An integrating factor of the differential equation $(1 - x^2)\frac{dy}{dx} - xy = 1$ is

If $x \log x \frac{dy}{dx} + y = \log x^2$ and $y(e) = 0$,then $y(e^2) = $

If the general solution of $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$ is $x = f(y) + c e^{-\operatorname{Tan}^{-1} y}$, then $f(y) =$

If the solution of the differential equation $x y^{\prime}=y+x^2 \sin x$ subject to the condition $y(\pi)=0$ is $y=f(x)$ and $f(x)$ has an extreme value at $x=\alpha$, then

Let $y=y(x)$ be the solution of the differential equation $\sec^2 x dx + (e^{2y} \tan^2 x + \tan x) dy = 0$,where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{4}) = 0$. If $y(\frac{\pi}{6}) = \alpha$,then $e^{8\alpha}$ is equal to:

The integrating factor of the differential equation $(2x + 3y^2) dy = y dx$ $(y > 0)$ is