The solution of the differential equation $y' = y \tan x - 2 \sin x$ is

The general solution of the differential equation $\frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x} y = \frac{\cos x}{1 + \tan x}$ is

The solution of $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$ is

Suppose $y=y(x)$ is the solution curve to the differential equation $\frac{dy}{dx}-y=2-e^{-x}$ such that $\lim_{x \rightarrow \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x$- and $y$-intercepts of the tangent to the curve at $x=0$,then the value of $a-4b$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $(\tan x)^{1/2} dy = (\sec^3 x - (\tan x)^{3/2} y) dx$,where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{4}) = \frac{6\sqrt{2}}{5}$. If $y(\frac{\pi}{3}) = \frac{4}{5}\alpha$,then $\alpha^4$ equals . . . . . . .

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 $f(y) =$