Let $y = y(x)$ be the solution curve of the differential equation

$(1 + \sin x)\dfrac{dy}{dx} + (y + 1)\cos x = 0,\ y(0) = 0.$ If the curve $y = y(x)$ passes through the point $\left( \alpha , \dfrac{-1}{2} \right)$,

then a value of $\alpha$ is:

Let $y = y(x)$ be the solution of the differential equation $x \frac{dy}{dx} - y = x^2 \cot x$, $x \in (0, \pi)$. If $y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$, then

$6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right)$ is equal to :

Let $y=y(x)$ be the solution of the differential equation

$$ x \frac{d y}{d x}-\sin 2 y=x^3\left(2-x^3\right) \cos ^2 y, x \neq 0 . $$

If $y(2)=0$, then $\tan (y(1))$ is equal to

Let $y=y(x)$ be the solution of the differential equation $x^4 \mathrm{~d} y+\left(4 x^3 y+2 \sin x\right) \mathrm{d} x=0, x>0, y\left(\frac{\pi}{2}\right)=0$.

Then $\pi^4 y\left(\frac{\pi}{3}\right)$ is equal to :