The solution of the differential equation $ydx - (x + 2y^2)dy = 0$ is $x = f(y)$. If $f(-1) = 1$,then $f(1)$ is equal to

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

The general solution of the differential equation $(y^2 - x^3) dx - xy dy = 0, (x \neq 0)$ is (where $c$ is a constant of integration)

If $x dy + (y + y^2 x) dx = 0$ and $y = 1$ at $x = 1$,then

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

Let $f:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function. If $6 \int_{1}^{x} f(t) dt = 3xf(x) + x^{3} - 4$ for all $x \ge 1$, then the value of $f(2) - f(3)$ is