The general solution of ${x^2}\frac{{dy}}{{dx}} = 2$ is

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

The solution of the differential equation $(x + y - 1)dx + (2x + 2y - 3)dy = 0$ is

Find a particular solution satisfying the given condition: $\frac{dy}{dx} = y \tan x$; $y = 1$ when $x = 0$.

The general solution of the differential equation $\frac{1}{x} \frac{dy}{dx} = \tan^{-1} x$ is

If $y(x)$ is the solution of the differential equation $x dy - (y^2 - 4y) dx = 0$ for $x > 0$ and $y(1) = 2$,and the slope of the curve $y = y(x)$ is never zero,then the value of $10y(\sqrt{2})$ is . . . .