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

Find the general solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ where $x \neq 0$.

If the solution of the differential equation $x \frac{dy}{dx} + y = x e^x$ is $xy = e^x \phi(x) + C$, then $\phi(x)$ is equal to

The general solution of the differential equation $(1+y^2) dx = (\tan^{-1} y - x) dy$ 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

Let $y = y(x)$ be the solution of the differential equation $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$, where $x \in (0, 1)$ and $\lim_{x\to 1^-} y(x) = 1$. Then $y\left(\frac{1}{2}\right)$ equals: