(\frac{1}{x}\right) - \frac{\pi}{6}$.Taking $\cos$ on both sides: $\frac{1}{y} = \cos \left(\cos^{-1} \frac{1}{x} - \frac{\pi}{6}\right) = \cos \left(\cos^{-1} \frac{1}{x}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(\cos^{-1} \frac{1}{x}\right) \sin \left(\frac{\pi}{6}\right)$.$\frac{1}{y} = \frac{1}{x} \cdot \frac{\sqrt{3}}{2} + \sqrt{1 - \frac{1}{x^2}} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2x} + \frac{1}{2} \sqrt{1 - \frac{1}{x^2}}$.Comparing this with $STATEMENT-2$,we see $STATEMENT-2$ is False.

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

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