The slope of the normal at any point $(x, y), x > 0, y > 0$ on the curve $y=y(x)$ is given by $\frac{x^{2}}{x y-x^{2} y^{2}-1}$. If the curve passes through the point $(1, 1)$,then $e \cdot y(e)$ is equal to

If $y$ is a function of $x$,then $\frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0$. If $x$ is a function of $y$,then the equation becomes:

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:

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:

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

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 $\frac{\pi}{4}$.