Question

The curve passing through the point $(1,1)$ satisfies the differential equation $\frac{dy}{dx}+\frac{\sqrt{(x^2-1)(y^2-1)}}{xy}=0$. If the curve passes through the point $(\sqrt{2},k)$, then the largest value of $|\left[k\right]|$ is
(where, $[.]$ represents the greatest integer function)

Answer 1

$\frac{dy}{dx}+\frac{\sqrt{(x^2-1)(y^2-1)}}{xy}=0$
$\Rightarrow \frac{y}{\sqrt{y^2-1}}dy=-\frac{\sqrt{x^2-1}}{x}dx$
$\Rightarrow \int \frac{y}{\sqrt{y^2-1}}dy=-\int \frac{x^2-1}{x\sqrt{x^2-1}}dx$

Let $y^2-1=t^2\Rightarrow 2ydy=2tdt$
$\therefore \int dt=-\int \frac{x}{\sqrt{x^2-1}}dx+\int \frac{1}{x\sqrt{x^2-1}}dx$
$\Rightarrow t=-\sqrt{x^2-1}+{\sec }^{-1}x+C$
$\Rightarrow \sqrt{y^2-1}=-\sqrt{x^2-1}+{\sec }^{-1}x+C$

The curve passes through $(1,1)$.
$\therefore C=0$
Hence, the curve is $\sqrt{y^2-1}=-\sqrt{x^2-1}+{\sec }^{-1}x$
Also, $(\sqrt{2},k)$ lies on the curve.
$\Rightarrow \sqrt{k^2-1}=-1+\frac{\pi }{4}$
$\Rightarrow k^2={(\frac{\pi }{4}-1)}^2+1$
$\Rightarrow k^2=m+1,m\in (0,1)$
$\Rightarrow k=\pm \sqrt{m+1}$
$\Rightarrow \left[k\right]=1\text{or}-2$
$\therefore$ The largest value of $|\left[k\right]|=2$