The differential equation $\cot y \, dx = x \, dy$ has a solution of the form

The solution of the differential equation $\frac{dy}{dx} = 2xy$ is

The solution of the equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$ is

The solution of $\frac{dy}{dx} + 1 = e^{x+y}$ is

The solution of the equation $\frac{dy}{dx} = e^{x - y} + x^2 e^{-y}$ is

Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2x-2y=0$,then its equation is