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

$A$ solution curve of the differential equation $(x^2+xy+4x+2y+4) \frac{dy}{dx}-y^2=0, x>0$,passes through the point $(1,3)$. Then the solution curve

The solution of the differential equation $\frac{dy}{dx} = \frac{\sin y + e^x}{\ln y - x \cos y}$ 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:

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

If $y = \frac{x}{\ln |c x|}$ (where $c$ is an arbitrary constant) is the general solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,then the function $\phi \left( \frac{x}{y} \right)$ is: