Let $y=y(x)$ be the solution of the differential equation $\sec x \, dy + \{2(1-x) \tan x + x(2-x)\} \, dx = 0$ such that $y(0)=2$. Then $y(2)$ is equal to :

If the solution of the differential equation $(1 + x^3) \frac{dy}{dx} + 6x^2y = 1 + x^2$ is $y = \frac{1}{(1 + x^3)^s} [x + \frac{x^p}{p} + \frac{x^q}{q} + \frac{x^r}{r} + c]$, then the $LCM$ of $p, q, r$ and $s$ is...

If $y=y(x)$ is the solution curve of the differential equation $x^{2} dy + (y - \frac{1}{x}) dx = 0$ for $x > 0$ and $y(1) = 1$,then $y(\frac{1}{2})$ is equal to:

Find a particular solution satisfying the given condition: $\frac{dy}{dx} + 2y \tan x = \sin x$; $y = 0$ when $x = \frac{\pi}{3}$.

$A$ function $y = f(x)$ satisfying the differential equation $\frac{dy}{dx} \sin x - y \cos x + \frac{\sin^2 x}{x^2} = 0$ is such that $y \rightarrow 0$ as $x \rightarrow \infty$. Which of the following statements is correct?

Let the solution curve $x=x(y), 0 < y < \frac{\pi}{2}$,of the differential equation $(\log_e(\cos y))^2 \cos y dx - (1+3x \log_e(\cos y)) \sin y dy = 0$ satisfy $x(\frac{\pi}{3}) = \frac{1}{2 \log_e 2}$. If $x(\frac{\pi}{6}) = \frac{1}{\log_e m - \log_e n}$,where $m$ and $n$ are co-prime,then $mn$ is equal to $.....$.