If $\frac{dy}{dx} + y \tan x = \sin 2x$ and $y(0) = 1$,then $y(\pi)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x$,where $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right)=0$,then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to:

If the solution curve of the differential equation $(\tan^{-1} y - x) dy = (1 + y^2) dx$ passes through the point $(1, 0)$,then the abscissa of the point on the curve whose ordinate is $\tan(1)$ is

The solution of $y' - y = 1, y(0) = -1$ is given by $y(x) = $

Integrating factor of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ 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: