If $x \frac{dy}{dx} + y = \frac{x f(xy)}{f'(xy)}$, then $|f(xy)|$ is equal to

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

If the curve $y = y(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx$, $x > 0$, then $y(e)$ is equal to:

The general solution of the differential equation $\sec(x-y+1) dy = dx$ is

$A$ particle moves in a straight line with a velocity given by $\frac{dx}{dt} = x + 1$ (where $x$ is the distance covered). The time taken by the particle to traverse a distance of $99 \ m$ is:

The particular solution of the differential equation $\frac{dy}{dx} = \sec y$ with the initial condition $y(0) = 0$ is: