The solution of the equation $(1 + x^2)\frac{dy}{dx} = 1$ is

The substitution $\frac{dy}{dx}=z$ reduces the differential equation $\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$ to a differential equation whose solution is $z=$

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

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

The solution of $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$ is

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