Find the general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.

The general solution of the differential equation $\frac{dy}{dx} = e^{y+x} + e^{y-x}$ is, where $c$ is an arbitrary constant.

$A$ 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 particular solution of the differential equation $y(1+\log x) = (\log x^x) \frac{dy}{dx}$,given $y(e) = e^2$,is

The general solution of the differential equation $2x \frac{dy}{dx} - y = 3$ is a family of

$A$ curve passing through $(2, 3)$ and satisfying the differential equation $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$ for $x > 0$ is