The general solution of the differential equation $\frac{dy}{dx} = e^{x-y}$ is . . . . . .

Solve $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$

The solution of the differential equation $(1 - x^2)dy + xydx = xy^2dx$ is

Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ given that $y = 1$ when $x = 0$.

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx, x > 0$, then $f(e)$ is equal to:

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.