The general solution of the differential equation $(x+2y^3) \frac{dy}{dx} - y = 0, y > 0$ is

Suppose $f(x)$ is a differentiable real function such that $f(x) + f'(x) \le 1$ for all $x$ and $f(0)=0$. The largest possible value of $f(1)$ is

If a curve passes through the point $(1, -2)$ and has the slope of the tangent at any point $(x, y)$ on it as $\frac{x^2 - 2y}{x}$,then the curve also passes through the point:

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)((y+1)e^{x^{2}/2}-x)$,with $y(2)=0$. Then $y'(1)$ is equal to . . . .

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 $y(x)$ be the solution of the differential equation $2 x^{2} dy + (e^{y} - 2x) dx = 0$,$x > 0$. If $y(e) = 1$,then $y(1)$ is equal to: