If the general solution of the differential equation $(y-x+1) dy - (y+x+2) dx = 0$ is $f(x, y, c) = 0$,then the value of $c$ such that $f(1, 1, c) = 0$ is

Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$,where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to . . . . . . .

Let $y$ be the solution of the differential equation $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$ satisfying $y(1) = 1$. Then,$y$ satisfies:

The solution of $(1+xy)y \, dx + (1-xy)x \, dy = 0$ is:

Statement $-1$: The slope of the tangent at any point $P$ on a parabola,whose axis is the $x$-axis and vertex is at the origin,is inversely proportional to the ordinate of the point $P$.
Statement $-2$: The system of parabolas $y^2 = 4ax$ satisfies a differential equation of degree $1$ and order $1$.

The equation of the curve passing through $(3, 4)$ and satisfying the differential equation $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$ can be: