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:

If the curve $y = y(x)$ represented by the solution of the differential equation $(2xy^2 - y)dx + xdy = 0$ passes through the intersection of the lines $2x - 3y = 1$ and $3x + 2y = 8$,then $|y(1)|$ is equal to ...... .

Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^2 dx=dy$,$y(0)=-2$. Let the maximum and minimum values of the function $y=y(x)$ in $\left[0, \frac{\pi}{3}\right]$ be $\alpha$ and $\beta$,respectively. If $(3\alpha+\pi)^2+\beta^2=\gamma+\delta\sqrt{3}$,where $\gamma, \delta \in Z$,then $\gamma+\delta$ equals:

The solution of the differential equation $\cos^2 x \frac{d^2y}{dx^2} = 1$ is

The curve amongst the family of curves represented by the differential equation,$(x^2 - y^2)dx + 2xy\, dy = 0$ which passes through $(1, 1)$,is

The curve that satisfies the differential equation $x y \, dy - (1 + y^2) \, dx = 0$ passes through $(1, 0)$ and intersects the curve $x^2 + 3y^2 = 3$ at an angle $\theta$. Then $\frac{2\theta}{\pi} =$