The integrating factor of the differential equation $(\tan ^{-1} y-x) dy = (1+y^2) dx$ is . . . . . . .

The solution of the differential equation,$x^2 \frac{dy}{dx} \cos \frac{1}{x} - y \sin \frac{1}{x} = -1,$ where $y \rightarrow -1$ as $x \rightarrow \infty$ is

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}$,$x \in\left(0, \frac{\pi}{2}\right)$ satisfying the condition $y\left(\frac{\pi}{4}\right)=2$. Then,$y\left(\frac{\pi}{3}\right)$ is

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx} + 2y \tan x = \sin x$ with the condition $y(\frac{\pi}{3}) = 0$,then the maximum value of the function $y(x)$ over $\mathbb{R}$ is equal to:

Let $y=y(x), y>0$,be a solution curve of the differential equation $(1+x^2) dy = y(x-y) dx$. If $y(0)=1$ and $y(2\sqrt{2})=\beta$,then

The Integrating Factor $(I.F.)$ of the differential equation $\frac{dy}{dx} - \frac{3x^2y}{1+x^3} = \frac{\sin^2(x)}{1+x}$ is: