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

If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x=2 x \sec x$ and $y(0)=0$,then which of the following is true?

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

Let $y = y(x)$ be the solution of the differential equation,$x\frac{dy}{dx} + y = x \ln x$,for $x > 1$. If $2y(2) = \ln 4 - 1$,then $y(e)$ is equal to

Let $f$ be a real-valued differentiable function on $\mathbb{R}$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$,then the value of $f(-3)$ is equal to

If the solution curve of the differential equation $(y-2 \ln x) dx + (x \ln x^2) dy = 0, x > 1$ passes through the points $(e, \frac{4}{3})$ and $(e^4, \alpha)$,then $\alpha$ is equal to $................$.