Find the general solution of the differential equation: $\sec^{2} x \tan y \, dx + \sec^{2} y \tan x \, dy = 0$

Let $x=x(t)$ and $y=y(t)$ be solutions of the differential equations $\frac{dx}{dt}+ax=0$ and $\frac{dy}{dt}+by=0$ respectively,where $a, b \in R$. Given that $x(0)=2$,$y(0)=1$,and $3y(1)=2x(1)$,the value of $t$ for which $x(t)=y(t)$ is:

The general solution of the differential equation $(y \sin x + y) \frac{dy}{dx} - \cos^2 x = 0$ is:

Let $y=y(x)$ be a solution curve of the differential equation,$(1-x^2 y^2) dx = y dx + x dy$. If the line $x = 1$ intersects the curve $y = y(x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y(x)$ at $y = \alpha$,then a value of $\alpha$ is

The general solution of the differential equation $\tan x \tan y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$ is

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