The solution of the differential equation $x \, dy - y \, dx = 0$ represents:

The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ with the condition $y(1) = 0$ represents:

$A$ particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $(x, 0)$ is given by the formula $\frac{dx}{dt} = \cos^2(\pi x)$. Then the particle never reaches the point on:

The general solution of the differential equation $\frac{dy}{dx} = e^{x+y} + x^2 e^{x^3+y}$ is (where $C$ is a constant of integration):

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

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation $y(1 + xy)dx = xdy$,then $f(-\frac{1}{2}) = $ . . . . .