Question

Assertion :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 $y\left(2\right)=2/\sqrt{3}$

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

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

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

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

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

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

At $x=2,y=\frac{2}{\sqrt{3}}$

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

Now $y=\sec ({\sec }^{-1}x-\frac{\pi }{6})=\cos [{\cos }^{-1}\frac{1}{x}-{\cos }^{-1}\frac{\sqrt{3}}{2}]$

$=\cos \left[{\cos }^{-1}(\frac{\sqrt{3}}{2x}+\sqrt{1-\frac{1}{x^2}}\sqrt{1-\frac{3}{4}})\right]$

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

Here the assertion is correct however, the reason is incorrect.