The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes through the point $(4, 3)$. The equation of the curve is

Find the general solution of the differential equation: $y \log y \, dx - x \, dy = 0$.

The particular solution of the differential equation $x dy + 2y dx = 0$,when $x = 2, y = 1$ 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

The solution of the differential equation $\frac{dy}{dx} = 1 - \cos(y-x) \cot(y-x)$ is

Find the equation of a curve passing through the point $(0, -2)$ given that at any point $(x, y)$ on the curve,the product of the slope of its tangent and the $y$-coordinate of the point is equal to the $x$-coordinate of the point.