The solution of the differential equation $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$ at $x=0, y=1$ is

The solution of the differential equation $x dy = (y + xy^3 (1 + \log_e x)) dx$ is (where $C$ is an arbitrary constant):

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

$A$ continuously differentiable function $\phi (x)$ in $(0, \pi)$ satisfying $y' = 1 + y^2$ and $y(0) = 0 = y(\pi)$ is

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