The integrating factor of the linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$ is:

Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)=2-\frac{f(x)}{x}$ for all $x \in(0, \infty)$ and $f(1) \neq 1$. Then

Solve the differential equation $\frac{dy}{dx} = \frac{1+y^2}{(\tan^{-1} y) - x}$.

If $y=y(x)$ is the solution curve of the differential equation $\frac{dy}{dx} + y \tan x = x \sec x$,$0 \leq x \leq \frac{\pi}{3}$,with $y(0)=1$,then $y\left(\frac{\pi}{6}\right)$ is equal to

If $x \, dy = y(dx + y \, dy)$,$y(1) = 1$,$y(x) > 0$,then $y(-3)$ is

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$,$x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,such that $y(0) = 1$. Then