Find the integrating factor of the differential equation $(1+x^{2}) dt = (\tan^{-1} x - t) dx$.

The solution of $(x+y+1) \frac{dy}{dx} = 1$ is

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-y=x^{2}\cot x, x\in(0,\pi)$. If $y(\frac{\pi}{2})=\frac{\pi}{2}$, then $6y(\frac{\pi}{6})-8y(\frac{\pi}{4})$ is equal to :

Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{dy}{dx}+y \cos x=4x, x \in(0, \pi)$. If $y\left(\frac{\pi}{2}\right)=0$,then $y\left(\frac{\pi}{6}\right)$ is equal to

Let $g$ be a differentiable function such that $\int_0^x g(t) dt = x - \int_0^x tg(t) dt$ for $x \geq 0$. Let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x)$ for $x \in [0, \frac{\pi}{2})$. If $y(0) = 0$,then $y(\frac{\pi}{3})$ is equal to

The general solution of the differential equation $\frac{dy}{dx} + y g'(x) = g(x) g'(x)$ is