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:

Find the family of curves such that the angle between the tangent at any point $(x, y)$ and the tangent to the curve $xy = c^2$ at the point of intersection is $\frac{\pi}{4}$.

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

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 ...... .

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x-3y+4}{3x+2y-7}$ is

Let $S = (0, 2 \pi) - \left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. Let $y = y(x)$,$x \in S$,be the solution curve of the differential equation $\frac{dy}{dx} = \frac{1}{1 + \sin 2x}$ with $y\left(\frac{\pi}{4}\right) = \frac{1}{2}$. If the sum of abscissas of all the points of intersection of the curve $y = y(x)$ with the curve $y = \sqrt{2} \sin x$ is $\frac{k \pi}{12}$,then $k$ is equal to: