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 value of [k] is (where [.] represents greatest integer function)

Answer 1

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

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

let $y^2-1=t^2\Rightarrow 2ydy=2tdt$

$\therefore \int \frac{t}{t}dt=-\int \frac{x^2-1}{x\sqrt{x^2-1}}dx$

$\therefore t=-\int \frac{x}{\sqrt{x^2-1}}dx+\int \frac{1}{x\sqrt{x^2-1}}dx$

$\therefore \sqrt{y^2-1}=-\sqrt{x^2-1}+{\sec }^{-1}x+c$

Curve passes through the point $(1,1)$ then the value of $c=0$

Hence the curve is $\sqrt{y^2-1}=-\sqrt{x^2-1}+{\sec }^{-1}x$