The solution of the differential equation $\frac{dy}{dx} + 2y = e^{-x}$ is

Let $y=y(x)$ be the solution of the differential equation $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$,where $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{4}) = \frac{-1}{\ln 2}$,then $y(\frac{\pi}{6})$ 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

The solution of the differential equation $(1+y^2) + (x - e^{\tan^{-1} y}) \frac{dx}{dy} = 0$ 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:

Find the general solution of the differential equation: $y dx + (x - y^2) dy = 0$