Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$, $y > 1$, $x(e) = e$.

Then $x(e^2)$ is equal to:

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \\log_e x) dx$, $x > 0$, then $f(e)$ is equal to :

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 :