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

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

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1+x^2)(1-y^2)$, with the initial condition $y(0) = \frac{1}{2}$. Then the value of $(2y(1) - 1)$ is equal to:

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 . . . .

The general solution of the differential equation $\frac{dy}{dx} = 1 + x + y + xy$ is

The general solution of the differential equation $\frac{dy}{dx} = (9x + y + 5)^2$ is...